sm-principles-of-analog-electronics-1st-saggio

SOLUTIONS MANUAL FOR PRINCIPLES OF ANALOG ELECTRONICS by Giovanni Saggio SOLUTIONS MANUAL FOR PRINCIPLES OF ANALOG ELECTRONICS by Giovanni Saggio Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20140131 International Standard Book Number-13: 978-1-4665-8204-0 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. 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CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com CHAPTER 9 – DIODE CIRCUITS Exercise 1 A half-wave rectifer as in Figure 9.1 has an input voltage signal 𝑣𝑠 = 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡). Assuming ideal diode (𝑉𝛾 = 0, zero resistance in forward biasing condition), calculate the following:  The average output voltage 𝑉𝑜,𝐷𝐶 across the resistor  The average current 𝐼𝑜,𝐷𝐶 flowing throught the resistor  The RMS value of the output voltage 𝑉𝑜,𝑅𝑀𝑆 across the resistor  The RMS value of the current 𝐼𝑜,𝑅𝑀𝑆 flowing throught the resistor  The form factor 𝐹𝐹,1𝑠  The ripple factor 𝛾 ANSWER The output voltage function of the half-wave rectifier is represented in Figure 1 vo VM π 2π 3π 4π Figure 1: output voltage function of the half-wave rectifier ωt To obtain the requested values, each time we can start with the respective definitions. So, the average output (DC) voltage will be: 𝑉𝑜,𝐷𝐶 = 1 𝜋 1 𝜋 𝑉𝑀 � 𝑣𝑠 𝑑(𝜔𝑡) = � 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡) 𝑑(𝜔𝑡) = 𝜋 𝜋 0 𝜋 0 The average output current: 𝐼𝑜,𝐷𝐶 = 𝑉𝐷𝐶 𝑉𝑀 = 𝑅 𝜋𝑅 The same result can be equivalently obtained considering that this circuit does not produce any phase shift, so that the current can be written as 𝑖(𝜔𝑡) = 𝐼𝑀 𝑠𝑖𝑛(𝜔𝑡)). So: 𝐼𝑜,𝐷𝐶 = 𝜋 2𝜋 1 2𝜋 1 𝐼𝑀 [− 𝑐𝑜𝑠(𝜔𝑡)]𝜋0 + 0 � 𝐼𝑀 𝑠𝑖𝑛(𝜔𝑡) 𝑑(𝜔𝑡) = �� 𝐼𝑀 𝑠𝑖𝑛(𝜔𝑡) 𝑑(𝜔𝑡) + � 0𝑑(𝜔𝑡)� = 2𝜋 2𝜋 0 2𝜋 0 𝜋 𝐼𝑀 𝐼𝑀 —1 − 1 = = 2𝜋 𝜋 To determine the RMS value of the output voltage, we have: 𝑉𝑜,𝑅𝑀𝑆 = � 1 𝜋 2 1 𝜋 𝑉𝑀2 𝜋 1 − 𝑐𝑜𝑠(2𝜔𝑡) � 𝑣𝑠 𝑑(𝜔𝑡) = � � 𝑉𝑀2 𝑠𝑖𝑛2 (𝜔𝑡) 𝑑(𝜔𝑡) == � � � � 𝑑(𝜔𝑡) 2 2𝜋 0 2𝜋 0 2𝜋 0 𝜋 1 𝜔𝑡 𝑠𝑖𝑛(2𝜔𝑡) 𝑉𝑀 1 𝜋 = 𝑉𝑀 � � − � = 𝑉𝑀 � � � = 4 2 2𝜋 2 2𝜋 2 0 The RMS value of the output current: 𝐼𝑜,𝑅𝑀𝑆 = 𝑉𝑜,𝑅𝑀𝑆 𝑉𝑀 = 𝑅 2𝑅 The same expression can be equivalently found as: 1 𝜋 2 1 2 𝜋 𝐼𝑀 𝐼𝑜,𝑅𝑀𝑆 = � � 𝐼𝑀 𝑠𝑖𝑛2(𝜔𝑡) 𝑑(𝜔𝑡) = � 𝐼𝑀 = 2 2𝜋 0 2𝜋 2 which corresponds to the RMS value of the total current, that is the sum of DC and AC components. Now, the RMS value of the only AC component, 𝐼𝑟,𝑅𝑀𝑆 , that is the RMS value of the ripple, can be determined starting from the definition applied to the term (𝑖 − 𝐼𝐷𝐶 ): but 1 2𝜋 1 2𝜋 2 ]𝑑(𝜔𝑡) 𝐼𝑟,𝑅𝑀𝑆 = � � (𝑖 − 𝐼𝐷𝐶 )2 𝑑(𝜔𝑡) = � � [𝑖 2 − 2𝑖𝐼𝐷𝐶 + 𝐼𝐷𝐶 2𝜋 0 2𝜋 0 1 2𝜋 2 � � 𝑖 2 𝑑(𝜔𝑡) = 𝐼𝑅𝑀𝑆 2𝜋 0 so 1 2𝜋 � 𝑖𝑑(𝜔𝑡) = 𝐼𝐷𝐶 2𝜋 0 2 2 2 2 2 𝐼𝑟,𝑅𝑀𝑆 = �𝐼𝑅𝑀𝑆 − 2𝐼𝑜,𝐷𝐶 + 𝐼𝑜,𝐷𝐶 = �𝐼𝑅𝑀𝑆 − 𝐼𝑜,𝐷𝐶 The form factor is defined as the RMS and average ratio, 𝐹𝐹 ≝ 𝐹𝐹,1𝑠 = 0,5𝑉𝑀 ≅ 1,57 0,32𝑉𝑀 𝑉𝑜,𝑅𝑀𝑆 𝑉𝑜,𝐷𝐶 = 𝐼𝑜,𝑅𝑀𝑆 𝐼𝑜,𝐷𝐶 , so: Let’s see now how we can calculate the ripple. Since the pure AC voltage component is defined as the sinusoidal wave 𝑣𝑜 (= 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡) 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 ≠ 0 𝑓𝑜𝑟 0 < 𝜔𝑡 < 𝜋 𝑎𝑛𝑑 𝑧𝑒𝑟𝑜 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒) apart from its DC component 𝑉𝑜,𝐷𝐶 , that is 𝑣𝑜 − 𝑉𝑜,𝐷𝐶 , we have: 1 𝜋 1 𝜋 2 2 𝑉𝑟,𝑅𝑀𝑆 = � � �𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡) − 𝑉𝑜,𝐷𝐶 � 𝑑(𝜔𝑡) = � � �𝑉𝑀2 𝑠𝑖𝑛2(𝜔𝑡) + 𝑉𝑜,𝐷𝐶 − 2𝑉𝑜,𝐷𝐶 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡)�𝑑(𝜔𝑡) 𝜋 0 𝜋 0 but we already saw that 1 𝜋 𝑉𝑜𝑅𝑀𝑆 = � � 𝑉𝑀2 𝑠𝑖𝑛2(𝜔𝑡) 𝑑(𝜔𝑡) 𝜋 0 therefore replacing 2 𝑉𝑟,𝑅𝑀𝑆 = ��𝑉𝑜,𝑅𝑀𝑆 − 𝑉𝑜,𝐷𝐶 � as a consequence the ripple of the half-wave rectifier is 2 2 2 𝑉𝑟,𝑅𝑀𝑆 ��𝑉𝑜,𝑅𝑀𝑆 − 𝑉𝑜,𝐷𝐶 � 𝑉𝑜,𝑅𝑀𝑆 𝑉𝑀 ⁄2 𝑟= = = �� − 1 = �� − 1 = 1.21 𝑉𝑀 ⁄𝜋 𝑉𝑜,𝐷𝐶 𝑉𝑜,𝐷𝐶 𝑉𝑜,𝐷𝐶 The same result can be obtained solving respect to the current rather than to the voltage: 2 2 𝐼𝑀 2 𝐼𝑟,𝑅𝑀𝑆 �𝐼𝑜,𝑅𝑀𝑆 − 𝐼𝐷𝐶 𝐼𝑜,𝑅𝑀𝑆 𝑟= = = �� − 1 = � 2 − 1 = 1.21 𝐼𝑀 𝐼𝑜,𝐷𝐶 𝐼𝑜,𝐷𝐶 𝐼𝑜,𝐷𝐶 𝜋 Summing 𝑉𝑜,𝐷𝐶 = 𝑉𝑀 𝜋 ; 𝐼𝑜,𝐷𝐶 = 𝑉𝑀 𝜋𝑅 ; 𝑉𝑜,𝑅𝑀𝑆 = 𝑉𝑀 2 ; 𝐼𝑜,𝑅𝑀𝑆 = 𝑉𝑀 2𝑅 ; 𝐹𝐹,1𝑠 ≅ 1,57 ; 𝛾 = 1.21 Exercise 2 A half-wave rectifier circuit, as in Figure 9.1, has an input sinusoidal voltage source 𝑣𝑠 = 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡), with 𝑉𝑀 = 10𝑉, 𝑅 = 500Ω. Assuming an ideal diode (𝑉𝛾 = 0, zero resistance in forward biasing conditions) calculate the following:        The maximum current value 𝐼𝑀 DC component of current 𝐼𝐷𝐶 RMS value of current 𝐼𝑅𝑀𝑆 DC component of voltage in output 𝑉𝐷𝐶 DC component of power delivered to the load 𝑃𝑜,𝐷𝐶 Power value supplied by the source 𝑃𝑠 The ripple value 𝛾 ANSWER The maximum current value is 𝐼𝑀 = its DC component and its RMS value 𝐼𝐷𝐶 = 𝑉𝑀 = 20𝑚𝐴 𝑅 1 𝜋 𝑉𝑀 � 𝑣𝑠 𝑑(𝜔𝑡) = = 6.3𝑚𝐴 𝜋𝑅 𝜋𝑅 0 1 𝜋 𝑉𝑀 𝐼𝑅𝑀𝑆 = � � 𝑣𝑠2 𝑑(𝜔𝑡) = = 10𝑚𝐴 2𝑅 𝜋𝑅 0 The DC component of the output voltage 𝑉𝐷𝐶 = 𝑅𝐼𝐷𝐶 = 6.3𝑉 The power from the source and the power to the load are respectively 2 𝑃𝑠 = 𝑅𝐼𝑅𝑀𝑆 = 100𝑚𝑊 Observation 2 𝑃𝑜,𝐷𝐶 = 𝑅𝐼𝐷𝐶 = 39.7𝑚𝑊 To determine the electric power we consider the absolute values of current and/or voltage when DC but RMS values when AC. For the current exercise the output voltage is a periodic one so that to calculate the electric power we should utilize the RMS values but, on the contrary, it is commonly adopted the average ones. This is because the final scope of the circuit is to furnish a constant voltage across the load. Finally, the ripple value 𝛾 = �� 𝐼𝑅𝑀𝑆 2 � − 1 ≅ 1.23 𝐼𝐷𝐶 Summarizing: 𝐼𝑀 = 20𝑚𝐴 ; 𝐼𝐷𝐶 = 6.3𝑚𝐴 ; 𝐼𝑅𝑀𝑆 = 10𝑚𝐴 ; 𝑉𝐷𝐶 = 6.3𝑉 ; 𝑃𝑜,𝐷𝐶 = 39.7m𝑊 ; 𝑃𝑠 = 100𝑚𝑊; 𝛾 ≅ 1.23 Exercise 3 A half-wave rectifier has an input stage with a transformer, turns ratio 𝑛 = 20: 1, and a load resistance 𝑅 = 50Ω.. The input voltage source has a RMS value of 𝑉𝑠,𝑅𝑀𝑆 = 220𝑉 (Figure 9.23). + vg + + v t1 D R v t2 n1 n2 Figure 9.23: half-wave rectifier with a transformer as the first stage Let’s determine the following:  DC component of voltage in output 𝑉𝑜,𝐷𝐶  DC component of current in output 𝐼𝑜,𝐷𝐶  RMS value of voltage in output 𝑉𝑜,𝑅𝑀𝑆  RMS value of current in output 𝐼𝑜,𝑅𝑀𝑆  the ripple value 𝛾 ANSWER Across the secondary coil there is a RMS voltage equal to across the half-wave rectifier’s input port is 220 10 𝑉𝑀 = √2 ∗ 22 ≅ 31.12𝑉 = 22𝑉, so that the maximum voltage According to the previous exercises we know that the DC output voltage can be written as and the output current while their RMS values 𝑉𝑜,𝐷𝐶 = 𝐼𝑜,𝐷𝐶 = 𝑉𝑀 ≅ 9.9𝑉 𝜋 𝑉𝑜,𝐷𝐶 ≅ 198𝑚𝐴 𝑅 𝑉𝑀 ≅ 15.56𝑉 2 𝑉𝑜,𝑅𝑀𝑆 = ≅ 0.31𝐴 𝑅 𝑉𝑜,𝑅𝑀𝑆 = 𝐼𝑜,𝑅𝑀𝑆 Summarizing 𝑉𝑜,𝐷𝐶 ≅ 9.9𝑉 ; 𝐼𝑜,𝐷𝐶 ≅ 198𝑚𝐴 ; 𝑉𝑜,𝑅𝑀𝑆 ≅ 15.56𝑉 ; 𝐼𝑜,𝑅𝑀𝑆 ≅ 0.31𝐴 ; 𝛾 ≅ 1.21. Observation Note that this ripple value is quite high, so the half-wave rectifier is a poor AC to DC converter. Exercise 4 A half-wave rectifier network is used as a battery charger (Figure 9.24). i D=iR =i B + D R vs + VB - - Figure 9.24: a simple battery charger by means of a half-wave rectifier When the source voltage is higher than that of the battery, the current flows from the source to the battery, which charges. Reverse conditions are not possible because of the presence of the diode. Assuming a voltage source of 𝑣𝑠 = 150 𝑠𝑖𝑛(𝜔𝑡), an ideal diode (𝑉𝛾 = 0 and a short-circuit in forward bias condition, an open-circuit in reverse bias conditions), and a 75𝑉 battery with a charge current of 750𝑚𝐴, determine the resistance value of the resistor 𝑅. ANSWER The charging current can flow only when the diode is in “active” mode (ON), and since 𝑉𝐵 = 75𝑉 this corresponds to the a-b, a’-b’, a”-b”,.. intervals of 𝑣𝑠 , in green color of Figure 9.24b. vs 150 75 a b a’ b’ a’’ ωt Figure 9.24b: the current flows through the diode during the green part of vs To determine the value of 𝜔𝑡 corresponding to the point a, b, a’, b’, a’’ etc.., we can impose so 150 𝑠𝑖𝑛(𝜔𝑡) = 75 1� 𝜋 𝜔𝑡 = � 6 5� 𝜋 6 When the current differs from zero, it is equal to 𝑣𝑠 − 75 150 sin(𝜔𝑡) − 75 𝑖= = 𝑅 𝑅 but we have to consider the DC value of the current which charges the battery 𝐼𝐷𝐶 5 �6𝜋 5� 𝜋 1 150 sin(𝜔𝑡) − 75 1 [−150 cos(𝜔𝑡) − 75𝜔𝑡]1 6 = � 𝑑(𝜔𝑡) = �6𝜋 2𝜋 1� 𝜋 𝑅 2𝜋𝑅 6 1 5 √3 √3 75 �−150 �− � − 75 𝜋 + 150 + 𝜋� 2𝜋𝑅 6 6 2 2 Now, according to the request, we impose 𝐼𝐷𝐶 = 750𝑚𝐴, obtaining 𝑅 ≅ 22Ω = Observations A battery’s capacity 𝐶 [𝐴ℎ] refers to the stored electric charge that can be delivered in an ammount of time at room temperature (77°𝐹 or 25°𝐶). A 500[𝐴ℎ] rated battery can supply 1𝐴 for 500ℎ, or 5𝐴 for 100ℎ, or 10𝐴 for 50ℎ, or 100𝐴 for 5ℎ. But the capacity is not the perfect parameter to give a real measure for a battery, because it depends on the discharge conditions: the current’s value (not necessary constant), the value of the voltage, the temperature, the discharging rate. The C-rate (or charge-rate or hourly-rate) of a battery specifies the discharge rate, as a multiple of the capacity. So, for example, a battery with a capacity 𝐶 = 1.5[𝐴ℎ] and a 𝐶/10 rate, deliveres 1.5 10 = 0.15[𝐴]; A 1𝐶 rate means that the battery discharges entirely in 1[ℎ]. This is similar for the E-rate but referred to the power, not the current. Exercise 5 A battery with 𝐶 = 1200𝑚𝐴ℎ, 𝑉𝐵 = 5𝑉, and 𝐶 − 𝑟𝑎𝑡𝑒 = 10, must be charged by a half-wave rectifier with a transformer at its input port, having a turns ratio 𝑛 = 15. The AC voltage source 𝑣𝑠 = 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡) has a RMS voltage of 𝑉𝑠,𝑅𝑀𝑆 = 220𝑉 (Figure 9.25). i D=iR =i B + vs + + v t1 v t2 n1 D R + VB n2 - Figure 9.25: Half-wave rectifier with a transformer at its input port Assuming a diode with 𝑉𝛾 = 0.74𝑉, calculate the following:  The average charging current, 𝐼𝐵  The value of the resistance 𝑅 necessary to limit the current  The RMS current flowing throught the battery, 𝐼𝑅𝑀𝑆  The power 𝑃𝑅 dissipated by the resistor  The power 𝑃𝐵 delivered to the battery  The charging time, 𝑡𝐵  The efficiency of the half-wave rectifier η = 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑤𝑒𝑟 𝑃 = 𝑃𝐵 𝑡𝑜𝑡 ANSWER The average charging current is due to the features of the battery, so 𝐼𝐵 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 1.2 = = 0.12𝐴 𝐶 − 𝑟𝑎𝑡𝑒 10 The voltage across the primary coil is that of the voltage source 𝑣𝑠 , while the voltage across the secondary coil is 𝑣𝑡2 = 𝑣𝑡1 𝑛 , so that the RMS voltage furnished to the circuit is 𝑉𝑡2,𝑅𝑀𝑆 = and its maximum value 𝑉𝑠,𝑅𝑀𝑆 220 = = 14. 6𝑉 𝑛 15 𝑉𝑡2,𝑀 = √2𝑉𝑡2,𝑅𝑀𝑆 ≅ 20.74𝑉 As a consequence, the maximum value of the voltage downstream the diode is 𝑉𝑀,𝑑 = 𝑉𝑡2,𝑀 − 𝑉𝛾 ≅ 20𝑉 The charging current will flow only when the diode is “active” (ON), that is when �𝑉𝑡2,𝑀 − 𝑉𝛾 � ≥ 𝑉𝐵 , between each α-β segments reported in green in Figure 9.25b. vt2 -Vγ Vt2,M -Vγ VB α β α’ β’ α’’ ωt Figure 9.25b: The voltage useful to charge the battery is showed in green According to the Figure 9.25b, we can determine the α and β values as and 𝛼 = 𝑠𝑖𝑛−1 � 𝑉𝐵 � ≅ 0.25𝑟𝑎𝑑 ≅ 14.48° 𝑉𝑡2,𝑀 − 𝑉𝛾 𝛽 = 𝜋 − 𝛼 ≅ 2.89𝑟𝑎𝑑 ≅ 165.52° The average charging current can be determined as 𝐼𝐵 = 1 𝛽 𝑣𝑡2 − 𝑉𝛾 − 𝑉𝐵 1 𝛽 𝑉𝑡2,𝑀 𝑠𝑖𝑛(𝜔𝑡) − 𝑉𝛾 − 𝑉𝐵 � 𝑑(𝜔𝑡) = � 𝑑(𝜔𝑡) 𝑅 𝑅 2𝜋 𝛼 2𝜋 𝛼 1 𝛽 = �−𝑉𝑡2,𝑀 𝑐𝑜𝑠(𝜔𝑡) − 𝑉𝛾 (𝜔𝑡) − 𝑉𝐵 (𝜔𝑡)��𝛼 2𝜋𝑅 1 = �−𝑉𝑡2,𝑀 𝑐𝑜𝑠(𝛽) + 𝑉𝑡2,𝑀 𝑐𝑜𝑠(𝛼) − �𝑉𝛾 + 𝑉𝐵 �𝛽 + �𝑉𝛾 + 𝑉𝐵 �𝛼� 2𝜋𝑅 but 𝛽 = 𝜋 − 𝛼 and recalling that cos(𝑥) = − cos(𝜋 − 𝑥) we can write 𝐼𝐵 = 1 �2𝑉𝑡2,𝑀 𝑐𝑜𝑠(𝛼) + 2�𝑉𝛾 + 𝑉𝐵 �𝛼 − 𝜋�𝑉𝛾 + 𝑉𝐵 �� 2𝜋𝑅 ! from which, imposing that 𝐼𝐵 = ⏞ 1200𝑚𝐴, we obtain the requested resistance of the resistor 𝑅= 2�𝑉𝑡2,𝑀 − 𝑉𝛾 � 𝑐𝑜𝑠(𝛼) + 2𝑉𝐵 𝛼 − 𝜋𝑉𝐵 ≅ 34Ω 2𝜋𝐼𝐵 So, the RMS value of the current flowing through the resistor is 𝐼𝑅𝑀𝑆 2 1 𝛽 �𝑉𝑡2,𝑀 𝑠𝑖𝑛(𝜔𝑡) − 𝑉𝐵 � =� � 𝑑(𝜔𝑡) 𝑅2 2𝜋 𝛼 2 𝑠𝑖𝑛2 (𝜔𝑡) +𝑉𝐵2 − 2�𝑉𝑡2,𝑀 − 𝑉𝛾 �𝑉𝐵 𝑠𝑖𝑛(𝜔𝑡) 1 𝛽 𝑉𝑡2,𝑀 =� � 𝑑(𝜔𝑡) 𝑅2 2𝜋 𝛼 1 which can be solved considering that sin2 (𝑥) = 2 [1 − cos(2𝑥)], sin(−𝑥) = − sin(𝑥), cos(𝜋 − 𝑥) = − cos(𝑥), and splitting the integral into three parts (highlighted with three different colors) 𝛽 2 2 𝛽 1 𝑉𝑡2,𝑀 𝑉𝑡2,𝑀 − � [1 − 𝑐𝑜𝑠(2𝜔𝑡)]𝑑(𝜔𝑡) = 𝑠𝑖𝑛(2𝜔𝑡)�� 𝜔𝑡 2 2 𝛼 2 𝛼 2 2 𝑉𝑡2,𝑀 𝑉𝑡2,𝑀 (𝜋 − 2𝛼) + = 𝑠𝑖𝑛(2𝛼) 2 2 𝛽 2 � 𝑉𝑡2,𝑀 𝑠𝑖𝑛2(𝜔𝑡) 𝑑(𝜔𝑡) = 𝛼 𝛽 𝛽 � 𝑉𝐵2 𝑑(𝜔𝑡) = 𝑉𝐵2 [𝜔𝑡]|𝛼 = 𝑉𝐵2 (𝜋 − 2𝛼) 𝛼 𝛽 𝛽 � −2�𝑉𝑡2,𝑀 − 𝑉𝛾 �𝑉𝐵 𝑠𝑖𝑛(𝜔𝑡) 𝑑(𝜔𝑡) = −2�𝑉𝑡2,𝑀 − 𝑉𝛾 �𝑉𝐵 [− 𝑐𝑜𝑠(𝜔𝑡)]|𝛼 = −4�𝑉𝑡2,𝑀 − 𝑉𝛾 �𝑉𝐵 𝑐𝑜𝑠(𝛼) 𝛼 therefore, summing up 2 𝐼𝑅𝑀𝑆 = 2 2 𝑉𝑡2,𝑀 1 𝑉𝑡2,𝑀 (𝜋 𝑠𝑖𝑛(2𝛼) + 𝑉𝐵2 (𝜋 − 2𝛼) − 4�𝑉𝑡2,𝑀 − 𝑉𝛾 �𝑉𝐵 𝑐𝑜𝑠(𝛼)� − 2𝛼) + � 2 2𝜋𝑅 2 2 numerically 𝐼𝑅𝑀𝑆 ≅ 0.21𝐴. The power 𝑃𝑅 dissipated by the resistor is 2 𝑃𝑅 = 𝑅𝐼𝑅𝑀𝑆 ≅ 1.42𝑊 The power 𝑃𝐵 delivered to the battery is 𝑃𝐵 = 𝑉𝐵 𝐼𝐵 = 0.6𝑊 In order to determine the charging time 𝑡𝐵 , being 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 1200𝑚𝐴ℎ then 𝑡𝐵 = Finally, for the efficiency 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 1200𝑚𝐴 = = 10ℎ 𝐼𝐵 0.12𝐴 η= so that it is arond 30%. Observations 𝑃𝐵 𝑃𝐵 = ≅ 0.30 𝑃𝑡𝑜𝑡 (𝑃𝐵 + 𝑃𝑅 ) The charging current must be limited to avoid the battery to be damaged because of the gases which can be released during the charge. A costant voltage (C-V) charging method refers to the method by which a charger sources current into the battery to force its voltage to a preset value, named set(-point) voltage. When this value is reached, the charger sources only the current to maintain the battery at the same constant voltage level. It is the opposite for the constant current (C-I) charging method. Some rechargeable batteries need to be C-V while others C-I charged. Some chargers are CV others C-I type; that’s why generally a charger can be used to charge some batteries and not others. Universal chargers are those that have both