55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Homework Assignment 10
Question 1 (Short Takes) Two points each unless otherwise indicated.
1.
What is the 3-dB bandwidth of the amplifier shown below if 𝑟𝜋 = 2.5K, 𝑟𝑜 = 100K,
𝑔𝑚 = 40 mS, and 𝐶𝐿 = 1 nF?
(a)
(b)
(c)
(d)
2.
65.25 kHz
10 kHz
1.59 kHz
10.4 kHz
Answer: 𝐶𝐿 sees an equivalent resistance 𝑟𝑜 = 100K. (If one turns off 𝑉𝐼 , 𝑔𝑚 𝑣𝜋 = 0,
and the current source is effectively removed from the circuit.) The time-constant is
𝜏 = 𝑅𝐶 = 100 𝜇s. The bandwidth is 1⁄(2𝜋𝜏) = 1.59 kHz, so the answer is (c).
A 9-V dc power supply generates 10 W in a resistor. What peak-to-peak amplitude should
an ac source have to generate the same power in the resistor?
(a) 12.73 V
(b) 25.5 V
(c) 18 V
(d) 12.73 V
Answer: The ac source’s effective (or rms) value should also be 9 V. This measn the
peak value should be 9√2 V, so the peak-to-peak value should be 18√2 = 25.5 V, so the
answer is (b).
3.
In the circuit below, what is the maximum current that can flow through 𝑅𝐿 ? Make
reasonable assumptions.
Answer. Assume that for 𝑄2 , 𝑉𝐵𝐸(𝑂𝑁) = 0.7 V. Thus, 𝑄2 will turn on and starve 𝑄1 from
additional base current when the current through 𝑅1 (which is also the current through 𝑅𝐿 )
is 𝐼 = 0.7⁄1.5 = 0.47 A.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
4.
List (do not explain) 4 BJT parameters that determine the SOA.
Answer: maximum power dissipation, maximum allowable collector current, maximum
collector-emitter voltage, a secondary breakdown.
5.
True or false: broadly speaking, BJT technology has superior performance in power
application when compared to the modern MOS technology, which explains why BJTs are
still widely incorporated in power designs.
Answer: False BJTs suffer from second breakdown and thermal runaway, while
MOSFETs don’t
6.
7.
8.
9.
An engineer designs a class-AB amplifier to deliver 1.2 W (sinusoidal) signal power to an
8 Ω resistive load. Ignoring saturation in the output BJTs, what is the required peak-topeak voltage swing across the load?
2 ⁄
Answer: 𝑃 = 𝑉𝑟𝑚𝑠
𝑅, so that 𝑉𝑟𝑚𝑠 = 3.1 V, so that 𝑉𝑝𝑝 = 8.77 V
An engineer designs a class-AB amplifier to deliver 2 W (sinusoidal) signal power to an
8 Ω resistive load. Ignoring saturation in the output BJTs, what is the required peak-topeak voltage swing across the load?
2 ⁄
Answer: 𝑃 = 𝑉𝑟𝑚𝑠
𝑅, so that 𝑉𝑟𝑚𝑠 = 4 V, so that 𝑉𝑝𝑝 = 11.32 V
An engineer designs a MOSFET-based class-AB amplifier to deliver 6.25 W (sinusoidal)
signal power to a 4 Ω resistive load. What is the required peak-to-peak voltage swing
across the load? (2 points)
2 ⁄
Answer: 𝑃 = 𝑉𝑟𝑚𝑠
𝑅, so that 𝑉𝑟𝑚𝑠 = 5 V, so that 𝑉𝑝𝑝 = 14.14 V
An engineer designs a MOSFET-based class-AB amplifier to deliver 6.25 W (sinusoidal)
signal power to a 4 Ω resistive load. What is the required peak-to-peak voltage swing
across the load? (2 points)
(a)
(b)
(c)
(d)
(e)
10.
9.77 V
19.53 V
10 V
14.14 V
7.07 V
2 ⁄
Answer: 𝑃 = 𝑉𝑟𝑚𝑠
𝑅, so that 𝑉𝑟𝑚𝑠 = 5 V, so that 𝑉𝑝𝑝 = 14.14 V, so (d).
True or false: a power MOSFETs’ transconductance 𝑔𝑚 is less subject to changes in
temperature than a power BJT’s 𝛽’s is subject to changes in temperature.
