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CHAPTER 12 TRANSISTORS Exercise 65, Page 172 The answers to questions 1 to 7 can all be found from within the text of the chapter, sections 12.1 to 12.10 on pages 163 to 172. 8. A bipolar junction transistor operates with a collector current of 1.2 A and a base current of 50 mA. What will the value of emitter current be? Emitter current, IE = IB + IC = 50 mA + 1.2 A = 0.05 + 1.2 = 1.25 A 9. What is the value of common emitter current gain for the transistor in problem 8? Common emitter current gain = I C 1.2 = 24 I B 0.05 10. Corresponding readings of base current, IB, and base-emitter voltage, VBE, for a bipolar junction transistor are given in the table below: VBE (V) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 IB (A) 0 0 0 0 1 3 19 57 130 Plot the IB/VBE characteristic for the device and use it to determine (a) the value of IB when VBE = 0.65 V, (b) the static value of input resistance when VBE = 0.65 V, and (c) the dynamic value of input resistance when VBE = 0.65 V The characteristic is shown below. (a) From the characteristic, when VBE = 0.65 V, base current, IB = 32.5 μA (b) When VBE = 0.65 V, IB = 32.5 μA from part (a) Hence, the static value of output resistance = (c) The dynamic value of output resistance = VBE 0.65 = 20 k IB 32.5 106 VBE BC 0.76 0.58 0.18 = 3 k 6 I B AB (70 10) 10 60 106 © John Bird Published by Taylor and Francis 106 11. Corresponding readings of base current, IB, and collector current, IC, for a bipolar junction transistor are given in the table below: IB (A) 0 10 20 30 40 50 60 70 80 IC (mA) 0 1.1 2.1 3.1 4.0 4.9 5.8 6.7 7.6 Plot the IC/IB characteristic for the device and use it to determine the static value of common emitter current gain when IB = 45μA. The characteristic is shown below. When IB = 45μA, IC = 4.4 mA. Thus, the static value of current gain = IC 4.4 10 3 = 98 I B 45 10 6 © John Bird Published by Taylor and Francis 107 © John Bird Published by Taylor and Francis 108 Exercise 66, Page 179 The answers to questions 1 to 5 can all be found from within the text of the chapter, sections 12.11 to 12.15 on pages 172 to 179. 6. The output characteristics for a BJT are shown below. If this device is used in a common-emitter amplifier circuit operating from a 12 V supply with a base bias of 60 μA and a load resistor of 1 kΩ, determine (a) the quiescent values of collector-emitter voltage and collector current, and (b) the peak-peak collector voltage when an 80 μA peak-peak signal current is applied. (a) First we need to construct the load line. The two ends of the load line will correspond to VCC, the 12 V supply, on the collector-emitter voltage axis and 12 V/1 kΩ = 12 mA on the collector current axis. This is shown on the characteristic below. Next we locate the operating point (or quiescent point) from the point of intersection of the IB = 60 μA characteristic and the load line. Having located the operating point we can read off the quiescent values, i.e. the no-signal values, of collector emitter voltage (VCQ) and collector current (ICQ). Hence, VCQ = 5 V and ICQ = 7 mA (b) Next we can determine the maximum and minimum values of collector-emitter voltage by locating the appropriate intercept points on the characteristics. The maximum and minimum values of base current will be (60 μA + 40 μA) = 100 μA on positive peaks of the signal and © John Bird Published by Taylor and Francis 109 (60 μA − 40 μA) = 20 μA on negative peaks of the signal. The maximum and minimum values of VCE are, respectively, 9.7 V and 1.2 V. Hence, the output voltage swing = (9.7 V − 1.2 V) = 8.5 V peak-peak 7. The output characteristics of a JFET are shown below. If this device is used in an amplifier circuit operating from an 18 V supply with a gate-source bias voltage of −3 V and a load resistance of 900 Ω, determine (a) the quiescent values of drain-source voltage and drain current, (b) the peak-peak output voltage when an input voltage of 2 V peak-peak is applied, and (c) the voltage gain of the stage. © John