C-V and C-I characteristics. Summarizing 𝐼𝐵 = 0.12A ; 𝑅 ≅ 34Ω ; 𝐼𝑅𝑀𝑆 ≅ 0.21𝐴 ; 𝑃𝑅 ≅ 1.42𝑊 ; 𝑃𝐵 = 0.6𝑊 ; 𝑡𝐵 = 10ℎ ; η ≅ 0.30 Exercise 6 The bridge full-wave rectifier has better performances than the half-wave rectifier in terms of form factor and ripple. Consider a bridge full-wave rectifier with an input transformer, as in Figure 9.26: D + v t2 n1 n2 R C A B D4 3 + v t1 D D + vs + D2 1 C Figure 9.26: Bridge full-wave rectifier with input transformer Calculate the following:  DC component of voltage in output 𝑉𝑜,𝐷𝐶  DC component of current in output 𝐼𝑜,𝐷𝐶  RMS value of voltage in output 𝑉𝑜,𝑅𝑀𝑆 vo  RMS value of current in output 𝐼𝑜,𝑅𝑀𝑆  The form factor 𝐹𝐹,2𝑠  The ripple value 𝛾 ANSWER For the current rectifier, the voltage period is no more 2𝜋 as for the half-wave rectifier, but just 𝜋, so that the DC component of voltage in output 𝑉𝑜,𝐷𝐶 can be written as 𝑉𝑜,𝐷𝐶 = 1 𝜋 1 𝜋 2𝑉𝑀 � 𝑣𝑠 𝑑(𝜔𝑡) = � 𝑉𝑀 𝑠𝑖𝑛(𝜔𝑡) 𝑑(𝜔𝑡) = 𝜋 𝜋 0 𝜋 0 Accordingly, the DC component of current in output 𝐼𝑜,𝐷𝐶 is 𝐼𝑜,𝐷𝐶 = 𝑉𝑜,𝐷𝐶 2𝑉𝑀 = 𝑅 𝜋𝑅 The RMS value of voltage in output 𝑉𝑜,𝑅𝑀𝑆 can be easily obtained as 1 𝜋 1 𝜋 𝑉𝑀 𝑉𝑜,𝑅𝑀𝑆 = � � 𝑣𝑠2 𝑑(𝜔𝑡) = � � 𝑉𝑀2 𝑠𝑖𝑛2(𝜔𝑡) 𝑑(𝜔𝑡) = 𝜋 0 𝜋 0 √2 Such a result should not be surprising, because we could expect it equal to the one of the originating 𝑣𝑠 . The RMS value of current in output 𝐼𝑜,𝑅𝑀𝑆 can be easily obtained as The form factor 𝐹𝐹,2𝑠 is defined as the 𝐼𝑜,𝑅𝑀𝑆 = 𝑉𝑜,,𝑅𝑀𝑆 𝑉𝑜,𝐷𝐶 𝑉𝑜,𝑅𝑀𝑆 𝑉𝑀 = 𝑅 𝑅√2 ratio, therefore 𝐹𝐹,2𝑠 𝑉𝑀 = √2 ≅ 1,1 2𝑉𝑀 𝜋 We start from the definition also to determine the ripple value 2 2 𝑉𝑟,RMS 𝑉𝑜𝑅𝑀𝑆 𝑉𝑀 ⁄√2 𝛾= = �� − 1 = �� − 1 = 0.482 𝑉𝑜,𝐷𝐶 𝑉𝑜,𝐷𝐶 2𝑉𝑀 ⁄𝜋 Comparing this result with the one obtained for the half-wave rectifier, we can say that the ripple is halved (0.482 vs. 1.21). Observations Please, pay close attention to the fact that for the full-wave rectifier the ground is connected to one terminal of the voltage source, but it cannot be connected to the load as well. So, the bridge configuration cannot be adopted when we must have a grounded load. Unfortunately, the definition of ripple is not unambiguous. Sometimes it can be found otherwise. Summarizing 𝑉𝑜,𝐷𝐶 = 2𝑉𝑀 𝜋 Exercise 7 ; 𝐼𝑜,𝐷𝐶 = 2𝑉𝑀 𝜋𝑅 ; 𝑉𝑜,𝑅𝑀𝑆 = 𝑉𝑀 √2 𝑉 𝑀 ; 𝐼𝑜,𝑅𝑀𝑆 = 𝑅√2 ; 𝐹𝐹,2𝑠 ≅ 1,1 ; 𝛾 = 0.482. Given a bridge full-wave rectifier as in Figure 9.26, with diodes that have a threshold voltage 𝑉𝛾 = 0 and a resistance 𝑅𝑜𝑛 = 10Ω when “on”, sourced by a sinusoidal voltage signal of amplitude 𝑉𝑠𝑀 = 30𝑉, and loaded by a resistance 𝑅𝐿 = 1𝑘Ω, calculate the following:  Maximum, average and RMS values of the output current: 𝐼𝑀 , 𝐼𝑜,𝐷𝐶 , 𝐼𝑜,𝑅𝑀𝑆  Average and RMS values of the output voltage: 𝑉𝑜,𝐷𝐶 , 𝑉𝑜,𝑅𝑀𝑆  Average value of the output power, 𝑃𝑜,𝐷𝐶  Average value of the input power, 𝑃𝑖 ANSWER The maximum output current can be easily calculated applying the KCL to a half path of the bridge 𝐼𝑀 = 𝑉𝑀 30 = = 29.4𝑚𝐴 𝑅 + 2𝑅𝑜𝑛 1000 + 20 therefore its RMS value (see Table 1.3 in Section 1.3.1 of Chapter 1) is 𝐼𝑜,𝑅𝑀𝑆 = 𝐼𝑀 √2 ≅ 20.8𝑚𝐴 and its average value (as already discussed in Exercise 6) is 𝐼𝑜,𝐷𝐶 = 2𝐼𝑀 2 ∗ 0.0294 = ≅ 18.7𝑚𝐴 𝜋 𝜋 Average and RMS values of the output voltage are simply due to Ohm’s law 𝑉𝑜,𝐷𝐶 = 𝑅𝐼𝑜,𝐷𝐶 = 1000 ∗ 0.0187 = 18.7𝑉 𝑉𝑜,𝑅𝑀𝑆 = 𝑅𝐼𝑜,𝑅𝑀𝑆 = 1000 ∗ 0.0208 = 20.8𝑉 Finally, according to the Joule’s law (Section 4.4.8 in Chapter 4), we can write for the output and input powers respectively 2 𝑃𝑜,𝐷𝐶 = 𝑅𝐼𝑜,𝐷𝐶 ≅ 0.3𝑊 2 𝑃𝑖 = (𝑅 + 2𝑅𝑜𝑛 )𝐼𝑜,𝑒𝑓𝑓 ≅ 0.41𝑊 Summarizing 𝐼𝑀 = 29.4𝑚𝐴; 𝐼𝑜,𝑅𝑀𝑆 ≅ 20.8𝑚𝐴; 𝐼𝑜,𝐷𝐶 ≅ 18.7𝑚𝐴; 𝑉𝑜,𝐷𝐶 = 18.7𝑉; 𝑉𝑜,𝑅𝑀𝑆 = 20.8𝑉; 𝑃𝑜,𝐷𝐶 ≅ 0.3𝑊; 𝑃𝑖 ≅ 0.41𝑊 Exercise 8 A resistance load requires a DC voltage 𝑉𝐷𝐶 = 12𝑉 and current 𝐼𝐷𝐶 = 0.5𝐴. To this aim a bridge full-wave rectifier is utilized, sourced by a AC line with a frequency 𝑓 = 50𝐻𝑧. Calculate the value of the smoothing capacitance for a 10% ripple (Figure 9.27). D + v t1 v t2 n1 n2 R C A B vo D4 3 vs + D D + + D2 1 C Figure 9.27: Filtered bridge full-wave rectifier ANSWER To solve this exercise we need to do some considerations. Let’s call cutin the instant of time when diodes start to conduct current and cutoff the instant of time when the diodes end. VsM cutout cutin VRM vR vs ωt Figure 9.27b:input voltage function (in blue), output voltage function (in red), rectified voltage function (in green) We can usefully define as  𝑇: period of the input function 𝑇  2: period of the half-wave  𝑇1 : time during which the diode is “ON”  𝑇2 : time during which the diode is “OFF” (discarge of the capacitor) and we can admit an approximation of the green part of the sinusoidal function as a straight line (this is supposing the time of the capacitor discharge much greater of the period of the input function) as showed in Figure 9.27c. VsM Vγ VRM vR T vs ωt T/2 T1 T2 Figure 9.27c: the charge-discharge functions are approximated with straight lines Defining as 𝑉𝛾 the peak-to-peak value of the residual ripple content, it follows that the RMS value of such a ripple is 𝑉𝑅𝑀𝑆 = 𝑉𝛾 2√3 (please, refer to Table 1.3, Section 1.3.1 in Chapter 1). During 𝑇2 the capacitor gets discharged through the resistor 𝑅, losing an amount of charge 𝑄 equal to 𝑄 = 𝐶𝑉𝛾 but since 𝑖(𝑡) = it results 𝑇2 𝑑𝑄 𝑑𝑡 𝑄 = � 𝑖(𝑡)𝑑𝑡 = 𝐼𝐷𝐶 𝑇2 with 𝐼𝐷𝐶 the DC component of the current. But being 𝑇2 ≫ 𝑇1 : 0 so that 𝑇1 + 𝑇2 = and since 𝑉𝛾 = 𝐼𝐷𝐶 𝑇 𝐼𝐷𝐶 = 𝐶 2 2𝑓𝐶 𝐼𝐷𝐶 = then Recalling the ripple’s definition 𝛾 = 𝛾= The value of the load resistance is 𝑉𝑅𝑀𝑆 𝑉𝐷𝐶 𝑉𝑅𝑀𝑆 𝑉𝐷𝐶 𝑇 ≅ 𝑇2 2 𝑉𝛾 = 𝑉𝐷𝐶 𝑅𝐿 𝑉𝐷𝐶 2𝑓𝑅𝐶 , we can put 𝑉𝐷𝐶 𝑉𝛾 2𝑓𝑅𝐶 1 2√3 2√3 = = = 𝑉𝐷𝐶 𝑉𝐷𝐶 4√3𝑓𝑅𝐶 𝑅= 𝑉𝐷𝐶 = 24Ω 𝐼𝐷𝐶 so that, imposing a 10% ripple as the request of the exercise, and inverting the previous equation we have 1 𝐶= ≅ 1.2𝑚𝐹 0.1 ∗ 4√3 ∗ 50 ∗ 24 Observation The choice of the capacitance of the capacitor must satisfy two opposite requests. In fact, to guarantee a low ripple value we need a capacitance as high as possible, but to assure a low current peak we need a capacitance as low as possible. The trade-off usually is guaranteed by capacitance of the order of hundreds of thousands of 𝜇𝐹. Exercise 9 A battery provides a voltage 𝑉𝐵 = 12𝑉 across a positive clipper circuit configuration, as in Figure 9.28, with 𝑅1 = 47Ω and 𝑅2 = 22Ω. Assuming the diode to offer a direct resistance 𝑅𝐷,𝑜𝑛 = 15Ω when “ON” but with no threshold voltage 𝑉𝛾 = 0, calculate the current flowing through the diode. R1 + A R2 D VB B Figure 9.28: A clipper circuit configuration sourced by a constant voltage ANSWER It is useful to replace the current circuit with an equivalent one at the A - B terminals, according to the Thevenin’s theorem (Figure 9.28b). So, the Thevenin voltage 𝑉𝑇𝐻 is the open circuited one across the same terminals, the diode being removed 𝑅2 𝑉𝑇𝐻 = 𝑉 = 3.826𝑉 𝑅1 + 𝑅2 𝐵 and the Thevenin resistance 𝑅𝑇𝐻 is the one the battery being short circuited and the diode removed 𝑅1 𝑅2 𝑅𝑇𝐻 = 𝑅1 ||𝑅2 = ≅ 14.9Ω 𝑅1 + 𝑅2 RTH =R1 //R2 A + - D R VB V TH= 2 (R1+R2) I B Figure 9.28b: Thevenin’s equivalent of the clipper circuit configuration The current flowing through the diode is then: 𝑉𝑇𝐻 3.826 𝐼= ≅ ≅ 257𝑚𝐴 𝑅𝑇𝐻 + 𝑅𝐷,𝑜𝑛 14.9 Basi di elettronica 1. AMPLIFIERS 1.1. EXERCISES Exercise n. 1 There are several different types of microphone: carbon, dynamic, crystal, capacitive (electret or condenser). The latters are preferred in portable devices (in mobile phones, in tie clip, in consumer video camera, in computer sound cards) because of their small volume and low production cost. Consider a capacitive microphone with a very low voltage output, 𝑣𝑠 = 1𝑚𝑉, and very high output impedance (due to the electret element itself), 𝑅𝑠 = 1𝑀Ω. To amplify the signal a voltage amplifier is used with an input resistance 𝑅𝑖 = 10𝑀Ω, a voltage gain of 20𝑑𝐵, and an output resistance 𝑅𝑜 = 100Ω.. The resistance load is 𝑅𝐿 = 47𝑘Ω.. io ii + v s= 1mV RS = 1M Ω + vi Ro = 100 Ω + Ri = 10M Ω Avv i Av= 20dB + vo RL = 47k Ω Figure 1: voltage amplifier with a mic as source and a load resistor 𝑣 Calculate the overall voltage amplification 𝑣𝑜 . 𝑠 ANSWER Applying the voltage divider formula at the output loop, we have 𝑅𝐿 𝑣𝑜 = 𝐴 𝑣 𝑅𝐿 + 𝑅𝑜 𝑣 𝑖 while at the input one 𝑅𝑖 𝑣𝑖 = 𝑣 𝑅𝑖 + 𝑅𝑠 𝑠 Mixing the two previous equations 𝑣𝑜 𝑅𝐿 𝑅𝑖 = 𝐴𝑣 𝑣𝑠 𝑅𝐿 + 𝑅𝑜 𝑅𝑖 + 𝑅𝑠 From the dB value of the voltage amplification and finally Exercise n. 2 𝐴𝑣 = 10 𝐴𝑣,𝑑𝐵 20 20 = 1020 = 10 𝑣𝑜 ≅ 0.9979 ∗ 10 ∗ 0.9091 ≅ 9.1 𝑣𝑠 Basi di elettronica A current amplifier has a gain of 40𝑑𝐵, an input resistance 𝑅𝑖 = 10Ω and an output resistance 𝑅𝑜 = 22𝑘Ω.. Calculate the input signal current 𝑖𝑖 required to produce an output signal current 𝑖𝑜 = 1𝐴 in a load resistor 𝑅𝐿 = 50Ω.. ANSWER 𝑅𝑜 𝐴𝑖 𝑅𝑜 + 𝑅𝐿 𝑖 𝑖 𝑖𝑖 = 10𝑚𝐴 Observation: The input resistance is irrelevant, since no current divider is in input of the amplifier 𝑖0 = Exercise n. 3 If an amplifier genetares generates an output voltage 𝑣𝑜 = 10𝑉 , having an output internal resistance 𝑅𝑜 = 470Ω, which voltage could deliver on a load resistance 𝑅𝐿 = 50Ω ? ANSWER Naming 𝑣𝐿 the voltage delivered across the load resistor, and considering that 𝑣𝑜 is the Thevenin voltage source and 𝑅𝑜 the Thevenin resistance at output of the amplifier, we can write: 𝑅𝐿 𝑣 𝑣𝐿 = 𝑅𝐿 + 𝑅𝑜 𝑖 therefore 𝑉𝑜 ≅ 0.96𝑉 𝑣𝐿 ≅ 0.96𝑉 Exercise n. 4 An amplifier has a working frequency 𝑓 = 50𝑀𝐻𝑧 and furnishes a max peak-to-peak output voltage 𝑉𝑜,𝑝−𝑝 = 12𝑉 . Which is the minimum slew rate value necessary to assure a correct behaviour of the amplifier? ANSWER 𝑉𝑀 = Exercise n. 5 𝑉𝑜,𝑝−𝑝 2 𝑉 𝑉 𝑆𝑅 = 𝑉𝑀 ∗ 2𝜋𝑓 ≅ 1.88 ∗ 109 � � = 1880 � � 𝑠 𝜇𝑠 𝑉 Calculate the working frequency of amplifier with a slew rate 𝑆𝑅 = 35 �𝜇𝑠� and a peak output voltage 𝑉𝑀 = 6𝑉. ANSWER 𝑓= Exercise n. 6 𝑆𝑅 ≅ 0.93[𝑀𝐻𝑧] 𝑉𝑀 ∗ 2𝜋 Basi di elettronica The ammount of each harmonic of a nonlinear amplifier was measured by an harmonic analyzer resulting the fundamental having an amplitude of 𝑉𝑜1,𝑅𝑀𝑆 = 12𝑉, the second harmonic with an amplitude of 𝑉𝑜2,𝑅𝑀𝑆 = 0.6𝑉, and for the third harmonic it was measured 𝑉𝑜3,𝑅𝑀𝑆 = 0.2𝑉, while all the others resulting negiglile. Determine the harmonic distortion components in percentage, %𝐷2 , %𝐷3 , and the total harmonic distortion THD. ANSWER �𝑉 � �𝑉 � %𝐷2 = �𝑉𝑜2,𝑅𝑀𝑆 � ∗ 100% = 5%; %𝐷3 = �𝑉𝑜3,𝑅𝑀𝑆 � ∗ 100% = 1.66% 𝑜1,𝑅𝑀𝑆 in percentage 𝑇𝐻𝐷 = Observation The theorical equation 2 2 + 𝑉𝑜3,𝑅𝑀𝑆 �𝑉𝑜2,𝑅𝑀𝑆 𝑉𝑜1,𝑅𝑀𝑆 𝑜1,𝑅𝑀𝑆 = √0.62 + 0.22 ≅ 0.053 12 %𝑇𝐻𝐷 = 5.3% 𝑇𝐻𝐷 = 2 �∑𝑁 𝑛=2 𝑉𝑜𝑛,𝑅𝑀𝑆 𝑉𝑜1,𝑅𝑀𝑆 represents the known THD in relation to the fundamental (also termed 𝑇𝐻𝐷𝑓 ), and can result sometimes inconvenient for practical reasons, since the difficulties to measure the amplitude of the fundamental 𝑉𝑜1 alone, so the following working equation can be preferred: 𝑇𝐻𝐷𝑟 = 2 �∑𝑁 𝑛=2 𝑉𝑜𝑛,𝑅𝑀𝑆 𝑉𝑡𝑜𝑡,𝑅𝑀𝑆 that is the root of the sum of the squared rms value of the harmonics above fundamental, 2 compared against 𝑉𝑡𝑜𝑡,𝑅𝑀𝑆 = �∑𝑁 𝑛=1 𝑉𝑜𝑛,𝑅𝑀𝑆 that is the sum of the amplitude of all the 𝑁 harmonic components (in general, only the first 𝑁 = 5 harmonics are significant). In any case, normally the two equations will give very close results, since the very greatest part of the measured output energy (even more than 99%) is contained in the fundamental harmonic 𝑉𝑜1,𝑅𝑀𝑆 . As a rule of thumb, the 𝑇𝐻𝐷𝑟 value approximate the 𝑇𝐻𝐷 one within the 0.5% if the THD is lower than 10%. Exercise n. 7 An input signal of a non-linear amplifier is 𝑣𝑠 = 10 𝑠𝑖𝑛(𝜔𝑡) and its transfer function 𝑣𝑜 = 10𝑣𝑠 + 0.1𝑣𝑠2 . Determine the 2nd harmonic distortion D 2 of the amplifier as a function of input signal amplitude A on the range 0.1..10V , when k1 = 10 and k2 = 0.1. Basi di elettronica ANSWER �𝑉 � �𝑉 � 𝐷2 = �𝑉𝑜2,𝑅𝑀𝑆 �; 𝐷3 = �𝑉𝑜3,𝑅𝑀𝑆 � 𝑜1,𝑅𝑀𝑆 𝑇𝐻𝐷 = 2 2 + 𝑉𝑜3,𝑅𝑀𝑆 �𝑉𝑜2,𝑅𝑀𝑆 = 𝑉𝑜1,𝑅𝑀𝑆 𝑇𝐻𝐷(%) = 𝑜1,𝑅𝑀𝑆 2 2 + 𝑉𝑜3,𝑅𝑀𝑆 �𝑉𝑜2,𝑅𝑀𝑆 2 �𝑉𝑜1,𝑅𝑀𝑆 2 2 + 𝑉𝑜3,𝑅𝑀𝑆 �𝑉𝑜2,𝑅𝑀𝑆 𝑉𝑡𝑜𝑡,𝑅𝑀𝑆 = �𝐷22 + 𝐷32 ∗ 100 Observation The THD value alone hardly define the goodness of an amplifier. In fact, the THD takes into account the harmonic contents as is, reporting the deviation in the shape of the expected waveform. But the THD value is not necessarily related to how the ear perceives distortion. In fact even harmonics are generally considered much less dissonant then odd ones, and higher harmonics more dissonant with respect the lower ones. Curiosity The maximum tolerable value of the THD strictly depends on the particular application. Typical values can be < 10% for audio in telephones, < 5% for most video applications, < 2% for a reasonable audio ampliﬁer for an intercom system, < 1% to < 0.01% for high quality audio, < 0.1% for RF amplifiers. Let’s consider a non-linear amplifier as a sort of “black box” which is fed by a voltage source 𝑣𝑠 = 𝑉 sin(𝜔𝑡) and furnishes an output current equal to 𝑖𝑜 = 𝐼1 𝑣𝑠 + 𝐼2 𝑣𝑠2 + 𝐼3 𝑣𝑠3 . Calculate the harmonic distortions of the second 𝐷2 and third 𝐷3 harmonic frequency components. sin2 (𝜔𝑡) = 1−cos(2𝜔𝑡) 2 𝑖𝑜 = 𝐼1 𝑉 sin(𝜔𝑡) + 𝐼2 𝑉 2 sin2 (𝜔𝑡) + 𝐼3 𝑉 3 sin3 (𝜔𝑡) ; sin3(𝜔𝑡) = 3 sin(𝜔𝑡)−sin(3𝜔𝑡) 4 1 1 𝑖𝑜 = 𝐼1 𝑉 sin(𝜔𝑡) + 𝐼2 𝑉 2 [1 − cos(2𝜔𝑡)] + 𝐼3 𝑉 3 [3 sin(𝜔𝑡) − sin(3𝜔𝑡)] 2 4 1 3 1 1 𝑖𝑜 = 𝐼2 𝑉 2 + �𝐼1 𝑉 + 𝐼3 𝑉 3 � [sin(𝜔𝑡)] − 𝐼2 𝑉 2 [cos(2𝜔𝑡)] − 𝐼3 𝑉 3 [sin(3𝜔𝑡)] 2 4 2 4 𝐷2 = 1 �−𝐼2 𝑉 2 2� |𝐼1 𝑉| = 1 𝐼2 𝑉 2 𝐼1 Basi di elettronica 𝐷3 = 1 �−𝐼3 𝑉 3 4� |𝐼1 𝑉| = 1 𝐼3 2 𝑉 4 𝐼1 Basi di elettronica 1. AMPLIFIERS: BASIC BJT CONFIGURATIONS 1.1. EXERCISES Exercise n. 1 A sinusoidal voltage source 𝑣𝑠 drives a C.E. amplifier in its simplest configuration (Figure 11.62). VCC RC R1 CB1 IC