Answer: True
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
11.
An engineer designs a standard (no inductor, transformer) class-A amplifier to deliver
100 mW (sinusoidal) signal power to a resistive load. How much power should the power
supply be able to supply?
12.
Answer: At least 400 mW, because 𝜂 = 𝑃𝑠𝑖𝑔𝑛𝑎𝑙 ⁄𝑃𝑠𝑢𝑝𝑝𝑙𝑦 ≤ 25%
13.
Answer: At least 800 mW, because 𝜂 = 𝑃𝑠𝑖𝑔𝑛𝑎𝑙 ⁄𝑃𝑠𝑢𝑝𝑝𝑙𝑦 ≤ 25%
An engineer designs a standard (no inductor, transformer) class-A amplifier to deliver 200
mW (sinusoidal) signal power to a resistive load. How much power should the power
supply be able to supply?
True or false: for the amplifier below, the harmonic distortion increases as the amplitude
of the input increases.
Answer: False—in fact, the opposite is true.
14.
Write down one phrase/sentence that describes the purpose of the diodes and constant
current source in the amplifier below.
Answer: Reduction of cross-over distortion
15.
True or false: the efficiency of a class-AB amplifier is independent of the amplitude of the
input signal.
Answer: False
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
16.
True or false: for a MOSFET, 𝑔𝑚 decreases with increasing temperature, which explains
why MOSFETs are not prone to thermal runaway.
Answer: True
17.
What is the maximum theoretical efficiency for a class-B amplifier?
Answer: 78%
18.
What does “SOA” in the context of power transistors stand for?
Answer: Safe Operating Area
19.
True or false: in power transistors, the junction temperature can reach as high as 150 oC.
Answer: True.
20.
True or false: everything else being equal, BJTs have an order of magnitude more gain
than FETs.
Answer: True
21.
True or false: if class-A amplifiers are not carefully biased, they will suffer from
crossover distortion.
Answer: False
22.
The small-signal output resistance 𝑟𝑜 of a BJT biased at 𝐼𝐶 = 1 mA is100K. What is 𝑟𝜊
when the transistor is biased at 𝐼𝐶 = 5 mA?
23.
Answer: 𝑟𝜊 is inversely-proportional to 𝐼𝐶 (𝑟𝑜 = 𝑉𝐴 ⁄𝐼𝐶 ) so that 𝑟𝑜 will be 5 × smaller at 5
mA, or 𝑟𝑜′ = 20 K.
24.
Answer: 𝜃𝑑𝑒𝑣−𝑐𝑎𝑠𝑒 = 1.6 °C/W
A MOSFET has rated power of 50 W at an ambient temperature 𝑇𝐴 = 25o C and a
maximum specified junction temperature of 105oC. What is the thermal resistance
between the device case and the junction?
A power MOSFET has rated power of 1,250 W at an ambient temperature 𝑇𝐴 = 25o C and
a maximum specified junction temperature of 175oC. What is the thermal resistance
between the junction and device case?
Answer: 𝜃𝑑𝑒𝑣−𝑐𝑎𝑠𝑒 = (175 − 25)⁄1250 = 0.12 °C/W
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
25.
True or false: the turn-on voltages of Schottky diodes are less than that of Si diodes.
However, their reverse leakage/saturation currents are also higher. (1 point)
Answer: True
26.
True or false: The turn-on voltage of red LEDs is larger than the turn-on voltage of blue
LEDs. (1 point)
Answer: True
27.
Below is a depiction of an n-channel enhancement-mode MOSFET. Annotate the diagram
with a “p” or “n” to show the type of substrate material, and then indicate the body diode.
p-material
Body diode
substrate
28.
29.
A single-pole op-amp has an open-loop gain of 100 dB and a unity-gain bandwidth
frequency of 2 MHz. What is the open-loop bandwidth of the op-amp? (2 points)
Answer. A gain of 100 dB corresponds to 105 and the gain-bandwidth product is 2
MHz. Thus, the open-loop bandwidth is (2 MHz)⁄105 = 20 Hz
A single-pole op-amp has an open-loop gain of 100 dB and a unity-gain bandwidth
frequency 5 MHz. What is the open-loop bandwidth of the amplifier? The amplifier is
used as a voltage follower. What is the bandwidth of the follower?
