55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Homework Assignment 11
Question 1 (Short Takes) Two points each unless otherwise indicated.
1. What is the 3-dB bandwidth of the amplifier shown below if ππ = 2.5K, ππ = 100K, ππ = 40 mS, and πΆπΏ = 1 nF?
(a) 65.25 kHz (b) 10 kHz (c) 1.59 kHz (d) 10.4 kHz
Answer: πΆπΏ sees an equivalent resistance ππ = 100K. (If one turns off ππΌ, πππ£π = 0, and the current source is effectively removed from the circuit.) The time-constant is π = π πΆ = 100 πs. The bandwidth is 1 (2ππ) = 1.59 kHzβ , so the answer is (c).
2. What is the time constant of the circuit?
Answer: The resistance the capacitor sees is π ππ» = 10K||10K = 5K, so the time constant is π = π ππ»πΆπΏ = (5 Γ 103)(1 Γ 10β6) = 5 ms.
3. Many BJT datasheets do not list π½ explicitly, but list an equivalent h-parameter instead. What is this parameter?
Answer: πππ
4. 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?
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) 105β = 20 Hz.
5. An amplifier has a differential gain of -50,000 and a common-mode gain of 2. What is the common-mode rejection ratio?
(a) β87.96 dB (b) 44 dB (c) -44 dB (d) 87.96 dB Answer: CMMR = 20 log10|π΄π π΄πβ | = 20 log10|50 Γ 103 2β | = 87.96 dB, so the answer is (d).
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
6. For the following circuit, what is the numerical value for the two-port y-parameter π¦12?
Answer: Short port 1, then apply a voltage π£2 and determine the current that flows into port 1, and apply the definition above. Then π¦12 = β1 300Kβ
7. For the following circuit, what is the numerical value for the two-port h-parameter β21? The definition of the h-parameters are shown.(2 points)
π£1 = β11π1 + β12π£2 π2 = β21π1 + β22π£2
Answer: To determine β21we set π£2 = 0 by shorting port 2, then use the second equation above to find β21 = π2 π1β . However, π2 = βπ1, so that β21 = β1.
8. Consider the following circuit, which is the power output stage of an amplifier. (a) What is the name of the shaded sub-circuit around π1? (b) Write down one sentence/phrase that describes the purpose of the sub-circuit and constant current source.
Answers: (a) ππ΅πΈ multiplier (b) Reduction of cross-over distortion
9. Explain what the difference is between the units βmsβ and βmSβ.
Answer: βmsβ is millisecond and βmSβ is milli-Siemens.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
10. 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) 105β = 50 Hz. A unity follower will have a bandwidth of 5 MHz.
11. A constant gain-bandwidth amplifier has a 3-dB bandwidth of 1 MHz. By how much (πs) does it delay a 250-kHz sinusoidal signal? Answer: The amplifierβs phase is ΞΈ = β tanβ1(f 1 Γ 106β ) and at 250 kHz this is β14Β°. Further, the period of a 250-kHz sine wave is 4 Β΅s and the delay is therefore:
Ξπ‘ =14
360Γ 4 πs = 0.156 πs
12. 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? 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) 105β = 20 Hz
13. What is the impedance of a 0.1 πF capacitor at π = 1 kHz?
(a) β βπ1.6 Γ 103 Ξ©
(b) π10 Γ 103 Ξ© (c) β +π1.6 Γ 103 Ξ© (d) β1.6 Γ 103 Ξ© (e) 10K
Answer: ππΆ = βπ (2πππΆ)β = βπ (2π Γ 1 Γ 103 Γ 0.1 Γ 10β6) =β β π1.592K. Thus, (a) is the answer.