IB CB2 + + vs IE vo - Figure11.62: simplest C.E. configuration Assume 𝑉𝐶𝐶 = 20𝑉, 𝛽 = 50, and that the capacitive reactances negligible at the frequencies of operation. Find 𝑅1 and 𝑅𝐶 so that the Q-point is fixed in 𝐼𝐶𝑄 = 30𝑚𝐴 and 𝑉𝐶𝐸𝑄 = 10𝑉. ANSWER KVL applied around the output loop yields 𝑉𝐶𝐶 = 𝑅𝐶 𝐼𝐶𝑄 + 𝑉𝐶𝐸𝑄 so that 𝑉𝐶𝐶 − 𝑉𝐶𝐸𝑄 𝑅𝐶 = ≅ 330Ω 𝐼𝐶𝑄 KVL applied around the input loop yields 𝑉𝐶𝐶 = 𝑅1 𝐼𝐵𝑄 + 𝑉𝐵𝐸𝑄 but since 𝐼𝐶𝑄 ≅ 𝛽𝐼𝐵𝑄 and 𝑉𝐵𝐸𝑄 ≅ 0.7𝑉 (from Section 8.6.8.2 in Chapter 8), we have 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 𝑅1 = = ≅ 32𝑘Ω 𝐼𝐵𝑄 𝐼𝐶𝑄 ⁄𝛽 Summarizing 𝑅1 ≅ 32𝑘Ω; 𝑅𝐶 ≅ 330Ω. Exercise n. 2 For a common emitter amplifier in its simplest version as in Figure 11.62 of the previous exercise, we have a DC source 𝑉𝐶𝐶 = 12𝑉, a BJT’s Q-point such as 𝑉𝐵𝐸𝑄 = 0.7𝑉; 𝐼𝐵𝑄 = 60𝜇𝐴; 𝑉𝐶𝐸𝑄 = 2 Basi di elettronica 6𝑉 ; 𝐼𝐶𝑄 = 3𝑚𝐴 and, correspondingly ℎ𝑖𝑒 = 800Ω ; ℎ𝑟𝑒 = 4𝑥10−4 ; ℎ𝑓𝑒 = 60 ; ℎ𝑜𝑒 = 50𝑥10−6 Ω−1. Derive the overall values of current and voltage gains, 𝐴𝑖 and 𝐴𝑣 , and those of the input and output resistances, 𝑅𝑖 and 𝑅𝑜 . ANSWER Figure 11.62b shows the DC model useful to perform the DC analysis. VCC RC R1 ICQ IBQ IEQ Figure 11.62b: DC model of the simplest C.E. configuration In applying KVL to the input loop we have 𝑉𝐶𝐶 = 𝑅1 𝐼𝐵𝑄 + 𝑉𝐵𝐸𝑄 from which we can determine 𝑅1 = 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 = 188𝑘Ω 𝐼𝐵𝑄 In a similar way, KLV around the output loop furnishes 𝑉𝐶𝐶 = 𝑅𝐶 𝐼𝐶𝑄 + 𝑉𝐶𝐸𝑄 from which 𝑅𝐶 = 𝑉𝐶𝐶 − 𝑉𝐶𝐸𝑄 = 2𝑘Ω 𝐼𝐶𝑄 The AC analysis allows calculating the requested values starting from what we obtained from the DC analysis. Now, in order to perform the AC analysis we can usefully adopt the AC model as in Figure 11.62c is ic io ib h ie + + vb vs + h rev ce hfe ib 1 R1 Ri ho R iBJT + vce =v o RC R oBJT Ro Figure 11.62c: AC model of the simplest C.E. configuration 3 Basi di elettronica To solve the problem we can usefully adopt the two-equation system for the BJT when in C.E. configuration 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼 𝐶. 𝐸. ) � 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼𝐼 𝐶. 𝐸. ) It is often convenient to write equations starting from the end (the load) and ending to the begin (the source) of the circuit, as in the following. 𝑅𝑖 The output voltage results 𝑣𝑐𝑒 = −𝑅𝐶 𝑖𝑐 from which we can extract 𝑖𝑐 to insert into eq.II C.E. obtaining an expression for 𝑣𝑐𝑒 : 𝑣𝑐𝑒 (1 + ℎ𝑜𝑒 𝑅𝐶 ) = −ℎ𝑓𝑒 𝑅𝐶 𝑖𝑏 The value of 𝑣𝑐𝑒 can be usefully inserted into the eq.I C.E. to have: ℎ𝑓𝑒 𝑅𝐶 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 �− �𝑖 1 + ℎ𝑜𝑒 𝑅𝐶 𝑏 from which we can write 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 𝑅 1 + ℎ𝑜𝑒 𝑅𝐶 𝐶 Finally, by visual inspection of the Figure 11.62c 𝑅𝑖 = 𝑅1 ||𝑅𝑖𝐵𝐽𝑇 Numerically we have 𝑅𝑖𝐵𝐽𝑇 ≅ 756Ω and 𝑅𝑖 ≅ 753Ω, which are very similar, being 𝑅1 ≫ 𝑅𝑖𝐵𝐽𝑇 (188𝑘Ω ≫ 756Ω), therefore the DC bias resistor 𝑅1 does not play any meaningful rule now. We can replace the 𝑣𝑐𝑒 expression 𝑣𝑐𝑒 = −𝑅𝐶 𝑖𝑐 into the eq.II C.E. 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 − ℎ𝑜𝑒 𝑅𝐶 𝑖𝑐 to easily obtain ℎ𝑓𝑒 𝑖𝑐 𝐴𝑖𝐵𝐽𝑇 ≝ = 𝑖𝑏 1 + ℎ𝑜𝑒 𝑅𝐶 The overall current gain 𝐴𝑖 is defined as the output/source current ratio 𝑖𝑜 −𝑖𝑐 𝑖𝑏 𝑅1 𝐴𝑖 ≝ = = −𝐴𝑖𝐵𝐽𝑇 𝑖𝑠 𝑖𝑏 𝑖𝑠 𝑅1 + 𝑅𝑖𝐵𝐽𝑇 4 𝐴𝑖 Basi di elettronica Numerically 𝐴𝑖𝐵𝐽𝑇 ≅ 54.5; 𝐴𝑖 ≅ −54.3. These are two results very similar because 𝑅1 ≫ 𝑅𝑖𝐵𝐽𝑇 so that the current divider does not make any relevant influence. Anyway, please pay close attention to the difference in sign between 𝐴𝑖𝐵𝐽𝑇 and 𝐴𝑖 , and this is for the unequal versus of the output currents. In order to determine the voltage amplification value, let’s start from the output voltage ℎ𝑓𝑒 𝑣𝑐𝑒 = −𝑅𝐶 𝑖𝑐 = −𝑅𝐶 𝑖 1 + ℎ𝑜𝑒 𝑅𝐶 𝑏 in which we can replace 𝑖𝑏 obtained from the following 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 = ℎ𝑖𝑒 𝑖𝑏 − ℎ𝑟𝑒 𝑅𝐶 𝑖𝑐 = ℎ𝑖𝑒 𝑖𝑏 − ℎ𝑟𝑒 𝑅𝐶 ⇒ 𝐴𝑣 ℎ𝑓𝑒 𝑖 1 + ℎ𝑜𝑒 𝑅𝐶 𝑏 𝑣𝑏𝑒 (1 + ℎ𝑜𝑒 𝑅𝐶 ) = �ℎ𝑖𝑒 (1 + ℎ𝑜𝑒 𝑅𝐶 ) − ℎ𝑟𝑒 ℎ𝑓𝑒 𝑅𝐶 �𝑖𝑏 𝐴𝑣𝐵𝐽𝑇 ≝ ℎ𝑓𝑒 𝑅𝐶 𝑣𝑐𝑒 =− 𝑣𝑏𝑒 ℎ𝑖𝑒 + ∆ℎ𝑅𝐶 Now we can obtain the overall voltage gain 𝐴𝑣 ≝ 𝑣𝑜 𝑣𝑠 but since 𝑣𝑜 = 𝑣𝑐𝑒 and 𝑣𝑠 = 𝑣𝑏𝑒 , it follows 𝐴𝑣 = 𝐴𝑣𝐵𝐽𝑇 . This is because the current amplifier is without any input voltage divider since 𝑅𝑠 = 0. Numerically 𝐴𝑣𝐵𝐽𝑇 = 𝐴𝑣 ≅ −144.2. According to the definition of 2-port’s output resistance 𝑣𝑐𝑒 𝑅𝑜𝐵𝐽𝑇 ≝ � 𝑖𝑐 𝑣 =0=𝑣 𝑠 𝑅𝑜 𝑏𝑒 ! To solve the relationship between 𝑣𝑐𝑒 and 𝑖𝑐 when 𝑣𝑠 = ⏞ 0, we can start using the eq.I C.E. 0 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 , from which: ℎ𝑟𝑒 𝑖𝑏 = − 𝑣 ℎ𝑖𝑒 𝑐𝑒 ℎ which replaced into the eq.II C.E. 𝑖𝑐 = ℎ𝑓𝑒 �− ℎ𝑟𝑒 𝑣𝑐𝑒 � + ℎ𝑜𝑒 𝑣𝑐𝑒 furnishes 𝑖𝑒 5 Basi di elettronica 𝑅𝑜𝐵𝐽𝑇 = The overall output resistance is 1 ℎ𝑟𝑒 ℎ𝑓𝑒 ℎ𝑜𝑒 − ℎ𝑖𝑒 𝑅𝑜 = 𝑅𝐶 ||𝑅𝑜𝐵𝐽𝑇 Numerically 𝑅𝑜𝐵𝐽𝑇 = 50𝑘Ω; 𝑅𝑜 ≅ 1923Ω.. Summarizing 𝑅1 = 188𝑘Ω; 𝑅𝐶 = 2𝑘Ω. 𝑅𝑖𝐵𝐽𝑇 ≅ 756Ω; 𝑅𝑖 ≅ 753Ω. 𝐴𝑖𝐵𝐽𝑇 ≅ 54.5; 𝐴𝑖 ≅ −54.3. 𝐴𝑣𝐵𝐽𝑇 = 𝐴𝑣 ≅ −144.2. 𝑅𝑜𝐵𝐽𝑇 = 50𝑘Ω; 𝑅𝑜 ≅ 1923Ω. Exercise n. 3 If in the previous exercise the ℎ𝑟𝑒 value is ignored, what percentage error is made? ANSWER Let’s rewrite the previous equation but imposing ℎ𝑟𝑒 =0 𝑅𝑖 = 𝑅1 ||ℎ𝑖𝑒 𝐴𝑖 = − ℎ𝑓𝑒 𝑅1 1 + ℎ𝑜𝑒 𝑅𝐶 𝑅1 + 𝑅𝑖𝐵𝐽𝑇 𝐴𝑣 = − ℎ𝑓𝑒 𝑅𝐶 (1 + ℎ𝑜𝑒 𝑅𝐶 )ℎ𝑖𝑒 𝑅𝑜 = 𝑅𝐶 || 1 ℎ𝑜𝑒 The current amplification value does not depend on ℎ𝑟𝑒 , so that it is unchanged. Numerically for all the other values 𝐴𝑣 ≅ −136.2; 𝑅𝑖 ≅ 797Ω; 𝑅𝑜 ≅ 1818Ω. Obviously, there is not percentage error on 𝐴𝑖 , while for the others 𝐸%𝐴𝑣 = 136.2−144.2 144.2 100 ≅ 5.45%; 𝐸%𝑅𝑖 ≅ 5.74%; 𝐸%𝑅𝑜 ≅ 5.45%. Observation These errors are so low that can be in the range of variations of parameters due to thermal drifts. For C.E. and C.B. configurations, if we consider ℎ𝑟𝑒 ≅ 0 the relative error on the determined parameters can be quite low, and we have the advantage of mathematics made 6 Basi di elettronica easier, but pay close attention for C.C. configuration for which ℎ𝑟𝑐 ≅ 1. Summarizing no error at all about 𝐴𝑖 , while for the other parameters 𝐸%𝐴𝑣 ≅ 5.45%; 𝐸%𝑅𝑖 ≅ 5.74%; 𝐸%𝑅𝑜 ≅ 5.45%. Exercise n. 4 The Q-point of the simplest C.E. configuration, already represented in Figure 11.62, is fixed by the parameters 𝑉𝐶𝐸𝑄 = 5𝑉; 𝐼𝐶𝑄 = 2𝑚𝐴; 𝐼𝐵𝑄 = 5𝜇𝐴; 𝑉𝐵𝐸𝑄 = 0,7𝑉. For the amplifier, a BJT series BC107A is used which, in correspondence of that Q-point, has the following set of h-parameters: ℎ𝑖𝑒 = 3𝑘Ω ; ℎ𝑟𝑒 = 1.7 ∗ 10−4 ; ℎ𝑓𝑒 = 190; ℎ𝑜𝑒 = 13 ∗ 10−6 Ω−1 . In order to obtain a current gain for a BJT that is equal to 𝐴𝑖𝐵𝐽𝑇 = 180, which values are necessary for:  The collector resistance 𝑅𝐶 (here it works as a load too)  The bias resistor 𝑅1  The DC voltage supply 𝑉𝐶𝐶 With these parameters, determine also the overall current gain 𝐴𝑖 . ANSWER The collector resistance value 𝑅𝐶 can be determined forcing to be 180 the current amplification value ! ℎ𝑓𝑒 = ⏞ 180 𝐴𝑖𝐵𝐽𝑇 = − 1 + ℎ𝑜𝑒 𝑅𝐶 so that 𝑅𝐶 = ℎ𝑓𝑒 − 𝐴𝑖𝐵𝐽𝑇 = 4.27𝑘Ω ℎ𝑜𝑒 𝐴𝑖𝐵𝐽𝑇 KVL applied around the output loop yields 𝑉𝐶𝐶 = 𝑅𝐶 𝐼𝐶𝑄 + 𝑉𝐶𝐸𝑄 so that the voltage requested to the power source must be 𝑉𝐶𝐶 = 13.55𝑉. DC analysis of the input loop gives 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 𝑅1 = 𝐼𝐵𝑄 Numerically 𝑅1 = 2.57𝑀Ω. Recalling the BJT’s input resistance expression, we have ℎ𝑓𝑒 ℎ𝑟𝑒 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 − 𝑅 ≅ 2870Ω 1 + ℎ𝑜𝑒 𝑅𝐶 𝐶 therefore, the overall current amplification value, reduced because of the input current divider, is ℎ𝑓𝑒 𝑅1 𝐴𝑖 = − = −179.8 1 + ℎ𝑜𝑒 𝑅𝐶 𝑅1 + 𝑅𝑖𝐵𝐽𝑇 7 Basi di elettronica Observation Pay close attention to AiBJT and 𝐴𝑖 values. Why are they so similar? Summarizing 𝑅𝐶 = 4.27𝑘Ω, 𝑉𝐶𝐶 = 13.55𝑉, 𝑅1 = 2.57𝑀Ω., 𝐴𝑖 = −179.8. Exercise n. 5 Consider the collector-feedback bias amplifier in Figure 11.63. VCC RC IB IB +IC IC RF + + VBE VCE Figure 1: collector-feedback bias topology The circuital parameters are 𝑉𝐶𝐶 = 12𝑉; 𝑅𝐶 = 2𝑘Ω; 𝑅𝐹 = 120𝑘Ω, and the BJT has 𝛽 = 80. Considering (as usual) 𝑉𝐵𝐸𝑄 = 0.7𝑉 and 𝐼𝐶𝐵𝑂 negligible, determine the Q-point. ANSWER To find the quiescent or Q-point we have to determine the values of 𝐼𝐵𝑄 , 𝑉𝐶𝐸𝑄 and 𝐼𝐶𝑄 . KVL applied around the input loop yields 𝑉𝐶𝐶 = �𝐼𝐵𝑄 + 𝐼𝐶𝑄 �𝑅𝐶 + 𝐼𝐵𝑄 𝑅𝐹 + 𝑉𝐵𝐸𝑄 With the assumption 𝐼𝐶 ≅ 𝛽𝐼𝐵 , we can determine the base bias quiescent current 𝐼𝐵𝑄 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 𝐼𝐵𝑄 = (1 + 𝛽)𝑅𝐶 + 𝑅𝐹 Numerically 𝐼𝐵𝑄 ≅ 40𝜇𝐴 , so that the bias collector quiescent current will be 𝐼𝐶𝑄 = 𝛽𝐼𝐵𝑄 ≅ 3.2𝑚𝐴. Finally, the output voltage 𝑉𝐶𝐸𝑄 which is equal to 𝑉𝐶𝐸𝑄 ≅ 5.5𝑉. 𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶 − 𝑅𝐶 �𝐼𝐶𝑄 + 𝐼𝐵𝑄 � Summarizing 𝐼𝐵𝑄 ≅ 40𝜇𝐴; 𝐼𝐶𝑄 ≅ 3.2𝑚𝐴; 𝑉𝐶𝐸𝑄 ≅ 5.5𝑉 Exercise n. 6 8 Basi di elettronica Determine the stability factors 𝑆𝐼𝐶𝐵𝑂 , 𝑆𝑉𝐵𝐸𝑄 , 𝑆𝛽 of the amplifier of the previous exercise. ANSWER Input loop KVL 𝑉𝐶𝐸 = 𝑅𝐹 𝐼𝐵 + 𝑉𝐵𝐸 Output loop KVL 𝑉𝐶𝐶 = 𝑅𝐶 (𝐼𝐵 + 𝐼𝐶 ) + 𝑉𝐶𝐸 therefore 𝑉𝐶𝐶 = 𝑅𝐶 (𝐼𝐵 + 𝐼𝐶 ) + 𝑅𝐹 𝐼𝐵 + 𝑉𝐵𝐸 from which 𝐼𝐵 = 𝑉𝐶𝐶 − 𝑉𝐵𝐸 − 𝐼𝐶 𝑅𝐶 𝑅𝐹 + 𝑅𝐶 With this result, the generic collector current equation 𝐼𝐶 = 𝛽𝐼𝐵 + (1 + 𝛽)𝐼𝐶𝐵𝑂 becomes 𝐼𝐶 = 𝛽 𝐼𝐶 �1 + 𝑉𝐶𝐶 − 𝑉𝐵𝐸 − 𝐼𝐶 𝑅𝐶 + (1 + 𝛽)𝐼𝐶𝐵𝑂 𝑅𝐹 + 𝑅𝐶 (𝑉𝐶𝐶 − 𝑉𝐵𝐸 ) 𝛽𝑅𝐶 �=𝛽 + (1 + 𝛽)𝐼𝐶𝐵𝑂 𝑅𝐹 + 𝑅𝐶 𝑅𝐹 + 𝑅𝐶 We can then write the stability factors as 𝑆𝐼𝐶𝐵𝑂 ≝ 𝑆𝑉𝐵𝐸𝑄 ≝ Observation 𝑆𝛽 ≝ ∆𝐼𝐶𝑄 1+𝛽 = ∆𝐼𝐶𝐵𝑂 1 + 𝛽 𝑅𝐶 𝑅 +𝑅 𝐹 𝐶 ∆𝐼𝐶𝑄 𝛽 =− ∆𝑉𝐵𝐸𝑄 𝑅𝐹 + (1 + 𝛽)𝑅𝐶 ∆𝐼𝐶𝑄 𝑉𝐶𝐶 − 𝑉𝐵𝐸 − 𝑅𝐶 𝐼𝐶 + (𝑅𝐹 + 𝑅𝐶 )𝐼𝐶𝐵𝑂 = ∆𝛽 𝑅𝐹 + (1 + 𝛽)𝑅𝐶 It is worth pointing out the differences in stability factors of the current collector fixed bias configuration (Section 11.10.1) versus the fixed bias one (Section 11.10.2). We can perform the comparison considering the current resistor 𝑅𝐹 having replaced the resistor 𝑅1 . In such a view, the current fixed bias configuration offers extra terms highlithed in red in the three previous equations, so that we have improvements in stability in the current configuration respect to the fixed bias one. But, be aware as these improvements vanish if 𝑅𝐹 ≫ 𝑅𝐶 (on the other end the feedback effect will be nullified). Numerically 9 Basi di elettronica 𝑆𝐼𝐶𝐵𝑂 = 𝑆𝛽 = 𝑆𝑉𝐵𝐸𝑄 = − 1+𝛽 ≅ 35 𝑅𝐶 1+𝛽 𝑅𝐹 + 𝑅𝐶 𝛽 𝐴 ≅ −0.28 𝑅𝐹 + (1 + 𝛽)𝑅𝐶 𝑉 𝑉𝐶𝐶 − 𝑉𝐵𝐸 − 𝑅𝐶 𝐼𝐶 + (𝑅𝐹 + 𝑅𝐶 )𝐼𝐶𝐵𝑂 ≅ 17.3𝜇𝐴 𝑅𝐹 + (1 + 𝛽)𝑅𝐶 Summarizing 𝐴 𝑆𝐼𝐶𝐵𝑂 ≅ 35, 𝑆𝑉𝐵𝐸𝑄 ≅ −0.28 𝑉, 𝑆𝛽 ≅ 17.3𝜇𝐴. Exercise n. 7 Consider the collector-feedback bias amplifier as already schematized in Figure 11.63. Find the 𝐼𝐶𝑄 − 𝑉𝐶𝐸𝑄 pairs in quiescent conditions when 𝑉𝐶𝐶 = 12𝑉, 𝑅𝐶 = 2𝑘Ω, 𝑅𝐹 = 47𝑘Ω, for the two occurrences of 𝛽 = 80 and 𝛽 = 120. ANSWER To find the expressions for 𝐼𝐶𝑄 and 𝑉𝐶𝐸𝑄 we start applying KVL to the input loop 𝑉𝐶𝐶 = �𝐼𝐵𝑄 + 𝐼𝐶𝑄 �𝑅𝐶 + 𝐼𝐵𝑄 𝑅𝐹 + 𝑉𝐵𝐸𝑄 which, taking into account that 𝐼𝐶 ≅ 𝛽𝐼𝐵 can be rewritten as 𝐼𝐶𝑄 𝐼𝐶𝑄 𝑉𝐶𝐶 ≅ 𝑅𝐶 + 𝑅𝐶 𝐼𝐶 + 𝑅𝐹 + 𝑉𝐵𝐸𝑄 𝛽 𝛽 from which 𝐼𝐶𝑄 ≅ 𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑄 𝑅 𝑅𝐶 + 𝐹 1+𝛽 The expression of 𝐼𝐶𝑄 is useful to determine 𝑉𝐶𝐸𝑄 too. In fact KVL around the output loop gives 𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶 − �𝐼𝐵𝑄 + 𝐼𝐶𝑄 �𝑅𝐶 which, again with the assumption of 𝐼𝐶 ≅ 𝛽𝐼𝐵 can be rewritten as 𝐼𝐶𝑄 𝑉𝐶𝐸𝑄 ≅ 𝑉𝐶𝐶 − � + 𝐼𝐶𝑄 +� 𝑅𝐶 𝛽 Numerically, it results for both the assumptions of 𝛽 = 80 and 𝛽 = 120 respectively 𝐼𝐶𝑄 �𝛽=80 ≅ 4,38𝑚𝐴; 𝑉𝐶𝐸𝑄 �𝛽=80 ≅ 3,13𝑉 𝐼𝐶𝑄 �𝛽=120 ≅ 4,74𝑚𝐴; 𝑉𝐶𝐸𝑄 �𝛽=120 ≅ 2,46𝑉 Observation 10 Basi di elettronica Higher 𝛽 corresponds to higher 𝐼𝐶𝑄 but lower 𝑉𝐶𝐸𝑄 , so higher drop on 𝑅𝐶 . Exercise n. 8 A four-resistor bias for a BJT is given as in Figure 11.64. VCC R1 RC IBQ + VBQ + VBEQ R2 - ICQ + VCEQ - RE I EQ Figure 11.64: Four-resistor bias Determine the values of all the resistors when 𝑉𝐶𝐶 = 12𝑉, 𝐼𝐶𝑄 = 10𝑚𝐴, the output voltage drops 50% on 𝑉𝐶𝐸𝑄 , 40% on 𝑅𝐶 , and 10% on 𝑅𝐸 , and a maximum 𝐼𝐶𝑄 swing of ∆𝐼𝐶𝑄 𝐼𝐶𝑄 = 5% is requested with 𝛽 varying. Determine the dissipated powers in each resistors as well. For the BJT we are using type 2N2222 for which 𝛽𝑚𝑖𝑛 = 100, 𝛽𝑚𝑎𝑥 = 300, 𝛽𝐴𝑉𝐺 = 175. ANSWERS The exercise does not indicate the value of 𝛽 to choice. It is better than to consider the worst case, that is the lowest value 𝛽𝑚𝑖𝑛 = 100, so that in all the other occurrences the amplification value will be higher than the one we will obtain. So 𝐼𝐶𝑄 𝐼𝐵𝑄 = = 100𝜇𝐴 𝛽 𝛽 𝛼= ≅ 0.99 1+𝛽 𝐼𝐶𝑄 𝐼𝐸𝑄 = = 10.1𝑚𝐴 𝛼 The exercise does not suggest how to split the battery’s voltage value among the components of the output loop too. Therefore we will empirically admit the half of 𝑉𝐶𝐶 across the collectoremitter terminals (in such a way the Q-point will be roughly in the middle of the BJT’s active region), and the second half for the 80% across the collector resistor 𝑅𝐶 and for the remaining 20% across the emitter resistor 𝑅𝐸 (this is because the latter plays a rule only as a fixed bias resistor, see Section 11.4.1.4 in Chapter 11). 𝑉𝐶𝐶 𝑉𝐶𝐸 = = 6𝑉 2 11 Basi di elettronica 𝑉𝐶𝐶 = 4.8𝑉 2 𝑉𝐶𝐶 = 0.2 = 1.2𝑉 2 𝑉𝑅𝐶 = 0.8 𝑉𝑅𝐸 The resistor values around the output loop 𝑉𝑅𝐶 = 480Ω 𝐼𝐶 𝑉𝑅𝐸 𝑅𝐸 = ≅ 119Ω 𝐼𝐸 𝑅𝐶 = and their dissipated powers 𝑃𝑅𝐶 = 𝑅𝐶 𝐼𝐶2 = 48𝑚𝑊 𝑃𝑅𝐸 = 𝑅𝐸 𝐼𝐸2 = 12.1𝑚𝑊 Since we do not know if the relationship 𝛽𝑅𝐸 ≥ 10𝑅𝐵 is true or not (Section 11.4.1.5 in Chapter 11), it is better to conservatively proceed adopting the exact analysis, and verifying the possibility to adopt the approximate analysis later on. To determine the Thevenin equivalent, according with the requirement, we can set ∆𝐼𝐶𝑄 ! = ⏞ 0.05 𝐼𝐶𝑄 so that (Section 9.2.1 in Chapter 9) ! ∆𝐼𝐶𝑄 𝑅𝑇𝐻 ∆𝛽 1 = �1 + � = ⏞ 0.05 𝐼𝐶𝑄1 𝑅𝐸 𝛽1 𝛽2 1 + 1 𝑅𝑇𝐻 𝛽2 𝑅𝐸 ! 