Answer A gain of 100 dB corresponds to 105 and the gain-bandwidth product is 5 MHz.
Thus, the open-loop bandwidth is (5 MHz)⁄105 = 50 Hz. A unity follower will have a
bandwidth of 5 MHz.
30.
Consider a first-order RC low-pass filter with corner frequency 𝑓 = 1 kHz. What is the
phase shift in a degrees at 15 kHz? (3 points)
Answer: The phase shift at 1 kHz is 45° and increases at 45° / decade. 15 kHz is
log(15⁄1) = 1.18 decades higher than 1 kHz. Thus, the phase shift is 45 + 1.18 ×
45 = 97.9°. However, the phase shift for a first-order RC network must always be less
than 90°, so the answer is “90°”. A more accurate calculation gives the phase shift as
tan−1(15⁄1) = 86.2° .
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
31.
An AAA cell has a no-load voltage of 1.605 V. When a 100 Ω resistor is connected
across its terminals, the voltage drops to 1.595 V. What is the cell’s internal resistance?
a) ≈ 620 mΩ
b) ≈ 10 mΩ
c) Need additional information
32.
Answer: The current flowing through the load resistance is 𝐼𝐿 =
1.595⁄100 = 15.95 mA. The internal resistance is
𝑅𝑂 = Δ𝑉⁄Δ𝐼 = (1.605 − 1.595)⁄(15.95 × 10−3 ) = 0.627 Ω. Thus, (a) is the answer.
What is the magnitude of the current phase angle for a 5.6 𝜇F capacitor and a 50-Ω resistor
in series with a 1.1 kHz, 5 VAC source?
(a) 72.9°
(b) 62.7°
(c) 27.3°
(d) 17.1
Answer: The impedance of the RC circuit is = 𝑅 − 1⁄𝑗2𝜋𝑓𝐶 = 50 − 𝑗25.84 Ω. The
magnitude of the phase angle is |tan−1(−25.84⁄50)| = 27.3°. Thus, (c) is the answer.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 2 (N 5.36) The circuit below is sometimes used and thermometer. Assume that the
two transistors are identical. Writing emitter currents in the form 𝐼𝐸 = 𝐼𝐸0 𝑒 𝑉𝐵𝐸 ⁄𝑉𝑇 derive an
expression for the output voltage as a function of temperature. (10 points)
Solution
and
So that
𝐼𝐸1 = 𝐼𝐸𝑂 𝑒 𝑉𝐵𝐸1 ⁄𝑉𝑇
𝐼𝐸1
10𝐼
𝑉𝐵𝐸1 = 𝑉𝑇 ln � � = 𝑉𝑇 ln �
�
𝐼𝐸0
𝐼𝐸0
𝐼𝐸2 = 𝐼𝐸𝑂 𝑒 𝑉𝐵𝐸2 ⁄𝑉𝑇
𝐼𝐸2
𝐼
𝑉𝐵𝐸2 = 𝑉𝑇 ln � � = 𝑉𝑇 ln � �
𝐼𝐸0
𝐼𝐸0
𝑉𝑂 = 𝑉𝐵𝐸2 − 𝑉𝐵𝐸1
10𝐼
I
� − 𝑉𝑇 ln � �
= 𝑉𝑇 ln �
𝐼𝐸0
𝐼𝐸0
𝑘𝑇
= 𝑉𝑇 ln 10 =
ln 10
𝑒
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 3 Consider a BJT with a rated power of 20 W, and a maximum allowable junction
temperature 𝑇𝑗,max = 175 ℃. The transistor is mounted on a heat sink with parameters
𝜃case−sink = 1 ℃/W, and 𝜃sink−amb = 5 ℃/W. Determine how much power the BJT can safely
dissipate. Assume an ambient temperature of 𝑇𝐴 = 25 ℃.
Solution
The thermal resistance from the device/junction to the case is not given explicitly, so we need to
determine it before proceeding. The BJT is rated at 20 W at 𝑇𝑗,max = 175 ℃, and an ambient
temperature of 𝑇𝐴 = 25 ℃ is assumed. A thermal model and the calculation of 𝜃dev−case is then
𝑇𝑗 = 𝑇𝐴 + 𝑃𝐷 (𝜃dev−case )
175 = 25 + 20(𝜃dev−case )
𝜃dev−case = 7.5 ℃
Now we can determine the maximum allowable power dissipation when the BJT is mounted on
a heat skink with the given parameters. A thermal model for the problem is shown below.