14. A MOSFET has rated power of 50 W at an ambient temperature ππ΄ = 25oC and a maximum specified junction temperature of 105oC. What is the thermal resistance between the device case and the junction? Answer: ππππ£βπππ π = 1.6 Β°C/W
15. A power MOSFET has rated power of 1,250 W at an ambient temperature ππ΄ = 25oC and a maximum specified junction temperature of 175oC. What is the thermal resistance between the junction and device case? Answer: ππππ£βπππ π = (175 β 25) 1250β = 0.12 Β°C/W
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
16. What is the maximum theoretical efficiency for a class-B amplifier?
Answer: 78%
17. What is the purpose of π 3 in the circuit below, and what should the value be to be effective?
Answer: This compensates for the op-ampβs input bias current. The value should be π 1||π 2.
18. Assume that your SPICE simulation software (such as Micro-Cap SPICE) do not have a photodiode βpartβ. Explain in 1β2 sentences how you can nevertheless simulate a photodiode.
Answer One can model a photodiode with a current source.
Question 2 A constant GBP op-amp has an open loop gain of 100 dB, and a unity gain bandwidth of 5 MHz. The op-amp is used in a non-inverting configuration with a gain of 40 dB. By how much (ππ ) does the amplifier delay a 10 kHz sine wave? (6 points)
Solution The GBP is 5 MHz, and 40 dB is equivalent to a voltage gain of 100, so the bandwidth of the feedback amplifier is 5 Γ 106 100β = 50 kHz. The phase at 10-kHz is
π = β tanβ1 10 Γ 103
50 Γ 103 = β11.31Β°
The period of a 10-kHz sine wave is 100 πs so that β11.31Β° corresponds to a delay of
Ξπ‘ =11.3360
Γ 100 πs = 3.14 πs
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 3 Determine the h-parameters for the circuit below. (16 points)
Solution
We obtain two h-parameters by setting π£2 = 0, namely
β11 =π£1π1π£2=0
β21 =π2π1π£2=0
With π£2 = 0, the input voltage is equal to the voltage across the 20K resistor, and
π£1 = (20K)π1 β β11 =(20K)π1
π1= 20K
Use KCL to find π2:
π1 + 50π1 + π2 = 0 β π2 = β51π1 and β21 =β51π1π1
= β51 A/A
We obtain the other two h-parameters by setting π1 = 0, namely
β12 =π£1π£2π1=0
β22 =π2π£2π1=0
With π1 = 0, the current through the current source is 0, so that
π2 =π£2
200K β β22 =
π2π£2
=1
200K= 5 Γ 10β6 S
With π1 = 0, the current through the current source is 0, and there is no voltage drop across the 20K resistor, so π£1 = π£2, and
β12 =π£1π£2π1=0
= +1 V/V
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 4 Determine the y-parameters for the circuit shown in (a). The model that defines the y-parameters is in (b). (16 points)
(a) (b) Solution Determine parameters π¦11 and π¦21 by setting π£2 = 0, resulting in the circuit shown. KCL at node π₯ gives
βπ1 +π£π₯2K
+π£π₯4K
+ 2π1 = 0 Further, π£π₯ = π£1 β (8K)π1 so that
βπ1 +π£1 β (8K)π1
2K+π£1 β (8K)π1
4K+ 2π1 = 0
Solving yields π¦11 = π1 π£1β = 0.15 mS. KCL at node 2 gives π2 + (2K)π1 +
π£π₯4K
= 0
π2 + 2π1 +π£1 β (8K)π1
4K= 0
Solving yields π¦21 = π2 π£1 = β0.25 mSβ .
Determine parameters π¦21 and π¦22 by setting π£1 = 0, resulting in the circuit shown. KCL at node π₯ gives
βπ1 +π£π₯2K
+π£π₯ β π£2
4K+ 2π1 = 0
Substituting π£π₯ = β(8K)π1 gives
βπ1 β 4π1 β 2π1 βπ£24K
+ 2π1 = 0
Simplifying gives π¦12 = π1 π£2β = β1 20K = 50 πSβ .