𝑅𝐸 + 𝑅𝑇𝐻 ∆𝛽 𝛽2 𝑅𝐸 = ⏞ 0.05 𝑅𝐸 𝛽1 𝛽2 𝛽2 𝑅𝐸 + 𝑅𝑇𝐻 from which KVL around the input loop 𝑅𝑇𝐻 = (0.05𝛽1 𝛽2 − ∆𝛽)𝑅𝐸 ≅ 914Ω ∆𝛽 − 0.05𝛽1 𝑉𝑇𝐻 = 𝑅𝑇𝐻 𝐼𝐵𝑄 + 𝑉𝐵𝐸𝑄 + 𝑅𝐸 �𝐼𝐵𝑄 + 𝐼𝐶𝑄 � ≅ 1.99𝑉 The latter value, utilized in ⎧𝑉𝑇𝐻 = ⎨ 𝑅 ⎩ 𝑇𝐻 allows to determine the 𝑅1 and 𝑅2 values 𝑅2 𝑉 𝑅1 + 𝑅2 𝐶𝐶 𝑅1 𝑅2 = 𝑅1 + 𝑅2 𝑅2 𝑉 ⎧𝑉𝑇𝐻 = 𝑅1 𝑅2 𝐶𝐶 ⎪ 𝑅𝑇𝐻 ⎨ 𝑅 𝑅 ⎪𝑅1 + 𝑅2 = 1 2 ⎩ 𝑅𝑇𝐻 12 Basi di elettronica 𝑅1 = 𝑅𝑇𝐻 𝑅2 = 𝑉𝐶𝐶 ≅ 5507Ω 𝑉𝑇𝐻 𝑅𝑇𝐻 𝑉𝐶𝐶 ≅ 1096Ω 𝑉𝐶𝐶 − 𝑉𝑇𝐻 Finally, knowing that 𝑉𝐵𝑄 = 𝑉𝐵𝐸𝑄 + 𝑉𝑅𝐸 = 1.9𝑉, the dissipated powers are 𝑃𝑅1 2 �𝑉𝐶𝐶 − 𝑉𝐵𝑄 � = ≅ 18.5𝑚𝑊 𝑅1 𝑃𝑅2 2 𝑉𝐵𝑄 = ≅ 3.3𝑚𝑊 𝑅2 For the current exercise, since 𝛽𝑅𝐸 > 10𝑅2 is verified, we could admit the approximate analysis. Summarizing 𝑅𝐶 = 𝑉𝑅𝐶 𝐼𝐶 = 480Ω, 𝑅𝐸 = 𝑉𝑅𝐸 𝐼𝐸 ≅ 119Ω. 𝑃𝑅𝐶 = 48𝑚𝑊, 𝑃𝑅𝐸 = 12.1𝑚𝑊. 𝑅1 ≅ 5507Ω, 𝑅2 ≅ 1096Ω. 𝑃𝑅1 ≅ 18.5𝑚𝑊, 𝑃𝑅2 ≅ 3.3𝑚𝑊. Exercise n. 9 Design a voltage divider with emitter feedback bias as in Figure 11.64, so that the total variation of the bias current collector ∆𝐼𝐶𝑄 is within 12%, equally divided among the stability factors 𝑆𝐼𝐶𝐵𝑂 , 𝑆𝑉𝐵𝐸𝑄 , 𝑆𝛽 , for a temperature variation in the range 77-167°F (25-75°C). Utilize a BJT of type BC107 for which 𝐼𝐶𝐵𝑂 | 𝑇=25°𝐶 = 15𝑛𝐴, and when 𝐼𝐶 = 2𝑚𝐴, 𝑉𝐶𝐸 = 5𝑉 we get ℎ𝐹𝐸𝑚𝑖𝑛 = 110 , ℎ𝐹𝐸𝑚𝑎𝑥 = 450 , ℎ𝐹𝐸𝑡𝑦𝑝 = 120 ; 𝑉𝐵𝐸𝑜𝑛 𝑚𝑖𝑛 = 0.55𝑉 , 𝑉𝐵𝐸𝑜𝑛 𝑚𝑎𝑥 = 0.7𝑉 , 𝑉𝐵𝐸𝑜𝑛 𝑡𝑦𝑝 = 0.62𝑉. ANSWER Typical h-parameter values for the BC107 type transistor are reported in section 11.3.4. In particular the cut-off collector current 𝐼𝐶𝐵𝑂 | 𝑇=25°𝐶 = 15𝑛𝐴, and when 𝐼𝐶 = 2𝑚𝐴, 𝑉𝐶𝐸 = 5𝑉, we have the following  DC Current Gain ℎ𝐹𝐸𝑚𝑖𝑛 = 110, ℎ𝐹𝐸𝑚𝑎𝑥 = 450, ℎ𝐹𝐸𝑡𝑦𝑝 = 120  Base-emitter voltage in active region 𝑉𝐵𝐸𝑜𝑛 𝑚𝑖𝑛 = 0.55𝑉, 𝑉𝐵𝐸𝑜𝑛 𝑚𝑎𝑥 = 0.7𝑉, 𝑉𝐵𝐸𝑜𝑛 𝑡𝑦𝑝 = 0.62𝑉 It is also useful to report the expressions of the stability factors as determined in sections 11.10.3.1, 11.10.3.2, and 11.10.3.3 𝑅 𝑅 1 + 𝑅𝑇𝐻 1 + 𝑅𝑇𝐻 𝐸 𝐸 𝑆𝐼𝐶𝐵𝑂 ≅ = 1 𝑅𝑇𝐻 𝐾 𝛽 1+ 𝑅 𝛽 𝐸 13 Basi di elettronica 𝑆𝑉𝐵𝐸𝑄 ≅ − 1 1 =− 1 𝑅𝑇𝐻 𝐾𝛽 𝑅𝐸 1+ 𝑅 𝛽 𝑅𝐸 𝐸 𝑅 𝑅 �1 + 𝑅𝑇𝐻 � ∆𝛽 �1 + 𝑅𝑇𝐻 � ∆𝛽 ∆𝐼𝐶𝑄 𝐸 𝐸 = = 𝛽1 𝛽2 𝐼𝐶𝑄1 �1 + 1 𝑅𝑇𝐻 � 𝛽1 𝛽2 𝐾𝛽2 𝛽2 𝑅𝐸 In order to evaluate the previous equations it is necessary to determine the ratio hypothesis (verified in the following) that 𝐾𝛽2 = 1, results ∆𝐼𝐶𝑄 𝑅𝑇𝐻 ∆𝛽 ! = �1 + � = ⏞ 0.04 𝐼𝐶𝑄1 𝑅𝐸 𝛽1 𝛽2 𝑅𝑇𝐻 𝑅𝐸 which, in the where we imposed a 4% value as requested, so ∆𝛽 𝑅𝑇𝐻 0.04 − 𝛽1 𝛽2 = ≅ 4.82 ∆𝛽 𝑅𝐸 𝛽1 𝛽2 1 𝑅𝑇𝐻 2 𝑅𝐸 which implies that 𝐾𝛽2 = 1 + 𝛽 ≅ 1.01 so resulting the approximation acceptable. The value 𝑅𝑇𝐻 ⁄𝑅𝐸 ≅ 4.82 allows to determine 𝑅𝑇𝐻 and 𝑅𝐸 apart, taking into account that (see section 11.10)  Temperature effect: ∆𝑉𝐵𝐸𝑜𝑛 (𝑇) = −2.5 ∗ 10−3 (𝑇2 − 𝑇1 ) = −125𝑚𝑉  “Dispersion” effect: ∆𝑉𝐵𝐸𝑜𝑛 (𝑑𝑖𝑠𝑝) = 𝑉𝐵𝐸𝑜𝑛 𝑚𝑎𝑥 − 𝑉𝐵𝐸𝑜𝑛 𝑚𝑖𝑛 = 150𝑚𝑉 so, a total of ∆𝑉𝐵𝐸𝑜𝑛 = −275𝑚𝑉 which can be used in the expression ∆𝐼𝐶𝑄 ! ∆𝐼𝐶𝑄 ∆𝑉𝐵𝐸 1 ∆𝑉𝐵𝐸 = ⏞ 0.04 = =− 𝐼𝐶𝑄1 ∆𝑉𝐵𝐸 𝐼𝐶𝑄1 𝐾𝛽 𝑅𝐸 𝐼𝐶𝑄1 to obtain 𝑅𝐸 ≅ 3400Ω and, therefore 𝑅𝑇𝐻 ≅ 4.82𝑅𝐸 ≅ 16.4𝑘Ω The Thevenin voltage 𝑉𝑇𝐻 can be get from the KVL around the input loop where 𝐼𝐵𝑄 = 𝐼𝐶𝑄 𝛽 =ℎ 𝐼𝐶𝑄 𝐹𝐸𝑡𝑦𝑝 𝑉𝑇𝐻 = 𝑅𝑇𝐻 𝐼𝐵𝑄 + 𝑉𝐵𝐸𝑄 + 𝑅𝐸 �𝐼𝐵𝑄 + 𝐼𝐶𝑄 � , obtaining: The expressions of 𝑉𝑇𝐻 = 𝑅 𝑅2 𝑅2 1 +𝑅2 𝑉𝑇𝐻 ≅ 7.76𝑉 𝑅 𝑅 𝑉𝐶𝐶 and 𝑅𝑇𝐻 = 𝑅 1+𝑅2 allow to determine the resistances 𝑅1 and 1 𝑅1 = 2 𝑅1 + 𝑅2 𝑉𝐶𝐶 𝑅𝑇𝐻 = 𝑅 𝑅2 𝑉𝑇𝐻 𝑇𝐻 14 Basi di elettronica Numerically 𝑅2 = 𝑉𝑇𝐻 𝑉𝑇𝐻 (𝑅1 + 𝑅2 ) = 𝑅 𝑉𝐶𝐶 𝑉𝐶𝐶 − 𝑉𝑇𝐻 1 𝑅1 ≅ 25.4𝑘Ω 𝑅2 ≅ 46.4𝑘Ω Finally, the collector resistance 𝑅𝐶 comes from KVL around the output closed path 𝑉𝐶𝐶 − 𝑉𝐶𝐸𝑄 − 𝑅𝐸 𝐼𝐶𝑄 𝑅𝐶 ≅ 𝐼𝐶𝑄 Numerically 𝑅𝐶 ≅ 99Ω Summarizing 𝑅𝐸 ≅ 3400Ω, 𝑅1 ≅ 25.4𝑘Ω, 𝑅2 ≅ 46.4𝑘Ω, 𝑅𝐶 ≅ 99Ω. Observation These values satisfy the DC conditions. However, the designer must take into account the AC conditions as well; in particular, the 𝑅𝐶 value will be fundamental. Therefore, if the AC conditions suggest different values for 𝑅𝐶 , we have to change it, return to DC analysis and iterate the AC analysis until a good compromise is obtained. Exercise n. 10 Given the circuit in Figure 11.65, for which 𝑉𝐶𝐶 = 12𝑉; 𝑉𝐵𝐸 = 0.7𝑉; 𝛽 = 70; and 𝑅𝐶1 = 6𝑘Ω; ′ ′′ = 1𝑘Ω ; 𝑅𝐸2 = 2.2𝑘Ω determine R F so that 𝑉𝐶𝐸1 = 3𝑉 and 𝑅𝐶2 = 4.7𝑘Ω ; 𝑅𝐸1 = 2.2𝑘Ω ; 𝑅𝐸2 𝑉𝐶𝐸1 = 2𝑉. VCC R C2 R C1 T2 I B2 A I C2 T 1 R’E2 I C1 I B1 RF R E1 B R’’E2 Figure 11.65 Voltage divider bias with feedback effect. ANSWER 15 Basi di elettronica KVL around the output loop with 𝑇1 yields 𝑉𝐶𝐶 = 𝑅𝐶1 𝐼𝐶1 + 𝑉𝐶𝐸1 + 𝑅𝐸1 𝐼𝐸1 which, with the approximation 𝐼𝐶1 ≅ 𝐼𝐸1 becomes 𝑉𝐶𝐶 ≅ 𝑉𝐶𝐸1 + (𝑅𝐶1 + 𝑅𝐸1 )𝐼𝐶1 from which 𝐼𝐶1 ≅ 1.1𝑚𝐴. Similarly for 𝐼𝐶2 ′ ′′ )𝐼 𝑉𝐶𝐶 ≅ 𝑉𝐶𝐸2 + (𝑅𝐶2 + 𝑅𝐸2 + 𝑅𝐸2 𝐶2 so that 𝐼𝐶2 ≅ 1.3𝑚𝐴. To obtain the resistance 𝑅𝐹 we can consider the voltage drop across it, that is 𝑉𝐴 − 𝑉𝐵 𝑉𝐴 ≅ 𝑉𝐵𝐸1 + 𝑅𝐸1 𝐼𝐶1 so 𝑉𝐴 = 3.11𝑉, while ′′ 𝑉𝐵 ≅ 𝑅𝐸2 𝐼𝐶2 so 𝑉𝐵 ≅ 2.78𝑉, therefore 𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 = 0.33𝑉. In DC analysis, the current 𝐼𝐵1 comes through 𝑅𝐹 𝑉𝐴𝐵 𝑉𝐴𝐵 = 𝑅𝐹 = 𝐼𝐵1 𝐼𝐶1� 𝛽 The requested value is then 𝑅𝐹 ≅ 21𝑘Ω. Exercise n. 11 Determine the h-parameter ℎ𝑓𝑏 of a BJT with a C.B. configuration, as function of the hparameters of the same BJT with a C.E. configuration, that is ℎ𝑓𝑏 = 𝑓�ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 �. ANSWER ℎ𝑓𝑏 is defined as ℎ𝑓𝑏 ≝ 𝑖𝑐 � 𝑖𝑒 𝑣 𝑐𝑏 =0 so we have to take into account the BJT h-parameter model C.E. configured, as in Figure 11.65b for which 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼 𝐶𝐸) � 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼𝐼 𝐶𝐸) but forcing 𝑣𝑐𝑏 = 0. ib ic h ie + v be + h rev ce h fe ib 1 h oe 16 + v ce Basi di elettronica Figure 11.65b BJT h-parameter model C.E. configured KCL states that for a given node the algebraic sum of the currents is zero. In a broad sense a BJT can be considered as a node itself, and given the conventional versus of the currents in a two-port network (the output current enters into the network rather than exits), we have −𝑖𝑐 = 𝑖𝑏 + 𝑖𝑒 The latter equation inserted into the eq. II CE furnishes 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 = −ℎ𝑓𝑒 (𝑖𝑐 + 𝑖𝑒 ) + ℎ𝑜𝑒 𝑣𝑐𝑒 while into the eq. I CE yields 𝑣𝑏𝑒 = −ℎ𝑖𝑒 (𝑖𝑐 + 𝑖𝑒 ) + ℎ𝑟𝑒 𝑣𝑐𝑒 Now imposing 𝑣𝑐𝑏 = 0 as in the Figure 11.65c v cb=0 ib ic h ie B + v be C + h rev ce h feie 1 E + v ce h oe Figure 11.65c BJT h-parameter model C.E. configured with vcb=0 it follows that 𝑣𝑐𝑒 = 𝑣𝑏𝑒 and the previous equation can be rewritten as 𝑣𝑐𝑒 = −ℎ𝑖𝑒 (𝑖𝑐 + 𝑖𝑒 ) + ℎ𝑟𝑒 𝑣𝑐𝑒 therefore 𝑣𝑐𝑒 = − ℎ𝑖𝑒 (𝑖 + 𝑖𝑒 ) 1 − ℎ𝑟𝑒 𝑐 This expression of 𝑣𝑐𝑒 can be usefully replaced into the eq. II CE to obtain a link between the variables 𝑖𝑐 and 𝑖𝑒 ℎ𝑖𝑒 (𝑖 + 𝑖𝑒 ) 𝑖𝑐 = −ℎ𝑓𝑒 (𝑖𝑐 + 𝑖𝑒 ) − ℎ𝑜𝑒 1 − ℎ𝑟𝑒 𝑐 Finally ℎ𝑓𝑏 ≝ 𝑖𝑐 � 𝑖𝑒 𝑣 𝑐𝑏 =0 =− ℎ𝑖𝑒 ℎ𝑜𝑒 + (1 − ℎ𝑟𝑒 )ℎ𝑓𝑒 ℎ𝑖𝑒 ℎ𝑜𝑒 + (1 − ℎ𝑟𝑒 )�1 + ℎ𝑓𝑒 � Exercise n. 12 A four-resistor bias amplifier is supplied from a signal source with internal resistance 𝑅𝑠 = 1𝑘Ω.. The amplifier feeds a load resistor 𝑅𝐿 = 1𝑘Ω (Figure 11.66). 17 Basi di elettronica VCC RC R1 iL ig CB1 Rg CB2 + vi + RL R2 RE CE vg + vL Figure 11.66 C.E. amplifier Considering that the values of resistors are 𝑅1 = 40𝑘Ω, 𝑅2 = 10𝑘Ω, 𝑅𝐶 = 5𝑘Ω, 𝑅𝐿 = 1𝑘Ω, and the BJT’s hybrid parameters are ℎ𝑖𝑒 = 2𝑘Ω , ℎ𝑟𝑒 = 2 ∗ 10−3 , ℎ𝑓𝑒 = 120 , ℎ𝑜𝑒 = 2 ∗ 10−4 𝑆 , determine the following: 𝑖  The overall current amplifier 𝐴𝑖 = 𝑖𝐿 𝑣  The overall input resistance 𝑅𝑖 = 𝑖 𝑖 𝑔  The overall voltage amplifier 𝐴𝑣 =  The overall output resistance 𝑅𝑜 = 𝑔 𝑣𝐿 𝑣𝑖 𝑣𝐿 𝑖𝐿 ANSWER The equivalent AC model is represented in Figure 11.66b. ib ig Rg + vg ic iL h ie B + v i=vb R1 C + h rev ce h feie RC 1 R2 E RL + v o=vL h oe Figure 11.66b AC equivalent model of the CE amplifier For convenience we can combine 𝑅𝐶𝐿 = 𝑅𝐶 ||𝑅𝐿 ≅ 833Ω. and 𝑅12 = 𝑅1 ||𝑅2 = 8𝑘Ω. According the Sections 11.4.2.2, 11.4.2.3, 11.4.2.4 and 11.4.2.5, we have the partial results ℎ𝑓𝑒 𝐴𝑖𝐵𝐽𝑇 = 1 + ℎ𝑜𝑒 𝑅𝐶𝐿 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 − ℎ𝑟𝑒 18 ℎ𝑓𝑒 𝑅𝐶𝐿 1 + ℎ𝑜𝑒 𝑅𝐶𝐿 Basi di elettronica 𝐴𝑣𝐵𝐽𝑇 = − ℎ𝑓𝑒 𝑅𝐶𝐿 (1 + ℎ𝑜𝑒 𝑅𝐶𝐿 )ℎ𝑖𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 𝑅𝐶𝐿 𝑅𝑜𝐵𝐽𝑇 = 1 ℎ𝑓𝑒 ℎ𝑟𝑒 ℎ𝑜𝑒 − ℎ𝑖𝑒 + (𝑅𝑠 ||𝑅12 ) Numerically 𝐴𝑖𝐵𝐽𝑇 ≅ 103; 𝑅𝑖𝐵𝐽𝑇 ≅ 1829Ω; 𝐴𝑣𝐵𝐽𝑇 ≅ −46,9; 𝑅𝑜𝐵𝐽𝑇 ≅ 8553Ω. In order to obtain the complete results we have to consider the current and voltage dividers, so that but 𝑖𝐿 = − 𝑅 𝑅𝐶 𝐶 +𝑅𝐿 𝑖𝑐 and 𝑖𝑏 = 𝑅 𝑅12 𝐴𝑖 = 12 +𝑅𝑖𝐵𝐽𝑇 𝑖𝐿 𝑖𝑐 𝑖𝑏 = 𝛼𝑖𝑜 𝐴𝑖𝐵𝐽𝑇 𝛼𝑖𝑖 𝑖𝑐 𝑖𝑏 𝑖𝑔 𝑖𝑠 , so that 𝛼𝑖𝑜 = − 𝛼𝑖𝑖 = Therefore 𝐴𝑖 = 𝛼𝑖𝑜 𝐴𝑖𝐵𝐽𝑇 𝛼𝑖𝑖 = − 𝑅𝐶 𝑅𝐶 + 𝑅𝐿 𝑅12 𝑅12 + 𝑅𝑆 ℎ𝑓𝑒 𝑅𝐶 𝑅12 𝑅𝐶 + 𝑅𝐿 1 + ℎ𝑜 𝑅𝐶𝐿 𝑅12 + 𝑅𝑖𝐵𝐽𝑇 Numerically 𝛼𝑖𝑜 ≅ −0,833; 𝛼𝑖𝑖 ≅ 0,814; 𝐴𝑖 ≅ −69,8. 𝑅𝑖 = ℎ𝑓𝑒 𝑅𝐶𝐿 𝑣𝑖 = 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 = 𝑅12 || �ℎ𝑖𝑒 − ℎ𝑟𝑒 � 𝑖𝑠 1 + ℎ𝑜𝑒 𝑅𝐶𝐿 Numerically 𝑅𝑖 ≅ 1488Ω. but 𝑣𝑖 = 𝑣𝑏 = 𝑅 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 12 ||𝑅𝑖𝐵𝐽𝑇 +𝑅𝑆 𝐴𝑣 = 𝐴𝑖 𝑣𝐿 𝑣𝑖 𝑣𝑐 𝑣𝑖 𝑣𝑐 𝑣𝑏 = = = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 𝑣𝑖 𝑣𝑠 𝑣𝑖 𝑣𝑠 𝑣𝑏 𝑣𝑠 𝑣𝑠 so that 19 𝑅𝑖 𝐴𝑣 Basi di elettronica 𝛼𝑣 = therefore 𝐴𝑣 = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 = − 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 ℎ𝑓𝑒 𝑅𝐶𝐿 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 (1 + ℎ𝑜𝑒 𝑅𝐶𝐿 )ℎ𝑖𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 𝑅𝐶𝐿 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 Numerically 𝛼𝑣 ≅ 0,598; 𝐴𝑣 ≅ −28. 𝑅𝑜 = 𝑅𝑜𝐵𝐽𝑇 ||𝑅𝐶𝐿 𝑅𝑜 Summarizing 𝐴𝑖𝐵𝐽𝑇 ≅ 103; 𝑅𝑖𝐵𝐽𝑇 ≅ 1829Ω; 𝐴𝑣𝐵𝐽𝑇 ≅ −46,9; 𝑅𝑜𝐵𝐽𝑇 ≅ 8553Ω. 𝐴𝑖 ≅ −69,8; 𝑅𝑖 ≅ 1488Ω, 𝐴𝑣 ≅ −28; 𝑅𝑜 ≅ 759Ω. Observation Pay close attention to the input and output resistance values, 𝑅𝑖 ≅ 1488Ω and 𝑅𝑜 ≅ 759Ω. They are not suitable for an ideal voltage amplifier for which we require a high input resistance and a low output resistance. However, the obtained values can be of same interest for a transresistance amplifier, even if for the latter both resistances have to be as low as possible. Exercise n. 13 Design a C.E. amplifier as in Figure 11.67 such that the BJT works in its active region, the overall voltage gain must be 𝐴𝑣 = −200, and the DC supply 𝑉𝐶𝐶 = 12𝑉. Utilize an npn BJT with the following h-parameter set: ℎ𝑓𝑒 = ℎ𝐹𝐸 = 100, ℎ𝑖𝑒 = 1.5𝑘Ω,but ℎ𝑟𝑒 and ℎ𝑜𝑒 are negligible. Verify that the power collector 𝑃𝐶 = 𝑉𝐶𝐸𝑄 𝐼𝐶𝑄 is lower than 50mW. 20 Basi di elettronica VCC R1 I R1 RC VB I B R2 I R2 V RE E CE Figure 11.67 “Open” configuration of the CE amplifier Suggestions:  Choice a Q-point in the middle of the BJT’s active region.  𝑉𝑅𝐶 and 𝑉𝑅𝐸 voltage drops of 80% and 20% of  𝐼𝐵𝑄 = 0.1𝐼𝑅1 𝑉𝐶𝐶 2 respectively. ANSWER We can apply the superposition theorem, because the BJT has to work in its active region so that the amplifier behaves as a linear network. For this reason we can treat separately the DC and AC behavior of the circuit, the equivalent models being schematized in Figure 11.67b (a) and (b) respectively. VCC + VR1 + RC IRC VRC IR1 R1 - + ic + IR2 RE IRE VRE VR2 - ib R2 - (a) R1 R2 RC hie + vce hfe i b - (b) Figure 11.67b: equivalent circuit for (a) DC and (b) AC conditions The AC equivalent circuit shows a simplified h-parameter model of the BJT, because we know that ℎ𝑟𝑒 and ℎ𝑜𝑒 have practically zero values, so that we can write 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 (𝑒𝑞. 𝐼, 𝐶𝐸𝑠) � 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 (𝑒𝑞. 𝐼𝐼, 𝐶𝐸𝑠) Since 𝐴𝑣 = −200, taking into account the two aformentioned equations, we can impose 21 Basi di elettronica 𝐴𝑣 ≝ ! ℎ𝑓𝑒 𝑣𝑐𝑒 −𝑅𝐶 𝑖𝑐 𝑅𝐶 = =− 𝐴𝑖 = − 𝑅𝐶 = ⏞ − 200 𝑣𝑏𝑒 ℎ𝑖𝑒 𝑖𝑏 ℎ𝑖𝑒 ℎ𝑖𝑒 As a consequence we can determine the value of the collector resistor ℎ𝑖𝑒 𝑅𝐶 = 𝐴 = 3𝑘Ω ℎ𝑓𝑒 𝑣 In DC analysis we can easily determine some constant voltage values ! 𝑉𝐶𝐶 𝑉𝑅𝐶 = ⏞ 0.8 = 4.8𝑉 2 ! 𝑉𝑅𝐸 = ⏞ 0.2 𝑉𝐶𝐶 = 1.2𝑉 2 and, to guarantee the Q-point to be in the middle of the BJT active region, we set ! 𝑉 𝐶𝐶 𝑉𝐶𝐸𝑄 = ⏞ = 6𝑉 2 KVL around the output loop yields 𝑉𝐶𝐶 = 𝑅𝐶 𝐼𝐶𝑄 + 𝑉𝐶𝐸𝑄 + 𝑅𝐸 𝐼𝐸𝑄 ≅ (𝑅𝐶 + 𝑅𝐸 )𝐼𝐶𝑄 + 𝑉𝐶𝐸𝑄 but 𝑉𝑅𝐶 = 𝑅𝐶 𝐼𝐶𝑄 𝑉𝑅𝐸 = 𝑅𝐸 𝐼𝐸𝑄 ≅ 𝑅𝐸 𝐼𝐶𝑄 therefore 𝑅𝐸 = 𝑉𝑅𝐸 𝑉𝑅𝐸 𝑉𝑅𝐸 ≅ = = 1.6𝑚𝐴 𝐼𝐸𝑄 𝐼𝐶𝑄 𝑉𝑅𝐶 ⁄𝑅𝐶 To determine the values of the input loop resistances, 𝑅1 ed 𝑅2 , we can approach the problem with the Thevenin equivalent model, as in Figure 11.67c, RTH VB + + RE + VRE VTH - - Figure 11.67c: Thevenin equivalent model of the input loop for which 𝑉𝑇𝐻 = The BE-KVL gives us 𝑅2 𝑉 𝑅1 + 𝑅2 𝐶𝐶 𝑅𝑇𝐻 = 𝑅1 ||𝑅2 𝑉𝐵𝑄 = 𝑉𝑇𝐻 − 𝐼𝐵𝑄 𝑅𝑇𝐻 = 𝐼𝐶𝑄 𝑅1 𝑅2 𝑅2 𝑉𝐶𝐶 − 𝑅1 + 𝑅2 ℎ𝐹𝐸 𝑅1 + 𝑅2 22 Basi di elettronica but so that 𝑉𝐵𝑄 = 𝑉𝐵𝐸𝑄 + 𝑉𝑅𝐸 = 0.7 + 1.2 = 1.9𝑉 𝑅2 ≅ In addition 𝑅1 = 1.9 𝑅 10.1 1 𝑉𝐶𝐶 − 𝑉𝐵𝑄 𝑉𝐶𝐶 − �𝑉𝐵𝐸𝑄 + 𝑉𝑅𝐸 � = ≅ 63𝑘Ω 𝐼𝑅1 𝐼𝑅1 Finally, the dissipate power at the collector 𝑃𝐶𝑄 = 𝑉𝐶𝐸𝑄 𝐼𝐶𝑄 = 9.6𝑚𝑊 much lower than the mentioned 50mW. Summarizing 𝑅𝐶 = 3𝑘Ω; 𝑅𝐸 = 750Ω; 𝑅1 ≅ 63𝑘Ω; 𝑅2 ≅ 12𝑘Ω; 𝑃𝐶𝑄 = 9.6𝑚𝑊. Exercise n. 14 ′ ′ ′ ′ , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 of the scheme shown in Figure 11.68, as function Determine the h’-parameters ℎ𝑖𝑒 of the h-parameters of the only BJT, that is: ′ ℎ𝑖𝑒 = 𝑓�ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 � ′ = 𝑓�ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 � ℎ𝑟𝑒 ′ ℎ𝑓𝑒 = 𝑓�ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 � ′ ℎ𝑜𝑒 = 𝑓�ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 � RE Figure 11.68 BJT with emitter resistor ANSWER A series resistor 𝑅𝐸 positively influence the stability of the Q-point since, thanks to its presence, the DC component of the 𝐼𝐶 current becomes relatively independent from the variations of the temperature and, consequently, from the variations of the parameters of the BJT, starting from its β value. The resistor 𝑅𝐸 uses the current 𝐼𝐸 (very close to 𝐼𝐶 ), of the output loop, to feedback the voltage 𝑉𝑅𝐸 at the input loop which reduces the base voltage 𝑉𝐵 , so opposing the increase of the bias current. Figure 11.68b represents the AC equivalent model of the BJT including the emitter resistor 𝑅𝐸 . 