𝑇𝑗 = 𝑇𝐴 + 𝑃𝐷 (𝜃dev−case + 𝜃case−sink + 𝜃sink−amb )
𝑃𝐷,𝑚𝑎𝑥 =
=
𝑇𝑗,max − 𝑇𝐴
(𝜃dev−case + 𝜃case−sink + 𝜃sink−amb )
175 − 25
1 + 5 + 7.5
= 11.1 W
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 4 A power MOSFET has thermal characteristics given below. The device operates
without a heat sink and dissipates 0.2 W. What is the junction temperature if the ambient
temperature is 25 oC? Start by drawing and labeling a thermal model. (5 points)
Solution
𝜃dev−case = 1.75 ℃⁄W ,
𝜃case−amb = 50 ℃⁄W ,
𝑇𝑗,max = 150 ℃
𝑇𝐽 = 25 ℃ + 𝑃𝐷 (𝜃𝑑𝑒𝑣−𝑐𝑎𝑠𝑒 + 𝜃𝑐𝑎𝑠𝑒−𝑎𝑚𝑏 )
= 25 + 0.2(50 + 1.7)
= 35.34 ℃
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 5 (Midterm, 2007) A power MOSFET has thermal characteristics given below and
dissipates 25 W. Design, by specifying the thermal resistance, a heat sink that will ensure the
MOSFET does not overheat. The ambient temperature is 25 oC. Assume one can keep the
thermal resistance between the MOSFET case and heat sink (𝜃case−sink or 𝜃𝐶𝑆 ) below 1 oC/W.
(5 points)
Solution
𝜃juntion−case = 𝜃𝐽𝐶 = 1.75 ℃/W
𝜃case−ambient = 𝜃𝐶𝐴 = 50 ℃/W
𝑇𝑗,𝑚𝑎𝑥 = 150 ℃
\
𝑡𝐻𝐸𝑇𝐴
A thermal model that captures the information is shown below:
𝑇𝑗 = 𝑇𝐴 + 𝑃𝐷 �𝜃𝐽𝐶 + 𝜃𝐶𝑆 + 𝜃𝑆𝐴 �
150 = 25 + 25(1.75 + 1 + 𝜃𝑆𝐴 )
𝜃𝐴 = 2.25 ℃/W
The heat sink’s thermal resistance must be less than this value. Note: a number of students used
θCA in various ways (incorrectly) in their calculations. θCA is not part of the picture, since we
will replace it with a much lower θSA
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 6 A power MOSFET has thermal characteristics given below. The device operates
without a heat sink and dissipates 0.2 W. What is the junction temperature if the ambient
temperature is 25 oC? Start by drawing and labeling a thermal model. (5 points)
Solution
𝜃dev−case = 1.75 ℃⁄W ,
𝜃case−amb = 50 ℃⁄W ,
𝑇𝑗,max = 150 ℃
𝑇𝐽 = 25 ℃ + 𝑃𝐷 (𝜃𝑑𝑒𝑣−𝑐𝑎𝑠𝑒 + 𝜃𝑐𝑎𝑠𝑒−𝑎𝑚𝑏 )
= 25 + 0.2(50 + 1.7)
= 35.34 ℃
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 7 Consider a BJT with a rated power of 115 W at 𝑇case = 25 ℃, and a maximum
allowable junction temperature 𝑇𝑗,max = 200 ℃. The transistor is mounted on a heat sink with
parameters 𝜃case−sink = 1 ℃/W, and 𝜃sink−amb = 4 ℃/W. Determine how much power the
BJT can safely dissipate at an ambient temperature of 𝑇𝐴 = 25 ℃. (12 points)
Solution
The thermal resistance from the device/junction to the case is not given explicitly, so we need to
determine it before proceeding. The BJT is rated at 25 W at 𝑇𝑗,max = 200 ℃ and a thermal
model and the calculation of 𝜃dev−case is then
𝑇𝑗 = 𝑇𝐴 + 𝑃𝐷 (𝜃dev−case )
200 = 25 + 115(𝜃dev−case )
𝜃dev−case = 1.52 ℃⁄W
(4 points)
Now we can determine the maximum allowable power dissipation when the BJT is mounted on a
heat skink with the given parameters. A thermal model for the problem is shown below.