Bird Published by Taylor and Francis 110 (a) First we need to construct the load line. The two ends of the load line will correspond to VCC, the 18 V supply, on the collector-emitter voltage axis and 18 V/900 Ω = 20 mA on the collector current axis. This is shown on the characteristic below. Next we locate the operating point (or quiescent point) from the point of intersection of the VGS = - 3 V characteristic and the load line. Having located the operating point we can read off the quiescent values of drain-source voltage (VDSQ) and drain current (IDQ). Hence, VDSQ = 12.2 V and IDQ = 6.1 mA (b) Next we can determine the maximum and minimum values of collector-emitter voltage by locating the appropriate intercept points on the characteristics. The maximum and minimum values of gate-source voltage will be (3 V + 1 V) = 4 V on positive peaks of the signal and (3 V − 1 V) = 2 V on negative peaks of the signal. The maximum and minimum values of output voltage are, respectively, 14.7 V and 9.2 V. Hence, the output voltage swing = (14.7 V − 9.2 V) = 5.5 V peak-peak (c) Voltage gain = changein drain source voltage 5.5 = 2.75 changein gatesource voltage 2 © John Bird Published by Taylor and Francis 111 8. An amplifier has a current gain of 40 and a voltage gain of 30. Determine the power gain. Power gain = voltage gain current gain = 30 40 = 1200 9. The output characteristics of a transistor in common-emitter mode configuration can be regarded as straight lines connecting the following points. IB = 20 A 50 A 80 A VCE (v) 1.0 8.0 1.0 8.0 1.0 8.0 IC (mA) 1.2 1.4 3.4 4.2 6.1 8.1 Plot the characteristics and superimpose the load line for a 1 k load, given that the supply voltage is 9 V and the d.c. base bias is 50 A. The signal input resistance is 800 . When a peak input current of 30 A varies sinusoidally about a mean bias of 50 A, determine (a) the quiescent values of collector voltage and current (VCQ and ICQ), (b) the output voltage swing, (c) the voltage gain, (d) the dynamic current gain, and (e) the power gain. © John Bird Published by Taylor and Francis 112 The characteristics are shown plotted below. The two ends of the load line will correspond to VCC, the 9 V supply, on the collector-emitter voltage axis and 9 V/1 kΩ = 9 mA on the collector current axis. (a) The operating point (or quiescent point), X, is located from the point of intersection of the IB = 50 μA characteristic and the load line. Having located the operating point we can read off the quiescent values, i.e. the no-signal values, of collector emitter voltage (VCQ) and collector current (ICQ). Hence, VCQ = 5.2 V and ICQ = 3.7 mA (b) The maximum and minimum values of collector-emitter voltage may be determined by locating the appropriate intercept points on the characteristic. Note that the maximum and minimum values of base current will be (50 μA + 30 μA) = 80 μA on positive peaks of the signal and (50 μA − 30 μA) = 20 μA on negative peaks of the signal. The maximum and minimum values of VCE are, respectively, 7.6 V and 2.5 V. Hence, the output voltage swing = (7.6 V − 2.5 V) = 5.1 V peak-peak (c) Voltage gain = change in collector voltage change in base voltage © John Bird Published by Taylor and Francis 113 The change in collector voltage = 5.1 V from part (b). The input voltage swing is given by: i b R i , where i b is the base current swing = (80 – 20) = 60 μA and R i is the input resistance = 800 . Hence, input voltage swing = 60 106 800 = 48 mV = change in base voltage. Thus, voltage gain = changein collector voltage VC 5.1 = 106 changein base voltage VB 48 10 3 (d) Dynamic current gain, hfe I C I B From the characteristic, the output current swing, i.e. the change in collector current, I C = 6.5 – 1.3 = 5.2 mA peak to peak. The input base current swing, the change in base current, I B = 60 μA. IC 5.2 103 Hence, the dynamic current gain, hfe = 87 I B 60 106 (e) For a resistive load, the power gain is given by: power gain = voltage gain current gain = 106 87 = 9222 © John Bird Published by Taylor and Francis 114