23 Basi di elettronica ib ic hie + + B C + hrev ce hfei b v be vb E - + 1 v ce hoe - + i e=ib+ic RE ve - + vc - - Figure 11.68b AC equivalent model of the BJT with emitter resistor This model can be conveniently modified with the one in Figure 11.68c ib + ic h’ie B + h’revce C h’fei b v be 1 + v ce h’oe E Figure 11.68c ′ ′ ′ ′ with new hybrid parameters, ℎ𝑖𝑒 , ℎ𝑟𝑒 , ℎ𝑓𝑒 , ℎ𝑜𝑒 , which include the emitter resistor 𝑅𝐸 . Such new parameters are defined as ′ ℎ𝑖𝑒 ≝ 𝑣𝑏 � 𝑖𝑏 𝑣𝑐=0 𝑖 ′ ; ℎ𝑓𝑒 ≝ 𝑖𝑐 � 𝑏 𝑣𝑐=0 ′ ; ℎ𝑟𝑒 ≝ 𝑣𝑏 � 𝑣𝑐 𝑖𝑏=0 𝑖 ′ ; ℎ𝑜𝑒 ≝ 𝑣𝑐 � 𝑐 𝑖𝑏=0 . Considering that 𝑣𝑐𝑒 = 𝑣𝑐 − 𝑣𝑒 , we can write 𝑣𝑐 = (𝑣𝑐 − 𝑣𝑒 ) + 𝑣𝑒 = 𝑣𝑐𝑒 + 𝑣𝑒 ′ ′ In order to determine ℎ𝑖𝑒 e ℎ𝑓𝑒 we admit 𝑣𝑐 = 0, so that 𝑣𝑐𝑒 = −𝑣𝑒 = −𝑅𝐸 (𝑖𝑏 + 𝑖𝑐 ) which, together with the eq.II C.E. (Section 11.4.2) becomes from which 𝑣𝑐𝑒 = −𝑅𝐸 𝑖𝑏 − 𝑅𝐸 �ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 � 𝑣𝑐𝑒 = − 𝑅𝐸 𝑖𝑏 �1 + ℎ𝑓𝑒 � = −𝑣𝑒 1 + ℎ𝑜𝑒 𝑅𝐸 ′ We can usefully utilize the expression of 𝑣𝑐𝑒 to determine ℎ𝑖𝑒 . KVL around the input loop 𝑣𝑏 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 + 𝑣𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 − 𝑣𝑐𝑒 𝑣𝑏 = ℎ𝑖𝑒 𝑖𝑏 + (1 − ℎ𝑟𝑒 ) 24 𝑅𝐸 𝑖𝑏 �1 + ℎ𝑓𝑒 � 1 + ℎ𝑜𝑒 𝑅𝐸 ′ ℎ𝑖𝑒 Basi di elettronica 𝑣𝑏 (1 + ℎ𝑜𝑒 𝑅𝐸 ) = �ℎ𝑖𝑒 (1 + ℎ𝑜𝑒 𝑅𝐸 ) + (1 − ℎ𝑟𝑒 )𝑅𝐸 �1 + ℎ𝑓𝑒 ��𝑖𝑏 as a consequence ′ ℎ𝑖𝑒 ≝ (1 − ℎ𝑟𝑒 )�1 + ℎ𝑓𝑒 � 𝑣𝑏 � = ℎ𝑖𝑒 + 𝑅𝐸 𝑖𝑏 𝑣𝑐=0 1 + ℎ𝑜𝑒 𝑅𝐸 If we admit the approximations ℎ𝑟𝑒 ≪ 1 and ℎ𝑓𝑒 ≫ 1, it becomes ℎ𝑓𝑒 𝑣𝑏 ′ ℎ𝑖𝑒 ≝ � ≅ ℎ𝑖𝑒 + 𝑅 𝑖𝑏 𝑣𝑐=0 1 + ℎ𝑜𝑒 𝑅𝐸 𝐸 ′ ℎ𝑓𝑒 ′ We can usefully utilize the expression of 𝑣𝑐𝑒 to determine ℎ𝑓𝑒 too. In fact, together with the eq.II C.E. (Section 11.4.2) becomes 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 − ℎ𝑜𝑒 therefore 𝑅𝐸 𝑖𝑏 �1 + ℎ𝑓𝑒 � �1 + ℎ𝑓𝑒 �𝑅𝐸 = �ℎ𝑓𝑒 − ℎ𝑜𝑒 �𝑖 1 + ℎ𝑜𝑒 𝑅𝐸 1 + ℎ𝑜𝑒 𝑅𝐸 𝑏 ′ ℎ𝑓𝑒 ≝ ℎ𝑓𝑒 − ℎ𝑜𝑒 𝑅𝐸 𝑖𝑐 � = 𝑖𝑏 𝑣𝑐=0 1 + ℎ𝑜𝑒 𝑅𝐸 ′ ′ and ℎ𝑜𝑒 we have to admit 𝑖𝑏 = 0, so that the KVL around the To determine the parameters ℎ𝑟𝑒 input loop 𝑣𝑏 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 + 𝑣𝑒 which becomes 𝑣𝑏 = ℎ𝑟𝑒 𝑣𝑐𝑒 + 𝑣𝑒 Being 𝑖𝑏 = 0, the output current will be null, ℎ𝑓𝑒 𝑖𝑏 = 0, therefore 𝑖𝑐 𝑣𝑐𝑒 = ℎ𝑜𝑒 𝑣 from which 𝑖𝑐 = 𝑅𝑒 and 𝐸 But 𝑣𝑐𝑒 = 𝑣𝑐 − 𝑣𝑒 therefore 𝑣𝑒 = 𝑅𝐸 𝑖𝑐 𝑣𝑐𝑒 = 𝑣𝑒 ℎ𝑜𝑒 𝑅𝐸 𝑣𝑒 = 𝑣𝑐 − 𝑣𝑒 ℎ𝑜𝑒 𝑅𝐸 25 Basi di elettronica 𝑣𝑐 = �1 + 𝑣𝑒 = 1 �𝑣 ℎ𝑜𝑒 𝑅𝐸 𝑒 ℎ𝑜𝑒 𝑅𝐸 𝑣 1 + ℎ𝑜𝑒 𝑅𝐸 𝑐 ′ ℎ𝑟𝑒 ′ To determine ℎ𝑟𝑒 we can replace 𝑣𝑒 and 𝑣𝑐𝑒 values into the KVL equation around the input loop 𝑣𝑒 ℎ𝑟𝑒 ℎ𝑜𝑒 𝑅𝐸 ℎ𝑜𝑒 𝑅𝐸 ℎ𝑟𝑒 + ℎ𝑜𝑒 𝑅𝐸 𝑣𝑏 = ℎ𝑟𝑒 + 𝑣𝑒 = 𝑣𝑐 + 𝑣𝑐 = 𝑣 ℎ𝑜𝑒 𝑅𝐸 ℎ𝑜𝑒 𝑅𝐸 1 + ℎ𝑜𝑒 𝑅𝐸 1 + ℎ𝑜𝑒 𝑅𝐸 1 + ℎ𝑜𝑒 𝑅𝐸 𝑐 Therefore ′ ℎ𝑟𝑒 ≝ 𝑣𝑏 ℎ𝑟𝑒 + ℎ𝑜𝑒 𝑅𝐸 � = 𝑣𝑐 𝑖𝑏=0 1 + ℎ𝑜𝑒 𝑅𝐸 ′ ℎ𝑜𝑒 Utilizing the 𝑣𝑒 expression, we have ℎ𝑜𝑒 𝑅𝐸 𝑣𝑒 1 + ℎ𝑜𝑒 𝑅𝐸 𝑣𝑐 ℎ𝑜𝑒 𝑖𝑐 = = = 𝑣 𝑅𝐸 𝑅𝐸 1 + ℎ𝑜𝑒 𝑅𝐸 𝑐 therefore ′ ℎ𝑜𝑒 ≡ 𝑖𝑐 ℎ𝑜𝑒 � = 𝑣𝑐 𝑖𝑏=0 1 + ℎ𝑜𝑒 𝑅𝐸 Summarizing ′ ℎ𝑖𝑒 = ℎ𝑖𝑒 + (1−ℎ𝑟𝑒 )�1+ℎ𝑓𝑒 � so: 1+ℎ𝑜𝑒 𝑅𝐸 ′ 𝑅𝐸 , ℎ𝑓𝑒 = [𝐻 ′ ] = ′ ℎ𝑖𝑒 � ′ ℎ𝑓𝑒 ℎ𝑓𝑒 −ℎ𝑜𝑒 𝑅𝐸 1+ℎ𝑜𝑒 𝑅𝐸 ′ ℎ𝑟𝑒 ′ � ℎ𝑜𝑒 ′ , ℎ𝑟𝑒 = ℎ𝑟𝑒 +ℎ𝑜𝑒 𝑅𝐸 1+ℎ𝑜𝑒 𝑅𝐸 ℎ𝑓𝑒 𝑅𝐸 ⎡ℎ𝑖𝑒 + 1 + ℎ𝑜𝑒 𝑅𝐸 = ⎢⎢ ℎ − ℎ𝑜𝑒 𝑅𝐸 ⎢ 𝑓𝑒 ⎣ 1 + ℎ𝑜𝑒 𝑅𝐸 ℎ ′ , ℎ𝑜𝑒 = 1+ℎ𝑜𝑒 𝑅 𝑜𝑒 𝐸 ℎ𝑟𝑒 + ℎ𝑜𝑒 𝑅𝐸 ⎤ 1 + ℎ𝑜𝑒 𝑅𝐸 ⎥ ⎥ ℎ𝑜𝑒 ⎥ 1 + ℎ𝑜𝑒 𝑅𝐸 ⎦ Exercise n. 15 Consider a four-resistor emitter feedback biased C.E. amplifier as represented in Figure 11.69. Assume a DC voltage source 𝑉𝐶𝐶 = 10𝑉, an AC voltage source with internal resistance 𝑅𝑠 = 500Ω, and a load resistor 𝑅𝐿 = 10𝑘Ω.. Assume a Q-point for which 𝐼𝐵𝑄 = 10𝜇𝐴, 𝐼𝐶𝑄 = 1𝑚𝐴, 26 Basi di elettronica 𝑉 𝑉 𝑉𝐶𝐸𝑄 = 2𝐶𝐶 , 𝑉𝑅𝐶𝑄 = 0.8 2𝐶𝐶 . The voltage divider bias is such that 𝐼𝑅1 ≅ 10𝐼𝐵𝑄 . The BJT type at the Q-point presents the following set of h-parameters: ℎ𝑖𝑒 = 1.5𝑘Ω , ℎ𝑟𝑒 = 10−3 , ℎ𝑓𝑒 = 200, ℎ𝑜𝑒 = 10−3 Ω−1. VCC R1 is CB1 + CB2 + Rs vs ICQ+ic RC IBQ+ib vB R2 RE il v l RL CE IEQ + ie Figure 11.69 Four-resistor emitter feedback biased C.E. amplifier Determine for the values of each resistors 𝑅𝐶 , 𝑅𝐸 , 𝑅1 , 𝑅2 in DC conditions. Determine the values of BJT’s and overall input/output resistances and gains in AC conditions. ANSWER Because of the presence of the blocking capacitors, the DC equivalent model results as represented in Figure 11.69b. VCC R1 RC IBQ + VBQ + VBEQ R2 - ICQ + VCEQ - RE I EQ Figure 11.69b DC equivalent model of the four-resistor emitter feedback biased C.E. amplifier KVL around the output loop yields 𝑉𝐶𝐶 = 𝑅𝐶 𝐼𝐶 + 𝑉𝐶𝐸 + 𝑅𝐸 𝐼𝐸 ≅ (𝑅𝐶 + 𝑅𝐸 )𝐼𝐶 + 𝑉𝐶𝐸 with the reasonable hypothesis that 𝐼𝐸 ≅ 𝐼𝐶 . The resistance values around the output loop will be then 𝑉𝐶𝐶 𝑉𝑅𝐶 0.8 2 4 𝑅𝐶 = = = = 4𝑘Ω 𝐼𝐶𝑄 𝐼𝐶𝑄 0.001 27 Basi di elettronica 𝑉𝐶𝐶 𝑉𝑅𝐸 0.2 2 1 𝑅𝐸 = ≅ = = 1𝑘Ω 𝐼𝐸𝑄 𝐼𝐶𝑄 0.001 Named 𝑉𝐵 the base voltage, the resistance values around the input loop will be 𝑉𝐵 𝑉𝐵𝐸𝑄 + 𝑉𝑅𝐸 1.7 𝑅2 = = = ≅ 18.9𝑘Ω 𝐼𝑅2 𝐼𝑅1 − 𝐼𝐵𝑄 90 ∗ 10−6 𝑉𝐶𝐶 − 𝑉𝐵 𝑉𝐶𝐶 − �𝑉𝐵𝐸𝑄 + 𝑉𝑅𝐸 � 10 − (0.7 + 1) = = = 83𝑘Ω 𝐼𝑅1 𝐼𝑅1 10𝐼𝐵𝑄 To determine the overall input/output resistances and gains we have to draw the AC model as in Figure 11.69c. ic il is ib hie Rs B C + + + R1 R2 hfei b RC RL 1 vs hoe vl hrevce 𝑅1 = E Ri RoBJT RiBJT Ro Figure 11.69c AC equivalent model of the four-resistor emitter feedback biased C.E. amplifier For convenience, resistors were combined as 𝑅𝐶𝐿 = 𝑅𝐶 ||𝑅𝐿 ≅ 2.9𝑘Ω and 𝑅12 = 𝑅1 ||𝑅2 ≅ 15.4𝑘Ω. It is useful to recall, and later utilize, the hybrid parameter equations of the BJT in C.E. configurations 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼 𝐶𝐸) � 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 (𝑒𝑞. 𝐼𝐼 𝐶𝐸) 𝑅𝑖𝐵𝐽𝑇 ≝ 𝑣𝑏𝑒 𝑖𝑏 𝑅𝑖 ; 𝑣 The output voltage 𝑣𝑐𝑒 = −𝑅𝐶𝐿 𝑖𝑐 can be replaced into the eq. II CE to obtain − 𝑅𝑐𝑒 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 , from which we can extract 𝑣𝑐𝑒 and insert it into the eq. I CE yielding ℎ𝑟𝑒 ℎ𝑓𝑒 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 − 𝑅 1 + ℎ𝑜𝑒 𝑅𝐶𝐿 𝐶𝐿 𝐶𝐿 Finally 𝑅𝑖 = 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 Numerically 𝑅𝑖𝐵𝐽𝑇 ≅ 1350Ω, 𝑅𝑖 ≅ 1.24𝑘Ω. 𝐴𝑖 28 Basi di elettronica 𝑖 𝐴𝑖 ≝ 𝑖𝑐; 𝑠 The output voltage 𝑣𝑐𝑒 ℎ𝑜𝑒 𝑅𝐶𝐿 𝑖𝑐 , so that 𝑖𝑐 𝑖𝑏 = 𝐴𝑖𝐵𝐽𝑇 𝛼𝑖𝑖 𝑖𝑏 𝑖𝑠 = −𝑅𝐶𝐿 𝑖𝑐 can be replaced into the eq. II CE to obtain 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 − 𝐴𝑖 = ℎ𝑓𝑒 𝑖𝑐 = 𝑖𝑏 1 + ℎ𝑜 𝑅𝐶𝐿 The same result can be obtained from the current divider expression around the output loop 𝑖𝑐 = 1 ℎ𝑜𝑒 1 +𝑅𝐶𝐿 ℎ𝑜𝑒 𝐴𝑖𝐵𝐽𝑇 = ℎ𝑓𝑒 𝑖𝑏 . The expression of the current divider around the input loop is 𝑅12 𝛼𝑖𝑖 = ≅ 0.92 𝑅12 + 𝑅𝑖𝐵𝐽𝑇 therefore 𝐴𝑖 = 𝐴𝑣 ≝ 𝑣𝑐𝑒 𝑣𝑠 ℎ𝑓𝑒 𝑅12 ≅ 48 1 + ℎ𝑜 𝑅𝐶𝐿 𝑅12 + 𝑅𝑖𝐵𝐽𝑇 𝐴𝑣 ; 𝐴𝑣 = 1 𝑣𝑐𝑒 𝑣𝑏𝑒 = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 𝑣𝑏𝑒 𝑣𝑠 From the eq. I CE we can write 𝑖𝑏 = ℎ (𝑣𝑏𝑒 − ℎ𝑟𝑒 𝑣𝑐𝑒 ) which can be replaced into the eq. II CE obtaining 𝑖𝑐 = ℎ𝑓𝑒 ℎ𝑖𝑒 𝑖𝑒 𝑣 (𝑣𝑏𝑒 − ℎ𝑟𝑒 𝑣𝑐𝑒 ) + ℎ𝑜𝑒 𝑣𝑐𝑒 . But 𝑖𝑐 = − 𝑐𝑒 , so that 𝑣𝑐𝑒 �ℎ𝑖𝑒 + ℎ𝑖𝑒 ℎ𝑜𝑒 𝑅𝐶𝐿 − 𝑅 ℎ𝑓𝑒ℎ𝑟𝑒𝑅𝐶𝐿=−ℎ𝑓𝑒𝑅𝐶𝐿𝑣𝑏𝑒. Therefore 𝐴𝑣𝐵𝐽𝑇 = − 𝐶𝐿 ℎ𝑓𝑒 ℎ𝑖𝑒 + �ℎ𝑖𝑒 ℎ𝑜𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 �𝑅𝐶𝐿 The voltage divider expression around the input loop is 𝑣𝑏𝑒 = 𝑅 therefore 𝛼𝑣 = 𝐴𝑣 = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 == − 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 12 ||𝑅𝑖𝐵𝐽𝑇 +𝑅𝑆 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 ≅ −0.71 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 ℎ𝑓𝑒 𝑣𝑠 so that 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 ℎ𝑖𝑒 + �ℎ𝑖𝑒 ℎ𝑜𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 �𝑅𝐶𝐿 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 Numerically 𝐴𝑣𝐵𝐽𝑇 ≅ −110, 𝐴𝑣 ≅ −78. 29 Basi di elettronica 𝑅0 ≝ 𝑣𝑐𝑒 𝑖𝑙 𝑅𝑜 ; ! We have to force 𝑣𝑠 = ⏞ 0 , so that the KVL around the input loop yields ℎ𝑟𝑒 𝑣𝑐𝑒 = −(ℎ𝑖𝑒 + 𝑅𝑆||𝑅12𝑖𝑏, from which we can extract 𝑖𝑏 which can be replaced into the eq. II CE, obtaining ℎ𝑓𝑒 ℎ𝑟𝑒 𝑖𝑐 = − ℎ +𝑅 𝑣𝑐𝑒 + ℎ𝑜𝑒 𝑣𝑐𝑒 , therefore ||𝑅 𝑖𝑒 𝑆 12 𝑅𝑜𝐵𝐽𝑇 = Finally 1 ℎ𝑓𝑒 ℎ𝑟𝑒 ℎ𝑜𝑒 − ℎ𝑖𝑒 + 𝑅𝑆 ||𝑅12 𝑅𝑜 = 𝑅𝑜𝐵𝐽𝑇 ||𝑅𝐶𝐿 Please, pay attention to the fact that here we considered 𝑅𝐿 included in the output resistance. Numerically 𝑅𝑜𝐵𝐽𝑇 ≅ 1110Ω, 𝑅𝑜 ≅ 800Ω. Summarizing 𝑅𝐶 = 4𝑘Ω, 𝑅𝐸 = 1𝑘Ω, 𝑅1 = 83𝑘Ω, 𝑅2 ≅ 18.9𝑘Ω.. 𝑅𝑖𝐵𝐽𝑇 ≅ 1350Ω, 𝑅𝑖 ≅ 1.24𝑘Ω. 𝐴𝑖𝐵𝐽𝑇 ≅ 52, 𝐴𝑖 ≅ 47.7. 𝑅𝑜𝐵𝐽𝑇 ≅ 1110Ω, 𝑅𝑜 ≅ 800Ω. 𝐴𝑣𝐵𝐽𝑇 ≅ −110, 𝐴𝑣 ≅ −0.71. Exercise n. 16 The C.B. configuration as represented in Figure 11.70 has BJT h-parameters equal to ℎ𝑖𝑏 = 20Ω ; ℎ𝑟𝑏 = 30 ∗ 10−3 ; ℎ𝑓𝑏 = −0.98 ; ℎ𝑜𝑏 = 50 ∗ 10−6 𝑆 , and resistances 𝑅1 = 20𝑘Ω ; 𝑅2 = 20𝑘Ω; 𝑅𝐶 = 10𝑘Ω; 𝑅𝐸 = 7𝑘Ω;𝑅𝑠 = 1𝑘Ω; 𝑅𝐿 = 50𝑘Ω.. VCC RC R1 CC C CB B R2 E CE Rs RL + RE vs Figure 11.70 C.B. configuration 30 Basi di elettronica Determine within the amplifier’s bandwidth,  The BJT’s and overall current gains, 𝐴𝑖𝐵𝐽𝑇 and 𝐴𝑖  The BJT’s and overall input resistances, 𝑅𝑖𝐵𝐽𝑇 and 𝑅𝑖  The BJT’s and overall voltage gains, 𝐴𝑣𝐵𝐽𝑇 and 𝐴𝑣  The BJT’s and overall output resistances, 𝑅𝑜𝐵𝐽𝑇 and 𝑅𝑜 ANSWER Figure 11.70b represents the AC equivalent circuit of the C.B. configuration. Ri Ri BJT is E RS + vs + RE Ro BJT hib ie Ro io C + hrbvcb v eb B hfbi b 1 hob ic RC RL + vo Figure 11.70b AC equivalent circuit of the CB configuration We will utilize the BJT h-parameters for its C.B. configuration as follows 𝑣𝑒𝑏 = ℎ𝑖𝑏 𝑖𝑒 + ℎ𝑟𝑏 𝑣𝑐𝑏 (𝑒𝑞. 𝐼 𝐶𝐵) � 𝑖𝑐 = ℎ𝑓𝑏 𝑖𝑒 + ℎ𝑜𝑏 𝑣𝑐𝑏 (𝑒𝑞. 𝐼𝐼 𝐶𝐵) and the equivalent resistance 𝑅𝐶𝐿 = 𝑅𝐶 ||𝑅𝐿 . 𝑅𝑖 31 Basi di elettronica 𝑅𝑖𝐵𝐽𝑇 ≝ 𝑣𝑒𝑏 𝑖𝑒 , 𝑣𝑒𝑏 = ℎ𝑖𝑏 𝑖𝑒 + ℎ𝑟𝑏 𝑣𝑐𝑏 1 1 𝑣𝑐𝑏 = −ℎ𝑓𝑏 𝑖𝑒 � ||𝑅𝐶𝐿 � = −ℎ𝑓𝑏 𝑖𝑒 1 ℎ𝑜𝑏 + 𝑅𝐶𝐿 ℎ𝑜𝑏 We can replace the expression of 𝑣𝑐𝑏 into the one of 𝑣𝑒𝑏 𝑣𝑒𝑏 = ℎ𝑖𝑏 𝑖𝑒 − ℎ𝑟𝑏 therefore 𝑅𝑖𝐵𝐽𝑇 = But the overall input resistance is ℎ𝑓𝑏 1 ℎ𝑜𝑏 + 𝑅 𝐶𝐿 ℎ𝑟𝑏 ℎ𝑓𝑏 𝑣𝑒𝑏 = ℎ𝑖𝑏 − 1 𝑖𝑒 ℎ𝑜𝑏 + 𝑅 𝐶𝐿 𝑅𝑖 = 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 so 𝑅𝑖 = 𝑅𝐸 || �ℎ𝑖𝑏 − ℎ𝑟𝑏 ℎ𝑓𝑏 � 1 ℎ𝑜𝑏 + 𝑅 𝐶𝐿 Numerically 𝑅𝑖𝐵𝐽𝑇 ≅ 22Ω; 𝑅𝑖 ≅ 22Ω (It is evident that the emitter resistor does not influence the overall input resistance value). 𝐴𝑣 ≝ 𝑣𝑜 𝑣𝑠 𝐴𝑣 ; This equation can be rewritten as 𝐴𝑣 = 𝑣𝑜 𝑣𝑐𝑏 𝑣𝑐𝑏 𝑣𝑒𝑏 = = = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 𝑣𝑒𝑏 𝑣𝑒𝑏 𝑣𝑒𝑏 𝑣𝑠 𝐴𝑣𝐵𝐽𝑇 can be found looking for 𝑣𝑐𝑏 and 𝑣𝑒𝑏 value 1 𝑣𝑐𝑏 = −ℎ𝑓𝑏 𝑖𝑒 � ||𝑅 � ℎ𝑜𝑏 𝐶𝐿 therefore 𝑣𝑒𝑏 = ℎ𝑖𝑏 𝑖𝑒 + ℎ𝑟𝑏 𝑣𝑐𝑏 ⇒ 𝑖𝑒 = 𝑣𝑐𝑏 = − 𝑣𝑒𝑏 −ℎ𝑟𝑏 𝑣𝑐𝑏 ℎ𝑖𝑏 ℎ𝑓𝑏 1 (𝑣𝑒𝑏 − ℎ𝑟𝑏 𝑣𝑐𝑏 ) � ||𝑅 � ℎ𝑖𝑏 ℎ𝑜𝑏 𝐶𝐿 32 Basi di elettronica 𝑣𝑐𝑏 �1 − from which 𝐴𝑣𝐵𝐽𝑇 𝑣𝑐𝑏 = = 𝑣𝑒𝑏 𝐴𝑣𝐵𝐽𝑇 = − ℎ𝑓𝑏 ℎ𝑓𝑏 1 1 ℎ𝑟𝑏 � ||𝑅𝐶𝐿 �� = − 𝑣𝑒𝑏 � ||𝑅 � ℎ𝑖𝑏 ℎ𝑜𝑏 ℎ𝑖𝑏 ℎ𝑜𝑏 𝐶𝐿 ℎ𝑓𝑏 1 ℎ𝑓𝑏 𝑅𝐶𝐿 � ||𝑅𝐶𝐿 � � � − ℎ𝑖𝑏 ℎ𝑜𝑏 ℎ𝑖𝑏 1 + ℎ𝑜𝑏 𝑅𝐶𝐿 = ℎ𝑓𝑏 ℎ𝑓𝑏 1 𝑅𝐶𝐿 1− ℎ𝑟𝑏 � ||𝑅𝐶𝐿 � 1 − ℎ � � ℎ𝑖𝑏 ℎ𝑖𝑏 𝑟𝑏 1 + ℎ𝑜𝑏 𝑅𝐶𝐿 ℎ𝑜𝑏 − ℎ𝑓𝑏 𝑅𝐶𝐿 ℎ𝑖𝑏 (1 + ℎ𝑜𝑏 𝑅𝐶𝐿 ) � � ℎ𝑖𝑏 1 + ℎ𝑜𝑏 𝑅𝐶𝐿 ℎ𝑖𝑏 (1 + ℎ𝑜𝑏 𝑅𝐶𝐿 ) − ℎ𝑟𝑏 ℎ𝑓𝑏 𝑅𝐶𝐿 𝐴𝑣𝐵𝐽𝑇 = where ∆ℎ = ℎ𝑖𝑏 ℎ𝑜𝑏 − ℎ𝑟𝑏 ℎ𝑓𝑏 . ℎ𝑓𝑏 𝑅𝐶𝐿 ℎ𝑖𝑏 + �ℎ𝑖𝑏 ℎ𝑜𝑏 − ℎ𝑟𝑏 ℎ𝑓𝑏 �𝑅𝐶𝐿 = ℎ𝑓𝑏 𝑅𝐶𝐿 ℎ𝑖𝑏 + ∆ℎ𝑅𝐶𝐿 𝛼𝑣 can be found looking at the voltage divider around the input loop 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 𝑣 𝑣𝑒𝑏 = 𝑅𝑆 + 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 𝑠 therefore The overall voltage gain is then 𝛼𝑣 = 𝐴𝑣 = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 = Numerically 𝐴𝑣𝐵𝐽𝑇 ≅ −362; 𝐴𝑣 ≅ −7.9 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 𝑅𝑆 + 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 ℎ𝑓𝑏 𝑅𝐶𝐿 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 ℎ𝑖𝑏 + ∆ℎ𝑅𝐶𝐿 𝑅𝑆 + 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 If ℎ𝑟𝑏 and ℎ𝑜𝑏 could be negligible then 𝐴𝑣𝐵𝐽𝑇 ≅ ℎ𝑓𝑏 ℎ𝑖𝑏 𝑅𝐶𝐿 . 𝑅𝑜 33 Basi di elettronica 𝑅𝑜 = 𝑅𝐶𝐿 ||𝑅𝑜𝐵𝐽𝑇 ; ! As usual, the request is to kill the independent sources, so 𝑣𝑠 = ⏞ 0. But even if no independent source is in the input loop, the input current 𝑖𝑒 ≠ 0, thanks to the dependent voltage source ℎ𝑟𝑏 𝑣𝑐𝑏 . As a consequence a feed-forward effect will exits (see Section 11.3.2.2) by means of the dependent current source ℎ𝑓𝑏 𝑖𝑒 . So, in order to determine 𝑅𝑜𝐵𝐽𝑇 we have to take into account what in in 1 parallel with ℎ . 𝑜𝑏 𝑅𝑜𝐵𝐽𝑇 = 𝑣𝑐𝑏 𝑖𝑐 KVL around the input loop yields ℎ𝑟𝑏 𝑣𝑐𝑏 = −𝑖𝑒 (ℎ𝑖𝑏 + 𝑅𝑆 ||𝑅𝐸 ) and from the eq. II CB 𝑖𝑐 = ℎ𝑓𝑏 𝑖𝑒 + ℎ𝑜𝑏 𝑣𝑐𝑏 , so as a consequence ℎ𝑟𝑏 𝑖𝑐 = −ℎ𝑓𝑏 𝑣 + ℎ𝑜𝑏 𝑣𝑐𝑏 ℎ𝑖𝑏 + 𝑅𝑆 ||𝑅𝐸 𝑐𝑏 therefore 𝑅𝑜𝐵𝐽𝑇 = The overall output resistance value 𝑣𝑐𝑏 = 𝑖𝑐 𝑅𝑜 = 𝑅𝐶𝐿 || Numerically 𝑅𝑜𝐵𝐽𝑇 ≅ 1.2𝑀Ω; 𝑅𝑜 ≅ 8276Ω. 1 ℎ𝑓𝑏 ℎ𝑟𝑏 ℎ𝑜𝑏 − ℎ𝑖𝑏 + 𝑅𝑆 ||𝑅𝐸 1 ℎ𝑓𝑏 ℎ𝑟𝑏 ℎ𝑜𝑏 − ℎ𝑖𝑏 + 𝑅𝑆 ||𝑅𝐸 1 When ℎ𝑟𝑏 can be considered as negligible then 𝑅𝑜𝐵𝐽𝑇 ≅ ℎ . 𝑜𝑏 𝑖 𝐴𝑖 𝐴𝑖 ≝ 𝑖𝑜 ; 𝑠 The current gain can be rewritten as 𝐴𝑖 = 𝑖𝑜 𝑖𝑐 𝑖𝑒 𝑖𝑐 𝑖𝑒 𝑖𝑠 = 𝛼𝑖𝑜 𝐴𝑖𝐵𝐽𝑇 𝛼𝑖𝑖 To determine 𝐴𝑖𝐵𝐽𝑇 we can find the relationship between 𝑖𝑐 and 𝑖𝑒 . To this aim it is useful the current divider equation around the output loop 1 𝐺𝐶𝐿 ℎ𝑜𝑏 𝑖𝑐 = ℎ𝑓𝑏 𝑖𝑒 = ℎ𝑓𝑏 𝑖𝑒 1 𝐺𝐶𝐿 + ℎ𝑜𝑏 + 𝑅𝐶𝐿 ℎ𝑜𝑏 34 Basi di elettronica with 𝐺𝑥 the conductance value of the 𝑥 resistance, therefore ℎ𝑓𝑏 𝐴𝑖𝐵𝐽𝑇 = 1 + ℎ𝑜𝑏 𝑅𝐶𝐿 To determine 𝛼𝑖𝑖 we can consider the current divider equation around the input loop 𝐺𝑖𝐵𝐽𝑇 𝑅𝐸 𝑖𝑠 = 𝑖 𝑖𝑒 = 𝐺𝑖𝐵𝐽𝑇 + 𝐺𝐸 𝑅𝐸 + 𝑅𝑖𝐵𝐽𝑇 𝑠 therefore 𝛼𝑖𝑖 = 𝑅𝐸 𝑅𝐸 + 𝑅𝑖𝐵𝐽𝑇 To determine 𝛼𝑖𝑜 , let us consider the output loop, so 𝑅𝐶 𝑖𝑜 = − 𝑖 𝑅𝐶 + 𝑅𝐿 𝑐 therefore The overall current gain is then 𝐴𝑖 = − 𝛼𝑖𝑜 = − 𝑅𝐶 𝑅𝐶 + 𝑅𝐿 ℎ𝑓𝑏 𝑅𝐸 𝑅𝐶 𝑅𝐶 + 𝑅𝐿 1 + ℎ𝑜𝑏 𝑅𝐶𝐿 𝑅𝐸 + 𝑅𝑖𝐵𝐽𝑇 Numerically 𝐴𝑖𝐵𝐽𝑇 ≅ −0.976; 𝐴𝑖 ≅ 0.162. When we can assume ℎ𝑜𝑏 as negligible then 𝐴𝑖BJT ≅ ℎ𝑓𝑏 . Summarizing 𝑅𝑖𝐵𝐽𝑇 ≅ 22Ω; 𝑅𝑖 ≅ 22Ω 𝑅𝑜𝐵𝐽𝑇 ≅ 1.2𝑀Ω; 𝑅𝑜 ≅ 8276Ω 𝛼𝑣 ≅ 0.0219; 𝛼𝑖𝑖 ≅ 0.997; 𝛼𝑖𝑜 ≅ −0.1667; 𝐴𝑣𝐵𝐽𝑇 ≅ −362; 𝐴𝑣 ≅ −7.9 𝐴𝑖𝐵𝐽𝑇 ≅ −0.976; 𝐴𝑖 ≅ 0.162 Exercise n. 17 With respect to the previous exercise, with the same parameter values, determine the same gains and resistances but considering the absence of the base capacitor 𝐶𝐵 , as represented in Figure 11.71. 35 Basi di elettronica VCC RC R1 CC C B E R2 CE Rs RL + RE vs Figure 11.71 Configuration with the base terminal no longer grounded ANSWER The AC model of the current circuit s showed in Figure 11.71b. is ie ic hib E ++ RS C + hrbvcb hfbi e v eb RE ve B - + + ie vs + 1 v cb hob RC + RL v c=v o ve R12 - Ri io - - RiBJT RoBJT Ro Figure 11.71b AC model of the configuration with the base terminal no longer grounded For convenience, resistors can be combined as 𝑅𝐶𝐿 = 𝑅𝐶 ||𝑅𝐿 and 𝑅12 = 𝑅1 ||𝑅2. 