𝑇𝑗 = 𝑇𝐴 + 𝑃𝐷 (𝜃dev−case + 𝜃case−sink + 𝜃sink−amb )
𝑃𝐷,𝑚𝑎𝑥 =
=
𝑇𝑗,max − 𝑇𝐴
(𝜃dev−case + 𝜃case−sink + 𝜃sink−amb )
200 − 25
1.52 + 1 + 4
= 26.8W
(8 points)
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 8 (Heat sink, M 10.3) Two transistors each with 𝜃𝐽𝐶 = 1.4 ℃⁄W, 𝜃𝐶𝑆 = 0.3 ℃⁄W,
and a dissipation of 12 W, share the same heat sink. If the junction temperatures are 200 ℃, and
the ambient temperature is 10 ℃, use a thermal equivalent circuit to find the thermal resistance
of the heat sink. (10 points)
Solution
Shown is a thermal model for the problem. Form this diagram
10 + 24𝜃𝑆𝐴 + 12(0.3 + 1.4) = 200
Solving yields 𝜃𝑆𝐴 = 7.07 ℃⁄W.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 9
For the transistor 𝐼𝐶,max = 200 mA, and 𝑉𝑐𝑒(sus) = 50 V
(a) Determine the minimum transistor power rating so that the
transistor is always inside the SOA (6 points)
(b) What is the maximum average power this amplifier can
supply to the load with a sinusoidal input voltage? (3 points)
𝑅𝐸 = 200 Ω,
𝑉𝐴 = 200,
𝑉𝐶𝐶 = 10 V, β = 150
Solution
Part (a) The diagram is the dc load line with the 𝐼𝐶 and 𝑉𝑐𝑒 limits indicated. Maximum
transistor power dissipation occurs in the middle of the dc load line at 𝑣𝑐𝑒 = 10 V. At this
voltage 𝐼𝐶 = 50 mA, and the transistor power dissipation at this point is 500 mW. Thus, the
transistor should be rated to be able to dissipate at least 500 mW.
The SOA is the unshaded part
Part (b) From the Q-point the transistor can swing 𝑉𝑃 = 10 V without clipping against the
2 ⁄
supply rails. This is equivalent to 7.07 V rms. The power delivered is 𝑉𝑟𝑚𝑠
𝑅𝐸 =250 mW.
Alternate solution. The power supply supplies 𝐼𝐶𝑄 × (2𝑉𝑐𝑐 ) = 0.05 × 20 = 1 W. This is a
class-A amplifier for which the maximum efficiency is 25%, so the power to the load is 0.25 ×
1 = 0.25 W
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 10 The maximum current, voltage, and power ratings for a power MOSFET are 4 A,
40 V, and 30 W, respectively.
(a) Sketch and label the SOA for the MOSFET using linear voltage and current scales (4 points)
(b) For the amplifier above, determine 𝑅𝐷 and sketch the load line that produces maximum
power dissipation in the transistor for 𝑉𝐷𝐷 = 24 V (3 points)
(c) Determine the maximum possible drain current for 𝑉𝐷𝐷 = 24 V (3 points)
(d) Repeat (b), but now for 𝑉𝐷𝐷 = 40 V (6 points)
Solution
Part (a), (b) and (c) The SOA is shown below along with a load line that gets it the closest to
the maximum power ratings, yet still stay in the SOA. Clearly 𝑅𝐷 = 24⁄4 = 6 Ω. The
maximum drain current is 4 A.
SOA a load line for Part (a)–(c)
SOA and load line for Part (d)
Part (d) In this instance, the maximum power limit hyperbola determines the load line.
Maximum power dissipation occurs at 20 V. At 20 V, 𝐼 = 𝑃max ⁄20 = 1.5 A. By similar
triangles, the load line intersects the 𝐼𝐷𝑆 axis at 3 A.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 11 (Neaman TYU 8.1)
𝑅𝐷 = 20 Ω,
𝑉𝐷𝐷 = 24
Determine the maximum power ratings for the transistor. (10 points)
Solution
The maximum current that can flow is 𝑉𝐷𝐷 ⁄𝑅𝐷 = 24⁄20 = 1.2 A. This is also the maximum
drain current that the transistor must be able to handle. Thus, 𝐼𝐷(max) = 1.2 A
The maximum voltage across the transistor occurs when 𝐼𝐷 = 0 and this voltage is 24 V. Thus,
𝑉𝐷𝑆(max) = 24 𝑉.