𝑖 𝐴𝑖 ≝ 𝑖𝑜 ; 𝑠 𝐴𝑖 The overall current gain can be written as 𝑖𝑜 𝑖𝑐 𝑖𝑒 = 𝛼𝑖𝑜 𝐴𝑖𝐵𝐽𝑇 𝛼𝑖𝑖 𝑖𝑐 𝑖𝑒 𝑖𝑠 To determine 𝐴𝑖𝐵𝐽𝑇 it is useful to replace the dependent output current source with its Thevenin equivalent, so that we can use the KVL expression around the output loop. We replace then the ℎ current source ℎ𝑓𝑏 𝑖𝑒 and its parallel resistance 1�ℎ with an equivalent voltage source ℎ𝑓𝑏 𝑖𝑒 with 𝑜𝑏 𝑖𝑏 1 its series resistance �ℎ , as in Figure 11.71c, to obtain 𝐴𝑖 = 𝑜𝑏 36 Basi di elettronica ℎ𝑓𝑏 1 𝑖𝑒 = 𝑅12 (𝑖𝑒 + 𝑖𝑐 ) + � + 𝑅𝐶𝐿 � 𝑖𝑐 ℎ𝑖𝑏 ℎ𝑜𝑏 The latter is an expression of the variables 𝑖𝑒 and 𝑖𝑐 which furnishes ℎ𝑓𝑏 − ℎ𝑜𝑏 𝑅12 𝐴𝑖𝐵𝐽𝑇 = 1 + ℎ𝑜𝑏 (𝑅12 + 𝑅𝐶𝐿 ) ℎ Please, pay close attention to the fact that the previous exercise n.16 gave 𝐴𝑖𝐵𝐽𝑇 = 1+ℎ 𝑓𝑏𝑅 . Now 𝑜𝑏 𝐶𝐿 we have the added terms here highlighted in red. They are subtracted at the numerator, but ℎ𝑓𝑏 is negative itself, so the 𝐴𝑖𝐵𝐽𝑇 becomes greater in its absolute value; On the other end the denominator becomes now greater, which means a reduction of the absolute value of 𝐴𝑖𝐵𝐽𝑇 . The two aspects can be opposing. Taking into account the two attenuation values 𝛼𝑖𝑜 and 𝛼𝑖𝑖 the overall current gain will be 𝑅𝐶 𝑅𝐸 𝐴𝑖𝐵𝐽𝑇 𝐴𝑖 = − 𝑅𝐶 + 𝑅𝐿 𝑅𝐸 + 𝑅𝑖𝐵𝐽𝑇 Numerically 𝐴𝑖𝐵𝐽𝑇 ≅ −0.976; 𝛼𝑖𝑖 ≅ 0.9639; 𝛼𝑖𝑜 ≅ −0.1667; 𝐴𝑖 ≅ 0.157. is ie hib E ++ RS C + hrbvcb hfbi e hob v eb RE ve ie io + + v cb - + + Ri + B - vs - ic 1 hob RC RL v c=v o ve R12 - - RiBJT RoBJT Ro Figure 11.71c AC model with the Norton output loop replace with its Thevenin equivalent Observation The resistor 𝑅12 , by means of the current 𝑖𝑒 which flows through it, furnishes a voltage 𝑣𝑅12 at the input loop. This voltage surely outnumbers the one of the dependent voltage source ℎ𝑟𝑏 𝑣𝑐𝑏 , given the common values of ℎ𝑟𝑏 and 𝑖𝑒 . For this reason, in this and in ! similar cases, we can impose ℎ𝑟𝑏 = ⏞ 0 without making significant percentage errors. So, we will deal with the new equivalent AC network of Figure 11.71d. 37 Basi di elettronica is ie ic io hib E C ++ RS hfbi e v eb RE ve B - + ie v cb hob RC + RL v c=v o ve R12 - Ri 1 + vs + - - RiBJT RoBJT Ro Figure 11.71d AC equivalent model having hrb=0 From the eq. II CB, that is 𝑖𝑐 = ℎ𝑓𝑏 𝑖𝑒 + ℎ𝑜𝑏 𝑣𝑐𝑏 , we can write 𝑖𝑐 − ℎ𝑓𝑏 𝑖𝑒 𝑣𝑐𝑏 = ℎ𝑜𝑏 𝑅𝑖 which, taking into account the found expression of 𝐴𝑖𝐵𝐽𝑇 , becomes 𝑣𝑐𝑏 = 𝐴𝑖𝐵𝐽𝑇 𝑖𝑒 − ℎ𝑓𝑏 𝑖𝑒 �𝐴𝑖𝐵𝐽𝑇 − ℎ𝑓𝑒 �𝑖𝑒 = ℎ𝑜𝑏 ℎ𝑜𝑏 We can utilize this value of 𝑣𝑐𝑏 in the KVL around the input loop 𝑣𝑒 = ℎ𝑖𝑏 𝑖𝑒 + ℎ𝑟𝑏 𝑣𝑐𝑏 + 𝑅12 (𝑖𝑒 + 𝑖𝑐 ) obtaining 𝑣𝑒 = ℎ𝑖𝑏 𝑖𝑒 + ℎ𝑟𝑏 �𝐴𝑖𝐵𝐽𝑇 − ℎ𝑓𝑏 �𝑖𝑒 + 𝑅12 (𝑖𝑏 + 𝑖𝑐 ) ℎ𝑜𝑏 This (rearranged) expression can be used to find the input resistance 𝑅𝑖𝐵𝐽𝑇 𝑣𝑒 ℎ𝑟𝑏 𝑅𝑖𝐵𝐽𝑇 ≝ = ℎ𝑖𝑏 + �1 + 𝐴𝑖𝐵𝐽𝑇 �𝑅12 + �𝐴 − ℎ𝑓𝑏 � 𝑖𝑒 ℎ𝑜𝑏 𝑖𝐵𝐽𝑇 The previous exercise n.16 gave 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑏 − than the previous one. Finally, the overall input resistance is Numerically 𝑅𝑖𝐵𝐽𝑇 ≅ 262Ω; 𝑅𝑖 ≅ 252Ω. ℎ𝑟𝑏 ℎ𝑓𝑏 ℎ𝑜𝑏 + 1 𝑅𝐶𝐿 , so the current input resistance is greater 𝑅𝑖 = 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 38 Basi di elettronica 𝑅𝑜𝐵𝐽𝑇 ≝ 𝑣𝑐 𝑖𝑒 𝑅𝑜 . ! As already mentioned, we assume for simplicity ℎ𝑟𝑏 = ⏞ 0. We have 𝑣𝑒 = −(𝑅𝑆 ||𝑅𝐸 )𝑖𝑒 but 𝑣𝑒 = ℎ𝑖𝑏 𝑖𝑒 + 𝑅12 (𝑖𝑒 + 𝑖𝑐 ) so that −(𝑅𝑆 ||𝑅𝐸 )𝑖𝑒 = ℎ𝑖𝑏 𝑖𝑒 + 𝑅12 (𝑖𝑒 + 𝑖𝑐 ) from which [(𝑅𝑆 ||𝑅𝐸 ) + ℎ𝑖𝑏 + 𝑅12 ]𝑖𝑒 = −𝑅12 𝑖𝑐 𝑖𝑒 = − KVL around the output loop yields 𝑣𝑐 = − 𝑅12 𝑖 (𝑅𝑆 ||𝑅𝐸 ) + ℎ𝑖𝑏 + 𝑅12 𝑐 ℎ𝑓𝑏 1 𝑖𝑒 + 𝑖 + 𝑅12 (𝑖𝑒 + 𝑖𝑐 ) ℎ𝑜𝑏 ℎ𝑜𝑏 𝑐 in which we can replace the previous expression of 𝑖𝑒 , obtaining ℎ𝑓𝑏 𝑅12 1 𝑅12 𝑣𝑐 = 𝑖𝑐 + 𝑖𝑐 + 𝑅12 �− 𝑖 + 𝑖𝑐 � (𝑅𝑆 ||𝑅𝐸 ) + ℎ𝑖𝑏 + 𝑅12 𝑐 ℎ𝑜𝑏 (𝑅𝑆 ||𝑅𝐸 ) + ℎ𝑖𝑏 + 𝑅12 ℎ𝑜𝑏 Rearranging 𝑅𝑜𝐵𝐽𝑇 ≝ ℎ𝑓𝑏 𝑣𝑐 1 𝑅12 = + 𝑅12 + � − 𝑅12 � (𝑅𝑆 ||𝑅𝐸 ) + ℎ𝑖𝑏 + 𝑅12 𝑖𝑐 ℎ𝑜𝑏 ℎ𝑜𝑏 ! 1 In the previous exercise n.16, admitting ℎ𝑟𝑒 = ⏞ 0, we obtained 𝑅𝑜𝐵𝐽𝑇 = ℎ . The current output ℎ𝑓𝑏 𝑜𝑏 resistance can be then higher or lower, mainly depending on the term �ℎ − 𝑅12 �. 𝑜𝑏 Finally, the overall output resistance is 𝑅𝑜 = 𝑅𝐶𝐿 ||𝑅𝑜𝐵𝐽𝑇 Numerically 𝑅𝑜𝐵𝐽𝑇 ≅ 202𝑘Ω; 𝑅𝑜 ≅ 8𝑘Ω. 𝐴𝑣 𝑣 𝐴𝑣𝐵𝐽𝑇 ≝ 𝑣𝑐 ; 𝑒 The voltage gain can be easily found with few steps 𝑅𝐶𝐿 𝐴𝑖𝐵𝐽𝑇 𝑖𝑒 𝑣𝑐 −𝑅𝐶𝐿 𝑖𝑐 𝑅𝐶𝐿 𝐴𝑣𝐵𝐽𝑇 ≝ = =− =− 𝐴 𝑣𝑒 𝑣𝑒 𝑣𝑒 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇 39 Basi di elettronica In the previous exercise n.16 the result was 𝐴𝑣𝐵𝐽𝑇 = ℎ ℎ𝑓𝑏 𝑅𝐶𝐿 𝑖𝑏 +�ℎ𝑖𝑏 ℎ𝑜𝑏 −ℎ𝑟𝑏 ℎ𝑓𝑏 �𝑅𝐶𝐿 =ℎ ℎ𝑓𝑏 𝑅𝐶𝐿 𝑖𝑏 +∆ℎ𝑅𝐶𝐿 . In order to make an easy comparison, let us ignore ℎ𝑜𝑏 , admit 𝐴𝑖𝐵𝐽𝑇 ≅ ℎ𝑓𝑏 , and over-stimate 𝑅𝑖𝐵𝐽𝑇 ≅ ℎ𝑖𝑏 . Now, we have a lower voltage gain being the denominator higher of the term ∆ℎ𝑅𝐶𝐿 . Finally, the overall voltage gain 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 𝑣𝑜 𝑣𝑐 𝑣𝑐 𝑣𝑒 𝐴𝑣 = = = = 𝐴𝑣𝐵𝐽𝑇 𝛼𝑣 = 𝐴𝑣𝐵𝐽𝑇 𝑣𝑠 𝑣𝑠 𝑣𝑒 𝑣𝑠 𝑅𝐸 ||𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 Numerically 𝛼𝑣 ≅ 0.2015; 𝐴𝑣𝐵𝐽𝑇 ≅ 31; 𝐴𝑣 ≅ 6.3. Summarizing 𝑅𝑖𝐵𝐽𝑇 ≅ 262Ω; 𝑅𝑖 ≅ 252Ω 𝑅𝑜𝐵𝐽𝑇 ≅ 202𝑘Ω; 𝑅𝑜 ≅ 8𝑘Ω 𝛼𝑣 ≅ 0.2015; 𝛼𝑖𝑖 ≅ 0.9639; 𝛼𝑖𝑜 ≅ −0.1667; 𝐴𝑣𝐵𝐽𝑇 ≅ 31; 𝐴𝑣 ≅ 6.3 𝐴𝑖𝐵𝐽𝑇 ≅ −0.976; 𝐴𝑖 ≅ 0.157 Exercise n. 18 The C.C. configuration as represented in Figure 11.72 has BJT h-parameters equal to (BC107 type) ℎ𝑖𝑒 = 4𝑘Ω ; ℎ𝑟𝑒 = 2.2 ∗ 10−4 ; ℎ𝑓𝑒 = 250 ; ℎ𝑜𝑒 = 30 ∗ 10−6 𝑆 , and resistance equal to 𝑅1 = 140𝑘Ω; 𝑅2 = 140𝑘Ω; 𝑅𝐸 = 50𝑘Ω; 𝑅𝐿 = 50𝑘Ω; 𝑅𝑠 = 500Ω.. VCC R1 C CB1 B Rs R2 RE E CB2 RL + vs Figure 11.72 C.C. configuration Determine within the amplifier’s bandwidth,  The BJT’s and overall current gains, 𝐴𝑖𝐵𝐽𝑇 and 𝐴𝑖  The BJT’s and overall input resistances, 𝑅𝑖𝐵𝐽𝑇 and 𝑅𝑖  The BJT’s and overall voltage gains, 𝐴𝑣𝐵𝐽𝑇 and 𝐴𝑣  The BJT’s and overall output resistances, 𝑅𝑜𝐵𝐽𝑇 and 𝑅𝑜 40 Basi di elettronica ANSWER The exercise requires the dynamic parameters, so we draw the AC equivalent circuit in Figure 11.72b. We use BJT equivalent model with h-parameters in CE configuration, which are the one furnished by the exercise, even if we have to study a CC configuration. ic is ib hie B C ++ RS hfei b 1 v ce hoe - v be R12 vb E ie + RE i RE vs + RL iL - Ri + v e=v o - Ro RiBJT Figure 11.72b AC equivalent model with BJT h-parameters of the CE configuration For convenience we can combine 𝑅12 = 𝑅1 ||𝑅2 and 𝑅𝐸𝐿 = 𝑅𝐸 ||𝑅𝐿 . 𝑖 𝐴𝑖𝐵𝐽𝑇 ≝ 𝑖𝑒 ; 𝑏 𝐴𝑖 The 𝑒𝑞. 𝐼𝐼 𝐶𝐸 is 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 , but 𝑖𝑒 = 𝑖𝑏 + 𝑖𝑐 and 𝑣𝑐𝑒 = −𝑅𝐸𝐿 𝑖𝑒 , so that we can write 𝑖𝑒 − 𝑖𝑏 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 from which we can write 𝑖 𝐴𝑖𝐵𝐽𝑇 = 1 + ℎ𝑓𝑒 1 + ℎ𝑜𝑒 𝑅𝐸𝐿 The overall current gain is 𝐴𝑖 ≝ 𝑖𝐿, but it can be equivalently written as 𝐴𝑖 = 𝑠 𝑖𝐿 𝑖𝑒 𝑖𝑏 𝑖𝐿 𝑖𝑏 𝑅𝐸 𝑅12 = 𝐴𝑖𝐵𝐽𝑇 = 𝐴𝑖𝐵𝐽𝑇 𝑖𝑒 𝑖𝑏 𝑖𝑠 𝑖𝑒 𝑖𝑠 𝑅𝐸 + 𝑅𝐿 𝑅12 + 𝑅𝑖𝐵𝐽𝑇 Numerically 𝐴𝑖𝐵𝐽𝑇 = 143.4; 𝐴𝑖 ≅ 1.37 𝑅𝑖𝐵𝐽𝑇 ≝ 𝑣𝑏 𝑖𝑏 𝑅𝑖 ; From the definition 𝑣𝑏𝑒 ≝ 𝑣𝑏 − 𝑣𝑒 and from the 𝑒𝑞. 𝐼 𝐶𝐸 we can write 𝑣𝑏𝑒 = ℎ𝑖𝑒 𝑖𝑏 + ℎ𝑟𝑒 𝑣𝑐𝑒 41 Basi di elettronica ! (here we can consider ℎ𝑟𝑒 = ⏞ 0, according to the observation already made, see exercise n.17), so that 𝑣𝑏 − 𝑅𝐸𝐿 𝑖𝑒 = ℎ𝑖𝑒 𝑖𝑏 𝑣𝑏 − 𝑅𝐸𝐿 𝐴𝑖𝐵𝐽𝑇 𝑖𝑏 = ℎ𝑖𝑒 𝑖𝑏 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 + 𝑅𝐸𝐿 𝐴𝑖𝐵𝐽𝑇 𝑅𝑖 = 𝑅𝑆 + 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 Numerically 𝑅𝑖𝐵𝐽𝑇 ≅ 3.59𝑀Ω, 𝑅𝑖 ≅ 643Ω. 𝐴𝑣 𝑣𝑒 𝐴𝑣𝐵𝐽𝑇 ≝ 𝑣 ; Since 𝑏 it can be easily found that 𝑣𝑒 = 𝑅𝐸𝐿 𝑖𝑒 = 𝑅𝐸𝐿 𝐴𝑖𝐵𝐽𝑇 𝑖𝑏 ; 𝑣𝑏 = 𝑅𝑖𝐵𝐽𝑇 𝑖𝑏 𝐴𝑣𝐵𝐽𝑇 = But the overall voltage gain is 𝑅𝐸𝐿 𝐴 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇 𝐴𝑣 ≝ which can be written as 𝐴𝑣 = 𝑣𝑒 𝑣𝑠 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 𝑣𝑒 𝑣𝑏 𝑣𝑏 = 𝐴𝑣𝐵𝐽𝑇 = 𝐴𝑣𝐵𝐽𝑇 𝑣𝑏 𝑣𝑠 𝑣𝑠 𝑅𝑠 + 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 Numerically 𝐴𝑣𝐵𝐽𝑇 ≅ 0.998, 𝐴𝑣 ≅ 0.22. 𝑅𝑜𝐵𝐽𝑇 ≝ 𝑣𝑒 𝑖𝑒 𝑅𝑜 ; ! We have to impose 𝑣𝑠 = ⏞ 0 so that but therefore 𝑣𝑒 = [(𝑅𝑆 ||𝑅12 ) + ℎ𝑖𝑒 ]𝑖𝑏 𝑖𝑒 = 𝐴𝑖𝐵𝐽𝑇 𝑖𝑏 𝑅𝑜𝐵𝐽𝑇 = (𝑅𝑆 ||𝑅12 ) + ℎ𝑖𝑒 𝐴𝑖𝐵𝐽𝑇 42 Basi di elettronica The overall output resistance is then 𝑅𝑜 = 𝑅𝑜𝐵𝐽𝑇 ||𝑅𝐸𝐿 Numerically 𝑅𝑜𝐵𝐽𝑇 ≅ 31.35Ω, 𝑅𝑜 ≅ 31.3Ω. Summarizing 𝐴𝑖𝐵𝐽𝑇 = 143.4, 𝐴𝑖 ≅ 1.37. 𝑅𝑖𝐵𝐽𝑇 ≅ 3.59𝑀Ω, 𝑅𝑖 ≅ 643Ω.. 𝐴𝑣𝐵𝐽𝑇 ≅ 0.998, 𝐴𝑣 ≅ 0.22. 𝑅𝑜𝐵𝐽𝑇 ≅ 31.35Ω, 𝑅𝑜 ≅ 31.3Ω. Exercise n. 19 Let us consider a dual-output BJT amplifier that has two resistor loads, 𝑅𝐿1 and 𝑅𝐿2 , respectively, connected across the collector-ground and emitter-ground terminals (Figure 11.73). VCC RC R1 i l1 ig CB Rs + + vi CE R2 RE CL i l2 + RL1 + vl1 RL2 vl2 vs Figure 11.73 Dual-output BJT amplifier Here none of the BJT’s terminals are AC directly connected to the ground. Determine the expressions of  The BJT and overall input resistances, 𝑅𝑖𝐵𝐽𝑇 and 𝑅𝑖  The BJT and overall voltage gains at the collector and emitter terminals, respectively 𝐴𝑣𝐵𝐽𝑇,𝑐 , 𝐴𝑣𝐵𝐽𝑇,𝑒 , 𝐴𝑣,𝑐 , 𝐴𝑣,𝑒  The BJT current gains at the collector and emitter terminals, 𝐴𝑖𝐵𝐽𝑇,𝑐 , and 𝐴𝑖𝐵𝐽𝑇,𝑒  The BJT and overall output resistances at the collector and emitter terminals, 𝑅𝑜𝐵𝐽𝑇,𝑐 , 𝑅𝑜𝐵𝐽𝑇,𝑒 , 𝑅𝑜,𝑐 , 𝑅𝑜,𝑒 For simplicity, consider ℎ𝑟𝑏 = 0. 43 Basi di elettronica ANSWER For convenience we can combine 𝑅𝐿1𝐶 = 𝑅𝐿1 ||𝑅𝐶 ; 𝑅𝐿2𝐸 = 𝑅𝐿2 ||𝑅𝐸 ; 𝑅12 = 𝑅1 ||𝑅2 ; 𝑅12𝑆 = 𝑅𝑆 ||𝑅12. Figure 11.73b represents the AC equivalent model of the dual-output amplifier where, for ! ⏞ 0, according to the observation already made in the exercise simplicity reasons, we imposed ℎ𝑟𝑏 = n.17. ib is ic iL hie B ++ RS C + hrevce hfei b v be R12 vb + E ie RE vs + 1 v ce hoe RC RL1 + v L1 RL2 v L2 - Ri + - - RiBJT RoBJT Ro Figure 11.73b AC equivalent model of the dual-output amplifier KCL of the output loop gives 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 = ℎ𝑓𝑒 𝑖𝑏 − ℎ𝑜𝑒 [𝑅𝐿1𝐶 𝑖𝑐 + 𝑅𝐿2𝐸 (𝑖𝑐 + 𝑖𝑏 )] from which we can determine the partial voltage gain at the collector terminal ℎ𝑓𝑒 − ℎ𝑜𝑒 𝑅𝐿2𝐸 𝑖𝑐 𝐴𝑖𝐵𝐽𝑇,𝑐 ≝ = 𝑖𝑏 1 + ℎ𝑜𝑒 (𝑅𝐿1𝐶 + 𝑅𝐿2𝐸 ) If we admit the approximations ℎ𝑜𝑒 (𝑅𝐿1𝐶 + 𝑅𝐿2𝐸 ) ≪ 1 and ℎ𝑓𝑒 ≫ ℎ𝑜𝑒 𝑅𝐿2𝐸 , it can be reduced simply to 𝐴𝑖𝐵𝐽𝑇,𝑐 ≅ ℎ𝑓𝑒 . 𝑅𝑖 44 Basi di elettronica KVL around the input closed path yields 𝑣𝑏 = ℎ𝑖𝑒 𝑖𝑏 + 𝑅𝐿2𝐸 (𝑖𝑐 + 𝑖𝑏 ) = ℎ𝑖𝑒 𝑖𝑏 + 𝑅𝐿2𝐸 �1 + 𝐴𝑖𝐵𝐽𝑇,𝑐 �𝑖𝑏 so that 𝑅𝑖𝐵𝐽𝑇 ≝ 𝑣𝑏 = ℎ𝑖𝑒 + �1 + 𝐴𝑖𝐵𝐽𝑇,𝑐 �𝑅𝐿2𝐸 𝑖𝑏 If we admit the aforementioned approximations, it becomes 𝑅𝑖𝐵𝐽𝑇 ≅ ℎ𝑖𝑒 + ℎ𝑓𝑒 𝑅𝐿2𝐸 The overall input resistance being 𝑅𝑖 = 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 The voltage gain at the collector terminal is 𝑣𝐿1 𝑅𝐿1𝐶 𝑖𝑐 𝑅𝐿1𝐶 𝐴𝑣𝐵𝐽𝑇,𝑐 ≝ =− =− 𝐴 𝑣𝑏 𝑅𝑖𝐵𝐽𝑇 𝑖𝑏 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇,𝑐 𝐴𝑣 The voltage gain at the emitter terminal is 𝑣𝐿2 𝑅𝐿2𝐸 (𝑖𝑐 + 𝑖𝑏 ) 𝑅𝐿2𝐸 𝑅𝐿2𝐸 𝐴𝑣𝐵𝐽𝑇,𝑒 ≝ =− = �1 + 𝐴𝑖𝐵𝐽𝑇,𝑐 � ≅ 𝐴 𝑣𝑏 𝑅𝑖𝐵𝐽𝑇 𝑖𝑏 𝑅𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇,𝑐 which can be equivalently written as 𝐴𝑣𝐵𝐽𝑇,𝑒 = 𝑣𝑏 − ℎ𝑖𝑒 𝑖𝑏 ℎ𝑖𝑒 = 1− 𝑅𝑖𝐵𝐽𝑇 𝑖𝑏 𝑅𝑖𝐵𝐽𝑇 It is evident that this voltage gain is less the none, so we have a attenuation rather than a real gain. To determine the overall voltage gains we have to take into account the voltage divider effects due to the source resistor 𝑅𝑆 and to the BJT one 𝑅𝑖𝐵𝐽𝑇 : 𝑅𝑖𝐵𝐽𝑇 𝑣𝐿1 𝑣𝐿1 𝑣𝑏 𝑅𝐿1𝐶 𝐴𝑣,𝑐 ≝ = =− 𝐴𝑖𝐵𝐽𝑇,𝑐 𝑣𝑠 𝑣𝑏 𝑣𝑠 𝑅𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 𝐴𝑣,𝑒 ≝ 𝑅𝑖𝐵𝐽𝑇 𝑣𝐿2 𝑣𝐿2 𝑣𝑏 𝑅𝐿2𝐸 = = 𝐴𝑖𝐵𝐽𝑇,𝑐 𝑣𝑠 𝑣𝑏 𝑣𝑠 𝑅𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 + 𝑅𝑆 There is a 180° out of phase between the output voltage at the emitter and at the collector, so ! that if we admit 𝑅𝐿1𝐶 = ⏞ 𝑅𝐿2𝐸 we can realize a phase inverter: twin voltage output in module but 180° outphased. 45 Basi di elettronica Ai The current gain at the collector terminal is 𝑖𝐿1 𝑣𝐿1 𝑅𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 = = 𝐴 𝐴𝑖𝐵𝐽𝑇,𝑐 ≝ 𝑖𝑏 𝑅𝐿1 𝑣𝑏 𝑅𝐿1 𝑣𝐵𝐽𝑇,𝑐 The current gain at the emitter terminal is 𝑖𝐿2 𝑣𝐿2 𝑅𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 𝐴𝑖𝐵𝐽𝑇,𝑒 ≝ = = 𝐴 𝑖𝑏 𝑅𝐿2 𝑣𝑏 𝑅𝐿2 𝑣𝐵𝐽𝑇,𝑒 Ro ! To determine the output resistance, we have to kill the independent source, as usual, 𝑣𝑠 = ⏞ 0. Let us start calculating the output resistance at the collector terminal, 𝑅𝑜,𝑐 . Since 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝐿1 − [𝑅𝐿2𝐸 ||(ℎ𝑖𝑒 + 𝑅12𝑆 )]𝑖𝑐 we can make the voltage 𝑣𝐿1 explicit ℎ𝑓𝑒 𝑖𝑐 𝑣𝐿1 = − 𝑖 + [𝑅𝐿2𝐸 ||(ℎ𝑖𝑒 + 𝑅12𝑆 )]𝑖𝑐 ℎ𝑜𝑒 ℎ𝑜𝑒 𝑏 and since 𝑖𝑏 = −𝑖𝑐 𝑅𝐿2𝐸 𝑅𝐿2𝐸 + (ℎ𝑖𝑒 + 𝑅12𝑆 ) As a final result ℎ𝑓𝑒 𝑅𝐿2𝐸 𝑣𝐿1 1 1 𝑅𝑜𝐵𝐽𝑇,𝑐 ≝ = + + [𝑅𝐿2𝐸 ||(ℎ𝑖𝑒 + 𝑅12𝑆 )] ≅ �1 + ℎ𝑓𝑒 � 𝑖𝑐 ℎ𝑜𝑒 ℎ𝑜𝑒 ℎ𝑜𝑒 𝑅𝐿2𝐸 + (ℎ𝑖𝑒 + 𝑅12𝑆 ) The overall output resistance at the collector terminal is 𝑅𝑜,𝑐 = 𝑅𝑜𝐵𝐽𝑇,𝑐 ||𝑅𝐿1𝐶 Let us calculate now the output resistance at the emitter terminal, 𝑅𝑜,𝑒 . Since 𝑣𝐿2 𝑖𝑏 = − 𝑅12𝑆 + ℎ𝑖𝑒 and 𝑖𝑐 = ℎ𝑓𝑒 𝑖𝑏 + ℎ𝑜𝑒 𝑣𝑐𝑒 = − 𝑖𝑐 = − −(𝑖𝑏 + 𝑖𝑐 ) = � ℎ𝑓𝑒 𝑣𝐿2 − ℎ𝑜𝑒 (𝑣𝐿2 + 𝑅𝐿1𝐶 𝑖𝑐 ) 𝑅12𝑆 + ℎ𝑖𝑒 ℎ𝑓𝑒 𝑣𝐿2 ℎ𝑜𝑒 𝑣𝐿2 − (𝑅12𝑆 + ℎ𝑖𝑒 )(1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) (1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) ℎ𝑓𝑒 1 ℎ𝑜𝑒 + + �𝑣 (𝑅12𝑆 + ℎ𝑖𝑒 ) (𝑅12𝑆 + ℎ𝑖𝑒 )(1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) (1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) 𝐿2 The requested resistance is made by three parallel resistances 46 Basi di elettronica 𝑅𝑜𝐵𝐽𝑇,𝑒 = (𝑅12𝑆 + ℎ𝑖𝑒 ) || (𝑅12𝑆 + ℎ𝑖𝑒 )(1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) (1 + ℎ𝑜𝑒 𝑅𝐿1𝐶 ) || ℎ𝑓𝑒 ℎ𝑜𝑒 If we admit the hypothesis ℎ𝑜𝑒 𝑅𝐿1𝐶 ≪ 1, it becomes 𝑅12𝑆 + ℎ𝑖𝑒 𝑅𝑜𝐵𝐽𝑇,𝑒 ≅ ℎ𝑓𝑒 Finally 𝑅𝑜,𝑒 = 𝑅𝐿2𝐸 ||𝑅𝑜𝐵𝐽𝑇,𝑒 Summarizing With the expressions 𝑅𝐿1𝐶 = 𝑅𝐿1 ||𝑅𝐶 , 𝑅𝐿2𝐸 = 𝑅𝐿2 ||𝑅𝐸 , 𝑅12 = 𝑅1 ||𝑅2 , 𝑅12𝑠 = 𝑅𝑠 ||𝑅12 , the solutions can be written as: ℎ −ℎ𝑜𝑒 𝑅𝐿2𝐸 𝐴𝑖𝐵𝐽𝑇,𝑐 = 1+ℎ𝑓𝑒(𝑅 𝑜𝑒 , 𝐿1𝐶 +𝑅𝐿2𝐸 ) 𝑅𝑖𝐵𝐽𝑇 = ℎ𝑖𝑒 + �1 + 𝐴𝑖𝐵𝐽𝑇,𝑐 �𝑅𝐿2𝐸 , 𝑅𝑖 = 𝑅12 ||𝑅𝑖𝐵𝐽𝑇 𝑅 𝐴𝑣𝐵𝐽𝑇,𝑐 = − 𝑅 𝐿1𝐶 𝐴𝑖𝐵𝐽𝑇,𝑐 𝐴𝑣𝐵𝐽𝑇,𝑒 = 𝑖𝐵𝐽𝑇 𝑅𝐿2𝐸 𝑅𝑖𝑞 ℎ𝑖𝑒 𝐴𝑣𝐵𝐽𝑇,𝑒 = 1 − 𝑅 𝑅𝐿1𝐶 𝐴𝑣,𝑐 = − 𝑅 𝑅 𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇 𝐴𝑖𝐵𝐽𝑇,𝑐 𝑅 𝐴𝑣,𝑒 = 𝑅 𝐿2𝐸 𝐴𝑖𝐵𝐽𝑇,𝑐 𝑅 𝑖𝐵𝐽𝑇 𝑅 �1 + 𝐴𝑖𝐵𝐽𝑇,𝑐 � ≅ 𝑅 𝐿2𝐸 𝐴𝑖𝐵𝐽𝑇,𝑐 Observation 𝑖𝐵𝐽𝑇 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇 +𝑅𝑠 𝑅𝑖𝐵𝐽𝑇 𝑖𝐵𝐽𝑇 +𝑅𝑠 . The output collector and emitter voltages are out of phase by 180°. If we select components such that 𝑅𝐿1𝐶 = 𝑅𝐿2𝐸 , the two voltages have the same module. The amplifier realizes a phase inverter, also known as phase splitter. This network can be useful to provide anti-phase input to another stage. Please, pay close attention to the fact that the emitter resistor is not shunted by a capacitor. 