For a given 𝐼𝐷 , the transistor will dissipate
𝑃𝐷 = 𝐼𝐷 𝑉𝐷𝑆 = 𝐼𝐷 (𝑉𝐷𝐷 − 𝐼𝐷 𝑅𝐷 )
= (𝐼𝐷 )2 𝑅𝐷 + 𝐼𝐷 𝑉𝐷𝐷
= −20(𝐼𝐷 )2 + 24𝐼𝐷
This reaches its maximum where the derivative is zero. That is, where
𝑑𝑃𝐷
= −40𝐼𝐷 + 24 = 0
𝑑𝐼𝐷
⇒ where 𝐼𝐷 = 0.6 A
The power dissipation at this 𝐼𝐷 is
𝑃𝐷(max) = 𝐼𝐷 𝑉𝐷𝑆 = 0.6(24 − 0.6 × 20) = 7.2 W
Note, in general maximum power dissipation will occur when the transistor is operating at
𝑉𝐷𝑆 = 𝑉𝐶𝐶 ⁄2 and 𝐼𝐷 = 𝐼𝐷(max) ⁄2. Thus, one could directly write
𝑃𝐷(max) =
24 1.2
×
= 7.2 W
2
2
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 12
For the circuit above, make reasonable assumptions and then
(a) Determine the input resistance, without taking 𝑅4 and 𝑅3 into consideration. That is,
determine 𝑅𝑖 as shown in the figure (5 points)
(b) Determine the voltage gain 𝑣𝑜1 ⁄𝑣𝑠 (4 points)
Solution
Part (a). The simplest approach is to recognize that Q1, and Q2, form a complementary
Darlington pair, and treat it as a pnp transistor with gain 𝛽𝑐 = 5,000. 𝑅2 is the emitter resistor
and 𝑅1 is the collector resistor of the Darlington pair.
Then, using BJT impedance scaling, we can quickly estimate 𝑅𝑖 ≅ 𝛽𝑐 𝑅2 = 2.3 MΩ. A more
accurate method is to take 𝑟𝜋 of the Darlington pair into account.
and
𝑟𝜋𝑐
𝛽𝑐 𝑉𝑇 5,000 × 26 × 10−3
=
=
= 186 kΩ
𝐼𝐶𝑐
0.7 × 10−3
𝑅𝑖 = 𝑟𝜋𝑐 + (1 + 𝛽𝑐 )𝑅2 = 186 kΩ + (5,001)× 470 = 2.54 MΩ
Part(b). Again treat the transistor as single pnp transistor. 𝑅2 is the emitter resistor, and 𝑅1 is
the collector resistor. The gain of the amplifier is then
𝐴𝑣 =
𝑣𝑜
𝑅1
≅−
= 10
𝑣𝑠
𝑅2
17
55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 13 (Heath sink, own) A certain
manufacturer of a power transistor does not provide
the thermal resistance between the device junction
and the case (𝜃𝐽𝐶 ) explicitly. Rather, it provided
the power derating curve shown.
Use the plot to find 𝜃𝐽𝐶 . Be sure to supply the
proper units. (4 points)
Solution
If 𝑚 is the slope of the derating curve, then 𝜃𝐽𝐶 = − 1⁄𝑚. Thus,
𝜃𝐽𝐶 = −
175 − 25
≈ 1.6 ℃⁄W
20 − 115
18
55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 14 (Heath sink, own) The graph shown
is the power derating curve for a power transistor,
that the manufacture lists as a 250 W at a case
temperature of 25℃ . Use the graph to determine
the thermal resistance between the device junction
and the case (𝜃𝐽𝐶 ) explicitly. Be sure to supply the
proper units. (6 points)
Solution
If 𝑚 is the slope of the derating curve, then 𝜃𝐽𝐶 = − 1⁄𝑚. Using 𝑇𝐶 = 200℃ and 𝑇𝐶 = 40℃ :
𝜃𝐽𝐶 = −
40℃ − 200℃
= 0.711 ℃⁄W
(0.9)(250 W) − (0)(250 W)
19