𝐴𝑖𝐵𝐽𝑇,𝑐 = 𝐴𝑖𝐵𝐽𝑇,𝑒 = 𝑅𝑖𝐵𝐽𝑇 𝑅𝐿1 𝑅𝑖𝐵𝐽𝑇 𝑅𝐿2 1 𝐴𝑣𝐵𝐽𝑇,𝑐 𝐴𝑣𝐵𝐽𝑇,𝑒 ℎ 𝑅𝑜𝐵𝐽𝑇,𝑐 = ℎ + ℎ𝑓𝑒 𝑅 𝑜𝑒 𝑜𝑒 𝑅𝑜,𝑐 = 𝑅𝑜𝐵𝐽𝑇,𝑐 ||𝑅𝐿1𝐶 𝑅𝐿2𝐸 𝐿2𝐸 +(ℎ𝑖𝑒 +𝑅12𝑆 ) 𝑅𝑜𝐵𝐽𝑇,𝑒 = (𝑅12𝑆 + ℎ𝑖𝑒 ) || 𝑅𝑜,𝑒 = 𝑅𝐿2𝐸 ||𝑅𝑜𝐵𝐽𝑇,𝑒 1 + [𝑅𝐿2𝐸 ||(ℎ𝑖𝑒 + 𝑅12𝑆 )] ≅ ℎ �1 + ℎ𝑓𝑒 � 𝑜𝑒 (𝑅12s +ℎ𝑖𝑒 )(1+ℎ𝑜𝑒 𝑅𝐿1𝐶 ) ℎ𝑓𝑒 || (1+ℎ𝑜𝑒 𝑅𝐿1𝐶 ) ℎ𝑜𝑒 Exercise n. 20 Given the amplifier shown in Figure 11.74, determine its ideal current gain 𝐴𝑖 to realize a negative input resistance, 𝑅𝑖 < 0. 47 Basi di elettronica VCC RC R1 CC C B R2 E CE RL Rs is RE Figure 11.74 Current amplifier with current source ANSWER The AC equivalent model of the current amplifier with current source is drawn in Figure 11.74b. ie ic io hib E ++ is C + hrbvcb hfbi e v eb RS RE ve B - + ie - Ri + 1 v cb hob RC RL v c=v o ve R12 - - RiBJT + RoBJT Ro Figure 11.74b AC equivalent model of the current amplifier with current source If we omit the dependent voltage source at the input loop (the reason is already discussed in the exercise n.17), this model results similar to the one of a generic current amplifier (see Section 10.3.2 in Chapter 10), but having an additional feed-back resistor 𝑅𝐹 , in common between the input and the output loops, as in Figure 11.74c. 48 Basi di elettronica io i in + is (i) A i iin RS + Ro Ri vin i out vout RF - RL - Rin Figure 11.74c Equivalent model of a generic current amplifier with a feedback resistor This equivalent model can be modified replacing the Norton equivalent with its Thevenin counterpart, as in Figure 11.74d. i in io i out R + + is + A(i)R i o i o in RS Ri vin vout RF - RL - Rin Figure 11.74d Equivalent model of a generic current amplifier but with the Thevenin equivalent dependent source in the output loop KVL around the input loop yields 𝑣𝑖𝑛 = 𝑅𝑖 𝑖𝑖𝑛 + 𝑅𝐹 (𝑖𝑖𝑛 − 𝑖𝑜𝑢𝑡 ) and KVL around the output loop determines (𝑖) 𝐴𝑖 𝑅𝑜 𝑖𝑖𝑛 = (𝑅𝑜 + 𝑅𝐿 )𝑖𝑜𝑢𝑡 − 𝑅𝐹 (𝑖𝑖𝑛 − 𝑖𝑜𝑢𝑡 ) 𝑖𝑜𝑢𝑡 The two results can be mixed obtaining from which 𝑣𝑖𝑛 (𝑖) 𝐴 𝑅𝑜 + 𝑅𝐹 = 𝑖 𝑖 𝑅𝑜 + 𝑅𝐿 + 𝑅𝐹 𝑖𝑛 (𝑖) 𝐴 𝑅𝑜 + 𝑅𝐹 = 𝑅𝑖 𝑖𝑖𝑛 + 𝑅𝐹 �1 − 𝑖 �𝑖 𝑅𝑜 + 𝑅𝐿 + 𝑅𝐹 𝑖𝑛 𝑅𝑖𝑛 = 𝑅𝑖 + 𝑅𝐹 �1 − (𝑖) 𝐴𝑖 𝑅𝑜 + 𝑅𝐹 � 𝑅𝑜 + 𝑅𝐿 + 𝑅𝐹 For an ideal current amplifier we have 𝑅𝑖 = 0 (and 𝑅𝑜 = ∞), so that the latter equation becomes 49 Basi di elettronica (𝑖) 𝑅𝑖𝑛 = 𝑅𝐹 �1 − 𝐴𝑖 � To answer the question of the exercise: to have a negative value of the input resistance 𝑅𝑖𝑛 we (𝑖) have to simply admit 𝐴𝑖 > 1. 50 Exercise n. 1 Demonstrate that for a common source stage with source degeneration, i.e., that includes an unbypassed source resistor 𝑅𝑆 , the MOSFET’s voltage gain is 𝐴𝑣𝑀𝑂𝑆 = − 𝑔𝑚 𝑟𝑜 (𝑅𝐷 ||𝑅𝐿 ) . 𝑟𝑜 +(1+𝑔𝑚 𝑟𝑜 )𝑅𝑆 +(𝑅𝐷 ||𝑅𝐿 ) ANSWER Figure 12.22b shows the AC equivalent model for the current exercise. id is + Rs vs + vi R1 R2 G - D gmvgs + v gs ro + RD RL + vo v ds - S + - RS - v RS - - Figure 12.22b AC equivalent model of the C.S. configuration with un-bypassed source resistor For convenience we can combine 𝑅12 = 𝑅1 ||𝑅2 and 𝑅𝐷𝐿 = 𝑅𝐷 ||𝑅𝐿 . 𝑣𝑜 = −𝑅𝐷𝐿 𝑖𝑑 KVL applied around the output loop yields 𝑣𝑑𝑠 + 𝑣𝑅𝑆 = 𝑣𝑜 Ohm’s law applied to the previous equation yields �𝑖𝑑 − 𝑔𝑚 𝑣𝑔𝑠 �𝑟𝑜 + 𝑅𝑆 𝑖𝑑 = −𝑅𝐷𝐿 𝑖𝑑 but 𝑣𝑔𝑠 = 𝑣𝑖 − 𝑅𝑆 𝑖𝑑 so that [𝑖𝑑 − 𝑔𝑚 (𝑣𝑖 − 𝑅𝑆 𝑖𝑑 )]𝑟𝑜 + 𝑅𝑆 𝑖𝑑 = −𝑅𝐷𝐿 𝑖𝑑 therefore 𝑔𝑚 𝑣𝑖 𝑟𝑜 𝑖𝑑 = 𝑟𝑜 + 𝑔𝑚 𝑟𝑜 𝑅𝑆 + 𝑅𝑆 + 𝑅𝐷𝐿 𝑣𝑜 But, since 𝑖𝑑 = − 𝑅 , from the latter equation we can extract the following 𝐷𝐿 𝐴𝑣𝑀𝑂𝑆 ≝ 𝑣𝑜 𝑔𝑚 𝑟𝑜 𝑅𝐷𝐿 =− 𝑣𝑖 𝑟𝑜 + (1 + 𝑔𝑚 𝑟𝑜 )𝑅𝑆 + 𝑅𝐷𝐿 Exercise n. 2 For the amplifier shown in Figure 12.23 the resistance values for the resistors are 𝑅𝐷 = 4.7𝑘Ω, 𝑅𝐹 = 220𝑘Ω, while the MOSFET’s parameters are 𝑔𝑚 = 2𝑚𝑆, 𝑟𝑜 = 133𝑘Ω.. Determinate the input and output resistances, 𝑅𝑖 and 𝑅𝑜 ,and the voltage gain 𝐴𝑣 = 𝑣𝑜 𝑣𝑖 . VDD RD RF CB1 CB2 + vo + vi Ri Ro Figure 12.23 C.S. configuration with feedback resistor ANSWER Figure 12.23b shows the AC equivalent model of the C.S. configuration with feedback resistor. RF ii G + v gs id D gmvgs ro + R D v ds + vo S Ro Ri Figure 12.23b AC equivalent model of the C.S. configuration with feedback resistor The input resistance is defined as 𝑅𝑖 = 𝑣𝑜 𝑖𝑖 , so we look for a link between the output voltage 𝑣𝑜 and the input current 𝑖𝑖 . We can start considering that 𝑣𝑔𝑠 − 𝑣𝑜 = 𝑅𝐹 𝑖𝑖 but 𝑣𝑜 = �𝑖𝑖 − 𝑔𝑚 𝑣𝑔𝑠 �(𝑟𝑜 ||𝑅𝐷 ) The mix of the two equations yields 𝑣𝑔𝑠 − �𝑖𝑖 − 𝑔𝑚 𝑣𝑔𝑠 �(𝑟𝑜 ||𝑅𝐷 ) = 𝑅𝐹 𝑖𝑖 𝑣𝑔𝑠 [1 + 𝑔𝑚 (𝑟𝑜 ||𝑅𝐷 )] = [𝑅𝐹 + (𝑟𝑜 ||𝑅𝐷 )]𝑖𝑖 therefore To determine 𝑣𝑜 𝑣𝑔𝑠 𝑅𝐹 + (𝑟𝑜 ||𝑅𝐷 ) = = ≅ 22.3𝑘Ω 𝑖𝑖 𝑖𝑖 1 + 𝑔𝑚 (𝑟𝑜 ||𝑅𝐷 ) output resistance 𝑅𝑜 , we 𝑅𝑖 = the ! ⏞ 0 = 𝑣𝑔𝑠 , as schematize in Figure 12.23c. 𝑣𝑖 = RF ii G have to impose id D + v gs ro + R D v ds + vo S Ro Ri Figure 12.23c AC equivalent model with vi=0 As a consequence 𝑔𝑚 𝑣𝑔𝑠 = 0 𝑅𝑜 = 𝑅𝐹 ||𝑟𝑜 ||𝑅𝐷 ≅ 4.4𝑘Ω 𝑣 𝑣 To determine the voltage gain 𝐴𝑣 = 𝑣 𝑜 , we can recall the equation 𝑣𝑔𝑠 − 𝑣𝑜 = 𝑅𝐹 𝑖𝑖 = 𝑅𝐹 𝑅𝑜 so that 𝑔𝑠 𝑅𝐹 𝑣𝑔𝑠 �1 − � = 𝑣𝑜 𝑅𝑖 In this equation we can replace the expression of 𝑅𝑖 obtaining 𝑣𝑔𝑠 �1 − Finally 𝐴𝑣 = 𝑅𝐹 � = 𝑣𝑜 𝑅𝐹 + (𝑟𝑜 ||𝑅𝐷 ) 1 + 𝑔𝑚 (𝑟𝑜 ||𝑅𝐷 ) (𝑟𝑜 ||𝑅𝐷 )(1 − 𝑔𝑚 𝑅𝐹 ) 1 − 𝑔𝑚 𝑅𝐹 𝑣𝑜 = = ≅ −8.9 𝑅𝐹 (𝑟𝑜 ||𝑅𝐷 ) + 𝑅𝐹 𝑣𝑔𝑠 1+ 𝑟𝑜 ||𝑅𝐷 Observation The voltage gain can also be written as 1 1 𝑅𝐹 − 𝑔𝑚 𝑅𝐹 − 𝑔𝑚 𝐴𝑣 = = 1 1 1 𝑅𝐹 + 𝑟𝑜 ||𝑅𝐷 𝑟𝑜 ||𝑅𝐷 ||𝑅𝐹 1 and since its often easily results 𝑔𝑚 ≫ 𝑅 , then 𝐹 𝐴𝑣 ≅ −𝑔𝑚 (𝑟𝑜 ||𝑅𝐷 ||𝑅𝐹 ) 𝑖 Summarizing 𝑅𝑖 ≅ 22.3𝑘Ω, 𝑅𝑜 ≅ 4.4𝑘Ω, 𝐴𝑣 ≅ −8.9. Exercise n. 3 An E-MOSFET C.S. amplifier is reported as in Figure 12.24. VDD RD R1 CB1 CB2 + vi + vl R2 Figure 12.24 E-mosfet C.S. amplifier with voltage divider bias The bias network has 𝑅1 = 5𝑀Ω, 𝑅2 = 860𝑘Ω, 𝑅𝐷 = 2𝑘Ω, 𝑉𝐷𝐷 = 12𝑉. The MOSFET has 𝑉𝑡ℎ = 2𝑉 and a drain current 𝐼𝐷 = 0.1𝐴 when 𝑉𝐺𝑆 = 5𝑉. Calculate the Q-point: 𝑉𝐺𝑆𝑄 , 𝐼𝐷𝑄 , 𝑉𝐷𝑆𝑄 . ANSWER Voltage divider equation applied to the input loop yields 𝑅2 𝑉𝐺𝑆𝑄 = 𝑉 ≅ 1.76𝑉 𝑅1 + 𝑅2 𝐷𝐷 The parameter 𝐾𝑛 is commonly given, but not now, so we have to determine it. In general, this can be done referring to the parameters furnished by the manufacturer with the MOSFET datasheet, by considering a particular value named 𝐼𝐷(𝑜𝑛) at a specific value of 𝑉𝐺𝑆 . Since we do not have these data, accepting a certain error, we can proceed admitting 𝐼𝐷(𝑜𝑛) = 𝐼𝐷 , so that 𝐼𝐷(𝑜𝑛) 𝐾𝑛 ≅ = 0.01 (𝑉𝐺𝑆 − 𝑉𝑡ℎ )2 (the value of λ here is neglected). The value of 𝐾𝑛 is necessary to calculate the quiescent drain current Finally 2 𝐼𝐷𝑄 = 𝐾𝑛 �𝑉𝐺𝑆𝑄 − 𝑉𝑡ℎ � ≅ 0.63𝑚𝐴 𝑉𝐷𝑆𝑄 = 𝑉𝐷𝐷 − 𝑅𝐷 𝐼𝐷 ≅ 10.73𝑉 Summarizing VGSQ ≅ 1.76𝑉, IDQ ≅ 63.4𝑚𝐴, VDSQ ≅ 10.7𝑉. Exercise n. 4 An E-MOSFET is used in a common gate configuration for which the output signal is taken off the drain with respect to the ground, the gate is AC connected directly to the ground /Figure 12.25). VDD iD R1 RD il CB2 CG CB1 + vi R2 + Rs vl + vs RS RL Figure 12.25 C.G. configuration The given parameters are 𝑅1 = 1𝑀Ω, 𝑅2 = 1𝑀Ω, 𝑅𝑆 = 2.2𝑘Ω, 𝑅𝐷 = 4.7𝑘Ω, 𝑅𝐿 = 500Ω, 𝑅𝑠 = 500Ω, 𝑔𝑚 = 0.01𝑆, 𝑟𝑜 = 50𝑘Ω.. Calculate the following:  The MOSFET and overall input resistances, 𝑅𝑖𝑀𝑂𝑆 and 𝑅𝑖  The MOSFET and overall voltage gains, 𝐴𝑣𝑀𝑂𝑆 and 𝐴𝑣  The MOSFET and overall output resistances, 𝑅𝑜𝑀𝑂𝑆 and 𝑅𝑜 ANSWER For convenience we can combine 𝑅𝑠𝑆 = 𝑅𝑠 ||𝑅𝑆 ≅ 407Ω. and 𝑅𝐷𝐿 = 𝑅𝐷 ||𝑅𝐿 ≅ 2423Ω.. According to Section 12.4.2.1, we can write 𝑟𝑜 + 𝑅𝐷𝐿 1 + 𝑔𝑚 𝑟𝑜 𝑟𝑜 + 𝑅𝐷𝐿 𝑅𝑖 = 𝑅𝑖𝑀𝑂𝑆 ||𝑅𝑆 = ||𝑅 1 + 𝑔𝑚 𝑟𝑜 𝑆 Numerically 𝑅𝑖𝑀𝑂𝑆 ≅ 104.6Ω, 𝑅𝑖 ≅ 99.9Ω. According to Section 12.4.2.2, we can write 𝑅𝑜𝑀𝑂𝑆 = 𝑟𝑜 + (1 + 𝑔𝑚 𝑟𝑜 )𝑅𝑠𝑆 𝑅𝑖𝑀𝑂𝑆 = 𝑅𝑜 = 𝑅𝐷 ||𝑅𝑜𝑀𝑂𝑆 = 𝑅𝐷 ||[𝑟𝑜 + (1 + 𝑔𝑚 𝑟𝑜 )𝑅𝑠𝑆 ] Numerically 𝑅𝑜𝑀𝑂𝑆 ≅ 254.1𝑘Ω, 𝑅𝑜 ≅ 4.6𝑘Ω. According to Section 12.4.2.3, we can write 𝑅𝐷𝐿 𝐴𝑣𝑀𝑂𝑆 = (1 + 𝑔𝑚 𝑟𝑜 ) 𝑟𝑜 + 𝑅𝐷𝐿 𝑅𝑆 𝑅𝑆 𝑅𝐷 ||𝑅𝐿 𝐴𝑣 = 𝐴𝑣𝑀𝑂𝑆 = (1 + 𝑔𝑚 𝑟𝑜 ) 𝑅𝑆 + 𝑅𝑠 𝑟𝑜 + (𝑅𝐷 ||𝑅𝐿 ) 𝑅𝑆 + 𝑅𝑠 Numerically 𝐴𝑣𝑀𝑂𝑆 ≅ 23.16, 𝐴𝑣 ≅ 18.9. Summarizing 𝑅𝑖𝑀𝑂𝑆 ≅ 104.6Ω, 𝑅𝑖 ≅ 99.9Ω, 𝑅𝑜𝑀𝑂𝑆 ≅ 254.1𝑘Ω, 𝑅𝑜 ≅ 4.6𝑘Ω, 𝐴𝑣𝑀𝑂𝑆 ≅ 23.16, 𝐴𝑣 ≅ 18.9 Exercise n. 5 A C.S. configuration is provided with the gate DC directly connected to the power source, as represented in Figure 12.23. VDD RD ID + + VGS - VDS - RDS Figure 12.26 𝑚𝐴 If the power source is 𝑉𝐷𝐷 = 6𝑉 and the MOSFET parameters are 𝑉𝑡ℎ = 1𝑉 and 𝐾𝑛 = 1 𝑉 2 , which is the value to be give to the resistance 𝑅𝐷 to obtain a DC output voltage 𝑉𝐷𝑆 = 2𝑉? Determine the equivalent DC output resistance 𝑅𝐷𝑆 too. ANSWER Since 𝑉𝐷𝑆 = 2 < 5 = 𝑉𝐺𝑆 − 𝑉𝑡ℎ the MOS operates in its ohmic region. Therefore 𝐼𝐷 = 𝐾𝑛 �2(𝑉𝐺𝑆 − 𝑉𝑡ℎ )𝑉𝐷𝑆 − 𝑉𝐷𝑆 2 � = 16𝑚𝐴 The drain resistance is The DC output resistance is RD = 𝑉𝐷𝐷 − 𝑉𝐷𝑆 = 250Ω 𝐼𝐷 R DS = 𝑉𝐷𝑆 = 125Ω 𝐼𝐷 Exercise n. 6 Let us consider the amplifier shown in Figure 12.23, for which we already calculated its AC conditions. Determine now its DC conditions, that is, its Q-point established by the values of 𝑉𝐺𝑆𝑄 , 𝑉𝐷𝑆𝑄 and 𝐼𝐷𝑆𝑄 . Please,consider a battery 𝑉𝐷𝐷 = 10𝑉, a MOSFET transconductance parameter 𝑚𝐴 𝐾𝑛 = 0.2 𝑉 2 , and remember the value of the resistors 𝑅𝐷 = 4.7𝑘Ω, 𝑅𝐹 = 220𝑘Ω. ANSWER We want the MOSFET to work in its saturation region, so that 2 2 𝐼𝐷𝑆𝑄 = 𝐾𝑛 �𝑉𝐺𝑆𝑄 − 𝑉𝑡ℎ � �1 + λ𝑉𝐷𝑆𝑄 � ≅ 𝐾𝑛 �𝑉𝐺𝑆𝑄 − 𝑉𝑡ℎ � but, the Ohm’s law applied to the drain resistor yields 𝑉𝐷𝐷 − 𝑉𝐷𝑆𝑄 𝑉𝐷𝐷 − 𝑉𝐺𝑆𝑄 = 𝐼𝐷𝑆𝑄 = 𝑅𝐷 𝑅𝐷 so that we can write 𝑉𝐷𝐷 − 𝑉𝐺𝑆𝑄 2 𝐾𝑛 �𝑉𝐺𝑆𝑄 − 𝑉𝑡ℎ � = 𝑅𝐷 from which 2 2 (𝐾𝑛 𝑅𝐷 )𝑉𝐺𝑆𝑄 + (1 − 2𝐾𝑛 𝑅𝐷 𝑉𝑡ℎ )𝑉𝐺𝑆𝑄 + (𝐾𝑛 𝑅𝐷 𝑉𝑡ℎ − 𝑉𝐷𝐷 ) = 0 Two possible solutions of the previous equation are 𝑉𝐺𝑆𝑄1 ≅ 4𝑉 and 𝑉𝐺𝑆𝑄2 ≅ −2.1𝑉, but it is obvious to admit only 𝑉𝐺𝑆𝑄1 = 𝑉𝐺𝑆𝑄 ≅ 4𝑉. As a consequence 𝐼𝐷𝑆𝑄 = 1.27𝑚𝐴. Clearly we have 𝑉𝐷𝑆𝑄 = 𝑉𝐺𝑆𝑄 ≅ 4𝑉. Observation The feedback resistor 𝑅𝐹 plays no rule for the Q-point. Exercise n. 1 Figure 13.15 shows a pseudo-complementary Darlington configuration. VEE R1 IE2 IB2 IB1 IC1 T1 T2 R2 IC2 IE1 Figure 13.15 Pseudo-complementary Darlington configuration Given that it is supplied from a battery 𝑉𝐸𝐸 = 6𝑉, the resistance values are 𝑅1 = 4.7𝑀Ω and 𝑅2 = 4.7𝑘Ω, and that the current gains of the two transistors are 𝛽1 = 60 and 𝛽2 = 40, determine the BJT Q-points, {𝐼𝐵1 ; 𝐼𝐶1 ; 𝑉𝐶𝐸1 } and {𝐼𝐵2 ; 𝐼𝐶2 ; 𝑉𝐸𝐶2 } respectively. ANSWER Let us start considering the two transistors in their active regions. KVL-BE1 yields 𝑉𝐸𝐸 = 𝑅1 𝐼𝐵1 + 𝑉𝐵𝐸1 so that 𝐼𝐵1 = 1.13𝑛𝐴 and 𝐼𝐶1 ≅ 𝛽𝐼𝐵1 ≅ 67.7𝑛𝐴. KVL-EB2/CE1 yields 𝑉𝐸𝐸 = 𝑉𝐸𝐵2 + 𝑉𝐶𝐸1 so that 𝑉𝐶𝐸1 = 𝑉𝐸𝐸 − 𝑉𝐸𝐵2 = 6 − 0.7 = 5.3𝑉. Now 𝐼𝐵2 = 𝐼𝐶1 = 67.7𝑛𝐴 and KVL-CE2 yields 𝑉𝐸𝐸 = 𝑉𝐸𝐶2 + 𝑅2 𝐼𝐶2 but 𝐼𝐶2 ≅ 𝛽𝐼𝐵2 so that 𝑉𝐸𝐶2 = 𝑉𝐸𝐸 − 𝑅2 𝛽𝐼𝐵2 = −6.72𝑉 which is a wrong results for sure, since the emitter-collector voltage cannot be lower than zero. The consequence is that the transistor T 2 cannot be in its active region but it is in its saturation region where 𝑉𝐸𝐵2 ≅ 0.2𝑉 (Section 8.6.5 in Chapter 8). So 𝐼𝐶2 ⁄𝐼𝐵2 is lower than 𝛽2 and 𝑉𝐸𝐸 − 𝑉𝐸𝐶2 𝐼𝐶2 = = 1.23𝑚𝐴 𝑅2 Summarizing: {IB1 ; IC1 ; VCE1 } = {1.13𝑛𝐴; 67.7𝑛𝐴; 5.3𝑉}; {IB2 ; IC2 ; VEC2 } = {67.7𝑛𝐴; 1.23𝑚𝐴; 0.2𝑉} 2 Exercise n. 2 Figure 13.16 shows a Wilson current mirror network. VCC IR R A IC3=IO IB3 + VBE3 IC1 T1 T3 IE3 B IC2 +I VBE1 IB2 + VBE2 B1 T2 Figure 13.16 Wilson current source Demonstrate that the relationship which links the output current 𝐼𝑜 with the reference one 𝐼𝑅 is 𝐼𝑜 = 1+ 1 2 𝛽(𝛽 + 2) 𝐼𝑅 Please, consider 𝑇1 and 𝑇2 as twin BJTs so that 𝑉𝐵𝐸1 = 𝑉𝐵𝐸2 = 𝑉𝐵 , 𝐼𝐵1 = 𝐼𝐵2 = 𝐼𝐵 , 𝐼𝐶1 = 𝐼𝐶2 = 𝐼𝐶 , and all transistors having the same common-emitter current gain 𝛽1 = 𝛽2 = 𝛽3 = 𝛽. ANSWER KCL-node “A” yields KCL-node “B” yields 𝐼𝑅 = 𝐼𝐶1 + 𝐼𝐵3 = 𝐼𝐶 + 𝐼𝐵3 𝐼𝐸3 = 𝐼𝐵1 + 𝐼𝐵2 + 𝐼𝐶2 = 2𝐼𝐵 + 𝐼𝐶 = 2 from which we have 𝐼𝐶 = � 𝛽 �𝐼 2 + 𝛽 𝐸3 𝐼𝐶 2 + 𝐼𝐶 = �1 + � 𝐼𝐶 𝛽 𝛽 But since the current relationship of a generic BJT is 𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶 = previous equation can be rewritten as 𝛽 1+𝛽 1+𝛽 𝐼𝐶 = � �� 𝐼𝐶3 = 𝐼 2+𝛽 𝛽 2 + 𝛽 𝐶3 which, inserted into the equation KCL-node “A”, yields 3 𝐼𝐶 𝛽 1+𝛽 + 𝐼𝐶 = � 𝛽 � 𝐼𝐶 , the 1+𝛽 1+𝛽 𝐼𝐶3 1+𝛽 1 𝛽 2 + 2𝛽 + 2 𝐼𝐶3 + 𝐼𝐵3 = 𝐼𝐶3 + =� + � 𝐼𝐶3 = 𝐼 2+𝛽 2+𝛽 2+𝛽 𝛽 𝛽 2 + 2𝛽 𝐶3 𝛽 from which it follows 𝛽 2 + 2𝛽 𝐼 𝐼𝐶3 = 2 𝛽 + 2𝛽 + 2 𝑅 and, finally 1 𝐼𝐶3 = 𝐼𝑜 = 𝐼𝑅 2 1+ 𝛽(𝛽 + 2) 𝐼𝑅 = Exercise n. 3 The transformer-coupled common emitter class-A amplifier, shown in Figure 13.17, couples the AC load collector resistor from DC bias network, so that half of the DC supply power is no longer dissipated by the resistor load, as it is in a standard commone emitter class-A amplifier. There is no DC drop across the transformer so that the peak-to-peak output voltage will double. The battery supplies a voltage 𝑉𝐶𝐶 = 12𝑉, the load resistor is an 8Ω speaker, the emitter resistor values is 𝑅𝐸 = 100Ω, and the BJT collector quiescent current is 𝐼𝐶𝑄 = 30𝑚𝐴. Determine the power supplied from the battery 𝑃𝑖(𝐷𝐶) , the DC power dissipated on the transistor𝑃𝐵𝐽𝑇(𝐷𝐶) , on the emitter resistor 𝑃𝑅𝐸 (𝐷𝐶) , and the AC power used by the load 𝑃𝑜(𝐴𝐶) . Determine the conversion efficiency η = 𝑃𝑜(𝐴𝐶) 𝑃𝑖(𝐷𝐶) of the amplifier too. Consider an ideal loseless transformer. VCC R1 n1 n2 RC CB Rs R2 RE CE + vs Figure 13.17 Transformer-coupled CE class-A amplifier 4 ANSWER As usual, we can assume 𝐼𝐸𝑄 ≅ 𝐼𝐶𝑄 = 0.03𝐴. Consequently 𝑉𝐸𝑄 = 𝑅𝐸 𝐼𝐸𝑄 = 3𝑉 and 𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶 − 𝑉𝐸𝑄 = 9𝑉. The theoretical maximum swing of the output voltage across the BJT collector-emitter terminals towards lower values is ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑑𝑜𝑤𝑛 = 𝑉𝐶𝐸𝑄 − 𝑉𝐶𝐸(𝑆𝐴𝑇) = 9 − 0.2 = 8.8𝑉 and, to guarantee a symmetrical swing we can assume ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑢𝑝 = ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑑𝑜𝑤𝑛 = 8.8𝑉 Observation The theoretical total voltage swing across the BJT collector-emitter terminals is ∆𝑉𝐶𝐸,𝑚𝑎𝑥 = 𝑉𝐶𝐸𝑄 + ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑢𝑝 = 17.8𝑉, which is higher than 𝑉𝐶𝐶 . This is according to the BJT potentiality, but we have to consider the maximum available source power so that the current will be lower accordingly. The equivalent output resistance at the collector terminal 𝑅𝐶,𝑒𝑞 is ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑢𝑝 𝑅𝐶,𝑒𝑞 = ≅ 293Ω 𝐼𝐶𝑄 so that the transformer turn ratio is 𝑅𝐶,𝑒𝑞 𝑛=� ≅ 8.6 𝑅𝐶 The maximum voltage swing across the speaker ∆𝑣𝑅𝐶 is than ∆𝑉𝐶𝐸,𝑚𝑎𝑥−𝑢𝑝 ∆𝑣𝑅𝐶 = ≅ 1.03𝑉 𝑛 The DC power dissipated on the speaker is ∆𝑣𝑅2𝐶 𝑃𝑜(𝐴𝐶) = = 0.13𝑊 2𝑅𝐶 The power supplied from the battery is Pi(DC) = 𝑉𝐶𝐶 𝐼𝐶𝑄 = 0.36𝑊 The power dissipated on the emitter resistor is PRE (DC) = 𝑉𝐸𝑄 𝐼𝐸𝑄 = 0.09𝑊 The power dissipated on the BJT is PBJT(DC) = 𝑉𝐶𝐸𝑄 𝐼𝐶𝑄 = 0.27𝑊 Finally, the conversion efficiency is Po(AC) η= = 0.36 Pi(DC) The η(%) = 36% result is lower than the theoretical 50% value for this configuration. Please, take into account that there is a voltage drop across the emitter resistor and a non-zero value of 𝑉𝐶𝐸(𝑆𝐴𝑇) . 5 Exercise n. 4 Figure 13.18 shows an amplifier in a Cascode configuration. Consider the two BJTs being of BC107 type, and the resistor values as follows: 𝑅𝑠 = 500Ω, 𝑅1 = 1.5𝑘Ω, 𝑅2 = 1.5𝑘Ω, 𝑅3 = 1.5𝑘Ω, 𝑅𝐶 = 4.7𝑘Ω, 𝑅𝐿 = 220Ω. VCC RC R1 C3 C2 T2 R2 + + vo(CE) C1 Rs + vl T1 R3 RE RL CE vs Figure 13.18 Cascode amplifier Determine the overall voltage gain. ANSWER Transistor 𝑇1 and transistor 𝑇2 are configured in a common-emitter and common-base network respectively. From the conversion Table 11.5 (see Section 11.3.5 in Chapter 11) we can write ℎ ℎ𝑖𝑏 = 1+ℎ𝑖𝑒 ≅ 15.94Ω, ℎ𝑜𝑒 1+ℎ𝑓𝑒 𝑓𝑒 = 0.12 ∗ 10−6 Ω−1 . ℎ𝑟𝑏 = ℎ𝑖𝑒 ℎ𝑜𝑒 1+ℎ𝑓𝑒 − ℎ𝑟𝑒 = 25.8 ∗ 10−3 , ℎ ℎ𝑓𝑏 = − 1+ℎ𝑓𝑒 = −0.996, ℎ𝑜𝑏 = 𝑓𝑒 Figure 13.18b shows the AC equivalent model of the cascade amplifier. is i b,I Rs + vs + i c,I hie 1 hoe hrevo(CE) + R23 vi i e,II + v o(CE)= =vi(CB) hfe i b,I - 1 hob hrbvo(CE) + - E Ri(tot) Ri(CE) i c,II hib hfbi e,II io RC RL + vl - B Ro(CE) Ri(CB) Figure 13.18b AC equivalent model of the cascode amplifier 6 Ro(CB) Ro(tot) The input resistance of the CB stage (Section 11.7.2.1 in Chapter 11) is ℎ𝑖𝑏 + 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 ∆ℎ𝐶𝐵 𝑅𝑖(𝐶𝐵) = 1 + 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 ℎ𝑜𝑏 with ∆ℎ𝐶𝐵 = ℎ𝑖𝑏 ℎ𝑜𝑏 − ℎ𝑓𝑏 ℎ𝑟𝑏 and 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 = 𝑅𝐶 ||𝑅𝐿 . Numerically 𝑅𝑖(𝐶𝐵) ≅ 16Ω. Similarly, the input resistance of the CE stage (Section 11.4.2.2 in Chapter 11) is ℎ𝑖𝑒 + 𝑅𝑙𝑜𝑎𝑑,𝐼 ∆ℎ𝐶𝐸 𝑅𝑖(𝐶𝐸) = 1 + 𝑅𝑙𝑜𝑎𝑑,𝐼 ℎ𝑜𝑒 with ∆ℎ𝐶𝐸 = ℎ𝑖𝑒 ℎ𝑜𝑒 − ℎ𝑓𝑒 ℎ𝑟𝑒 and 𝑅𝑙𝑜𝑎𝑑,𝐼 = 𝑅𝑖(𝐶𝐵) . Numerically 𝑅𝑖(𝐶𝐸) ≅ 4𝑘Ω. The two previous results are necessary to determine the voltage gain of the two stages separately and of the overall network too. The voltage gain of the CB stage (Section 11.7.2.3 in Chapter 11) is ℎ𝑓𝑏 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 𝐴𝑣(𝐶𝐵) = − ℎ𝑖𝑏 + 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 ∆ℎ𝐶𝐵 where 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 = 𝑅𝐶 ||𝑅𝐿 . Numerically 𝐴𝑣(𝐶𝐵) ≅ 13.1. Similarly, the voltage gain of the CE stage (Section 11.4.2.4 in Chapter 11) is ℎ𝑓𝑒 𝑅𝑙𝑜𝑎𝑑,𝐼 𝐴𝑣(𝐶𝐸) = − ℎ𝑖𝑒 + 𝑅𝑙𝑜𝑎𝑑,𝐼 ∆ℎ𝐶𝐸 Numerically 𝐴𝑣(𝐶𝐸) ≅ −0.999. The overall input resistance is 𝑅𝑖(𝑡𝑜𝑡) = (𝑅2 ||𝑅3 )||𝑅𝑖(𝐶𝐸) Numerically 𝑅𝑖(𝑡𝑜𝑡) ≅ 630Ω. The overall voltage gain is then 𝐴𝑣(𝑡𝑜𝑡) = 𝛼𝑖 𝐴𝑣(𝐶𝐸) 𝐴𝑣(𝐶𝐵) = 𝑅𝑖(𝑡𝑜𝑡) 𝐴 𝐴 ≅ −7.3 𝑅𝑠 + 𝑅𝑖(𝑡𝑜𝑡) 𝑣(𝐶𝐸) 𝑣(𝐶𝐵) Observations The hybrid parameter determinant ∆ℎ can be neglected in many cases, because its common value is less than one. For instance, the values it assumes for some commercial BJTs are as in the following Table 13.1 (see Section 11.3.4 in Chapter 11 as reference). BJT BC107 BC107A BC107B BC108 BC108A BC108B BC108C BC846A BC846B BC846C hie 4,00E+03 3,00E+03 4,80E+03 5,50E+03 3,00E+03 4,80E+03 7,00E+03 2,70E+03 4,50E+03 8,70E+03 hre 2,20E-04 1,70E-04 2,70E-04 3,10E-04 1,70E-04 2,70E-04 3,80E-04 1,50E-04 2,00E-04 3,00E-04 hfe 250 190 300 370 190 300 500 220 330 600 hoe 3,00E-05 1,30E-05 2,60E-05 3,00E-05 1,30E-05 2,60E-05 3,40E-05 1,80E-05 3,00E-05 6,00E-05 ∆ 0,07 0,01 0,04 0,05 0,01 0,04 0,05 0,02 0,07 0,34 Table 13.1 Hybrid parameter values for some commercially available BJTs There is no voltage divider between the CE and CB stages. 7 The resistor R1 plays no rule in AC conditions having no AC voltage drop across it. Exercise n. 5 Figure 13.19 shows the scheme of a differential amplifier in a simplified version. VCC RC RC T1 v0 T2 + + vi1 vi2 RE -VEE Figure 13.19 Simple differential amplifier network Assuming two twin BJTs, and two twin collector resistors 𝑅𝐶 , determine the expression of the CMRR value. For convenience, neglect the hybrid parameter values of ℎ𝑟𝑒 and ℎ𝑜𝑒 . ANSWER 𝐴 The CMRR value is defined as 𝐶𝑀𝑅𝑅 ≝ �𝐴𝐶 �, with 𝐴𝐶 the common voltage gain and 𝐴𝐷 the 𝐷 differential voltage gain, being the total voltage output 𝑣𝑖1 + 𝑣𝑖2 𝑣𝑜 = 𝐴𝐷 (𝑣𝑖1 − 𝑣𝑖2 ) + 𝐴𝐶 � � 2 Let us start considering the expression for 𝐴𝐶 , to find which we can impose two twin common ! ! 𝑣 inputs as 𝑣𝑖1 = ⏞ 𝑣𝑖2 = ⏞ 𝑣𝑖,𝐶 , so that 𝐴𝐶 ≝ 𝑣 𝑜 . In such a way we can write 𝑖,𝐶 𝑣𝑖,𝐶 + 𝑣𝑖,𝐶 𝑣𝑜 = 𝐴𝐷 �𝑣𝑖,𝐶 − 𝑣𝑖,𝐶 � + 𝐴𝐶 � � = 𝐴𝐶 𝑣𝑖,𝐶 2 The CMRR value has to be determine in AC conditions, so Figure 13.19b (a) and (b) represent two AC equivalent models of the differential amplifier, with the only difference that in (b) the emitter resistor is conveniently split into two parts with respect to (a). In such a way, the entire network can be analyzed considering only its half-right part (being its left-part symmetrically the same). 8 RC T1 v0 RC RC T2 T1 + vi,C v0 + + vi,C RE RC T2 + vi,C vi,C 2RE 2RE (a) (b) Figure 13.19b AC equivalent model of the simple differential amplifier network with common voltage inputs, being in (a)a unique emitter resistor and in (b) the same resistor is spit into its equivalent parallel version Figure 13.19c shows the equivalent AC half-right part with the right BJT modelled by its hybrid parameter representation. ib ic hie B C + hfei b + v be v ce + v i,C + E - + ie 2RE RC vo ve - - Figure 13.19c AC equivalent model of the half-right part of the differential amplifier with the BJT replaced by its hybrid parameter equivalent representation The output voltage value is 𝑣𝑜 = −𝑅𝐶 𝑖𝑐 = −𝑅𝐶 ℎ𝑓𝑒 𝑖𝑏 and KVL applied to the input loop yields 𝑣𝑖,𝐶 = ℎ𝑖𝑒 𝑖𝑏 + 2𝑅𝐸 𝑖𝑒 = ℎ𝑖𝑒 𝑖𝑏 + 2𝑅𝐸 �𝑖𝑏 + ℎ𝑓𝑒 𝑖𝑏 � so that ℎ𝑓𝑒 𝑅𝐶 𝑣𝑜 𝐴𝐶 ≝ =− 𝑣𝑖,𝐶 ℎ𝑖𝑒 + 2�1 + ℎ𝑓𝑒 �𝑅𝐸 9 Observation To reduce the value of 𝐴𝐶 , we have to admit the value of the emitter resistor 𝑅𝐸 as high as possible. This is the reason to utilize a current source in place of 𝑅𝐸 (see Section 13.2). 𝑣 Now, we can determine the expression for 𝐴𝐷 ≝ 𝑣 𝑜 , being 𝑣𝑖,𝐷 the differential voltage input 𝑖,𝐷 applied halved and 180° out of phase at the two input terminals, 𝑣𝑖1 = schematized in Figure 13.19d. In such a way we can write 𝑣𝑖1 + 𝑣𝑖2 𝑣𝑜 = 𝐴𝐷 (𝑣𝑖1 − 𝑣𝑖2 ) + 𝐴𝐶 � � = 𝐴𝐷 𝑣𝑖,𝐷 2 RC T1 v i,D 2 𝑣𝑖,𝐷 2 and 𝑣𝑖2 = − 𝑣𝑖,𝐷 2 , as RC T2 v0 + + v i,D 2 Figure 13.19d AC equivalent model of the simple differential amplifier network with differential voltage inputs The study can be performed considering the AC equivalent model of the half-right side, as schematized in Figure 13.19e ib ic hie B + v i,D 2 C + hfei b + RC + v ce vo v be E Figure 13.19e AC equivaent model of the right-half side of the differential amplifier with one differential voltage input Voltage values across the output and input ports are 𝑣𝑜 = −𝑅𝐶 𝑖𝑐 = −𝑅𝐶 ℎ𝑓𝑒 𝑖𝑏 𝑣𝑖,𝐷 = ℎ𝑖𝑒 𝑖𝑏 − 2 10 so that −𝑅𝐶 ℎ𝑓𝑒 𝑖𝑏 1 ℎ𝑓𝑒 𝑣𝑜 = = 𝑅 𝑣𝑖,𝐷 −2ℎ𝑖𝑒 𝑖𝑏 2 ℎ𝑖𝑒 𝐶 Finally, we can write the requested CMRR value as 𝐴𝐶 1 ℎ𝑖𝑒 + 2�1 + ℎ𝑓𝑒 �𝑅𝐸 1 ℎ𝑖𝑒 + 2ℎ𝑓𝑒 𝑅𝐸 𝐶𝑀𝑅𝑅 ≝ � � = ≅ 𝐴𝐷 2 ℎ𝑖𝑒 2 ℎ𝑖𝑒 𝐴𝐷 ≝ 11 I Basi di elettronica Exercise n. 1 Given the network shown in Figure 14.22, we have, for the source 𝑣𝑠 = 10𝑚𝑉, 𝑅𝑠 = 200Ω,, for the amplifier 𝑅𝑖 = 1𝑘Ω, 𝐴𝑣 = 20, 𝑅𝑜 = 500Ω , and for the load 𝑅𝐿 = 100Ω.. Determine the value of the output voltage 𝑣𝑜 . Rs + Ri vi vs Ro RL Av vi + vo Figure 14.22 Scheme of a generic network including a voltage amplifier ANSWER The voltage divider effect around the input loop yields 𝑅𝑖 𝑣𝑖 = 𝑣 ≅ 8.3𝑚𝑉 𝑅𝑖 + 𝑅𝑠 𝑠 The amplifier can furnish an output voltage equal to 𝐴𝑣 𝑣𝑖 = 20𝑥8.3 ∗ 10−3 ≅ 0.16𝑉 which is reduced due to the loading effect to 𝑅𝐿 𝑣𝑜 = 𝐴 𝑣 ≅ 27.6𝑚𝑉 𝑅𝐿 + 𝑅𝑜 𝑣 𝑖 Exercise n. 2 (𝑖) A voltage amplifier as a voltage gain of 𝐴𝑣 = 300. It is used to supply a voltage of 𝑣𝐿 = 2𝑉 across a load 𝑅𝐿 = 5𝑘Ω . The available AC voltage source has a value of 𝑣𝑠 = 10𝑚𝑉 and an inner resistance of 𝑅𝑠 = 500Ω, and it is capable of a peak current 𝑖𝑠 = 1𝜇𝐴. Determine which input and output resistances are requested by the amplifier. ANSWER The requested network can be shematized as in Figure 14.22b. is ii R R s + vs o iL + + + (i) A v vi vi - io Ri Figure 14.22b generic voltage amplifier We want a peak current value of 𝑖𝑖 = 10𝜇𝐴, therefore ! 𝑣𝑠 𝑖𝑖 = 𝑖𝑠 = = ⏞ 1𝜇𝐴 𝑅𝑠 + 𝑅𝑖 from which 1 vo - RL Basi di elettronica 𝑣𝑠 − 𝑅𝑠 𝑖𝑠 𝑖𝑠 We can apply the current divider expressions to the input and output loops, obtaining 𝑅𝑖 𝑅𝐿 𝑅𝐿 (𝑖) (𝑖) 𝑣𝐿 = 𝐴𝑣 𝑣𝑖 = 𝐴𝑣 𝑣 𝑅𝐿 + 𝑅𝑜 𝑅𝐿 + 𝑅𝑜 𝑅𝑖 + 𝑅𝑠 𝑠 from which 𝑅𝑖 = (𝑖) 𝐴𝑣 𝑅𝑖 𝑅𝐿 𝑣𝑠 − 𝑅𝐿 (𝑅𝑖 + 𝑅𝑠 )𝑣𝐿 𝑅𝑜 = (𝑅𝑖 + 𝑅𝑠 )𝑣𝐿 Numerically 𝑅𝑖 = 9500Ω; 𝑅0 = 2125Ω. Exercise n. 3 Consider the double C.E.-swamped C.E. configuration as represented in Figure 14.23. VCC RC,I R1,I CB1 R2,I + CB3 CB2 Rs vs RC,II R1,II R2,II RE,I CE,I RE1,II RE2,II CE,II Figure 14.23 Double stage CE-swamped CE The parameters ℎ𝑟𝑒 and ℎ𝑜𝑒 , of both BJTs, are negligible. Determine the expressions of the following:  The input resistances of both stages and the overall input resistance  The voltage gains of both stages and the overall voltage gain ANSWER Figure 14.23b shows the AC equivalent model of the circuit where, for convenience, resistors are combined as 𝑅12,𝐼 = 𝑅1,𝐼 ||𝑅2,𝐼 and 𝑅12,𝐼𝐼 = 𝑅1,𝐼𝐼 ||𝑅2,𝐼𝐼 . 2 Basi di elettronica i b,I B vs i c,II hie,II hfe,Ii b,I + v be,I + i b,II i c,I hie,I Rs R12,I - + hfe,IIi b,II v be,II v ce,I E - B C + RC,I R12,II - + C v ce,II E + ie RE1,II Ri,I RC,II vo v e,II - Ri + - Ri,II Figure 14.23b AC equivalent model of the double stage CE-swamped CE For the first CE stage, according to results obtained in Section 11.4.2.2, and in Section 11.4.2.4, in Chapter 11, neglecting the hybrid parameters ℎ𝑟𝑒 and ℎ𝑜𝑒 as requested, we can write: 𝑅𝑖𝐵𝐽𝑇,𝐼 = ℎ𝑖𝑒,𝐼 ℎ𝑓𝑒,𝐼 𝐴𝑣𝐵𝐽𝑇,𝐼 = − 𝑅 ℎ𝑖𝑒,𝐼 𝑙𝑜𝑎𝑑,𝐼 where 𝑅𝑙𝑜𝑎𝑑,𝐼 = 𝑅𝐶,𝐼 ||𝑅𝑖,𝐼𝐼 = 𝑅𝐶,𝐼 ||𝑅12,𝐼𝐼 ||𝑅𝑖𝐵𝐽𝑇,𝐼𝐼 . Similarly, for the second swamped-CE stage, according to the results obtained in Section 11.5.2.2 and Section 11.5.2.3, in Chapter 11, we can write: 𝑅𝑖𝐵𝐽𝑇,𝐼𝐼 = ℎ𝑖𝑒,𝐼𝐼 + �1 + ℎ𝑓𝑒,𝐼𝐼 �𝑅𝐸1,𝐼𝐼 ≅ ℎ𝑖𝑒,𝐼𝐼 + ℎ𝑓𝑒,𝐼𝐼 𝑅𝐸1,𝐼𝐼 ℎ𝑓𝑒,𝐼𝐼 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 𝐴𝑣𝐵𝐽𝑇,𝐼𝐼 = − ℎ𝑖𝑒,𝐼𝐼 + ℎ𝑓𝑒,𝐼𝐼 𝑅𝐸1,𝐼𝐼 where 𝑅𝑙𝑜𝑎𝑑,𝐼𝐼 = 𝑅𝐶,𝐼𝐼 . The overall input resistance is 𝑅𝑖 = 𝑅12,𝐼 ||𝑅𝑖𝐵𝐽𝑇,𝐼 = 𝑅12,𝐼 ||ℎ𝑖𝑒,𝐼 The overall voltage gain has to take into account the input voltage divider effect, so that 𝐴𝑣 = 𝛼𝑖,𝐼 𝐴𝑣,𝐼 𝐴𝑣,𝐼𝐼 (no voltage divider effect is between the two stages). Therefore ℎ𝑓𝑒,𝐼𝐼 𝑅𝑖 ℎ𝑓𝑒,𝐼 𝐴𝑣 = �𝑅𝐶,𝐼 ||𝑅12,𝐼𝐼 ||�ℎ𝑖𝑒,𝐼𝐼 + ℎ𝑓𝑒,𝐼𝐼 𝑅𝐸1,𝐼𝐼 �� 𝑅 𝑅𝑖 + 𝑅𝑠 ℎ𝑖𝑒,𝐼 ℎ𝑖𝑒,𝐼𝐼 + ℎ𝑓𝑒,𝐼𝐼 𝑅𝐸1,𝐼𝐼 𝐶,𝐼𝐼 Exercise n. 4 Consider the double C.S.- C.S. configuration as represented in Figure 14.24. 3 Basi di elettronica VDD RD1 RD2 CB1 CB2 + CB3 CS1 CS2 vi + vo RS1 RG1 RS2 RG2 Figure 14.24 Double stage CS-CS Determine the expressions of the following:  The overall input resistance  The overall voltage gain  The overall output resistance ANSWER As usual, in two-stage transistor amplifier the voltage gain of the first stage is decreased by the load effect due to the input impedance of the second stage. However, the input impedance of the second stage is theoretically infinite, so that in the case of MOS type transistor amplifier the loading effect of the first stage is practically negigible. The loading effect of the second stage depends on the amplifier’s load. Figure 14.24b shows the AC equivalent model of the double stage CS-CS amplifier. i d,II i d,I + vi - Ri,I G RG,I gm,Ivgs,I + v gs,I - S D r o,I + v ds,I G RD,I RG,II + gm,IIvgs,II v gs,II - - S D + + r o,II v ds,II v o RD,II - Ri,II Ro,I Ro,II Figure 14.24b AC equivalent model of the double stage CS-CS amplifier For the first CS stage, according to results obtained in Section 12.2.2.1, in Section 12.2.2.2, and in Section 12.2.2.3 in Chapter 11, we can write: 𝑅𝑖,𝐼 = 𝑅𝐺,𝐼 𝑅𝑜,𝐼 = 𝑟𝑜,𝐼 ||𝑅𝐷,𝐼 𝐴𝑣,𝐼 = −𝑔𝑚,𝐼 �𝑟𝑜,𝐼 ||𝑅𝐷,𝐼 ||𝑅𝐺,𝐼𝐼 � Similarly, for the second CS stage, we can write 4 Basi di elettronica 𝑅𝑖,𝐼𝐼 = 𝑅𝐺,𝐼𝐼 𝑅𝑜,𝐼𝐼 = 𝑟𝑜,𝐼𝐼 ||𝑅𝐷,𝐼𝐼 𝐴𝑣,𝐼𝐼 = −𝑔𝑚,𝐼𝐼 �𝑟𝑜,𝐼𝐼 ||𝑅𝐷,𝐼𝐼 � The overall input resistance corrsponds to the one of the first stage 𝑅𝑖 = 𝑅𝑖,𝐼 The overall output resistance corresponds to the one of the second stage 𝑅𝑜 = 𝑅𝑜,𝐼𝐼 The overall voltage gain is 𝐴𝑣 = 𝐴𝑣,𝐼 𝐴𝑣,𝐼𝐼 = 𝑔𝑚,𝐼 �𝑟𝑜,𝐼 ||𝑅𝐷,𝐼 ||𝑅𝐺,𝐼𝐼 �𝑔𝑚,𝐼𝐼 �𝑟𝑜,𝐼𝐼 ||𝑅𝐷,𝐼𝐼 � since there is no voltage divider effect. 5 K18912 ISBN: 978-1-4665-8204-0 90000 w w w. c r c p r e s s . c o m 9 781466 582040

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