KCL at the node at the bottom of the 2K resistor gives
π2 =π£π₯2K
β π1
Further, π£π₯ = β(8K)π1 so that
π2 =β(8K)π1
2Kβ π1
Substituting π1 = π¦12π£2 and simplifying yields π¦21 = β5π¦12 = 250 πS.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 5 Determine the y-parameters for the circuit shown in (a). The model that defines the y-parameters is in (b). (12 points)
(a) (b)
Solution
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 6 You can assume that for all the transistors in the circuit below, π½ is large. Show that πΌ1 = 0.4 mA. Let π£1 = π£2 = 0 V and then determine VA, VB, VC, VD , ππΈ , and π£π . Assume
ππ΅πΈ(ππ) = 0.7 V for all the transistors. Further, note that π 3 is small: for the purposes on this hand-analysis, ignore its effect. That is, assume π 3 = 0. (14 Points)
Solution
Throughout we will assume ππ΅πΈ(ππ) = 0.7 V for all the transistors. Further, since π½ is large, we will ignore base currents and take πΌπΈ = πΌπΆ for all the transistors.
πΌ1 = (20 β 0.7) (48.25 kΞ©) = 0.4 mAβ , which is mirrored as πΌπ
πΌπ΅(π3) is very small, so πΌπΆ1 = πΌπ2 = 0.2 mA
Thus, ππ΄ = 10 β 0.2 Γ 20 = 6 V
ππ΅ = 6 β 1.4 = 4.6 V
Thus, πΌπΈ(π4) = 4.6 11.5 = 0.4 mAβ , which is practically the same current that flows through π 5
ππΆ = 10 β 0.4 Γ 5 = 8 V , and ππ· = 7.3 V
We are ignoring the effects of π 3, so that IC(Q9) is the mirror of πΌ1. That is πΌπΆ(π5) = πΌπΈ(π5) =0.4 mA.
πΌπΈ(π5) = 0.4 mA , so ππΈ = 7.3 β (0.4 mA)(16.5 kΞ©) = 0.7 V
Finally, π£π = ππΈ β 0.7 = 0 V
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 7 (diodes, load line) Consider the circuit below. Assume πππ = 3.5 V, and π =180 Ξ©. Also shown, are the LEDβs voltage-current characteristics. Draw the circuitβs dc load line on the characteristics and find πΌπ· and ππ· (6 points)
Solution. On the voltage axis, mark the supply voltage:3.5 V. On the current axis, mark the maximum current that can flow through the resistor: πΌ = 3.5 180β = 19.4 mA. Connect the two points to get the dc load line. The dc load line intersects the diode V-I curve at around πΌπ· β 6 mA and ππ· β 2.8 V.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 8
π π = 30K π π = 10K πΆπ = 10 πF πΆπ = 50 pF
(a) Determine the open-circuit constant associated with πΆπ, and short-circuit time constant associated with πΆπ (4 points)
(b) Determine the corner frequencies and magnitude of the transfer function π(π ) = ππ(π ) πΌπ(π )β at midband. (2 points)
(c) Sketch the Bode magnitude plot and Bode phase plot of the voltage transfer function. Be sure to add the proper units. (4 points)
Solution
Part (a) ππ = (π π + π π)πΆπ = 0.4 s ππ = (π πβπ π)πΆπ = 0.375 πs
Part (b)
ππΏ =1
2πππ= 0.398 Hz
ππ» =1
2πππ= 424 kHz
At midband, πΆπ β short, πΆπ β open. Thus ππ = πΌπ(π πβπ π). The transfer function is
π(π ) =ππ(π )πΌπ(π )
= (π πβπ π) = 7.5 kΞ©
Part (c)
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 9 Sketch the Bode magnitude plots of the following functions:
π(π ) =β10π
(π + 20)(π + 2000)
π(π ) =10(π + 10)(π + 100)
Solution
First transfer function
π(π ) =β10π
(π + 20)(π + 2000) = β10
(2)(2000) βπ
π 20 + 1 π 2000 + 1
= 250 Γ 10β6π
π 20 + 1 π 2000 + 1
The transfer function has zeros at π = 0, π = β rad/s, and has poles at π = 20, π = 2,000 rad/s
The zero at the origin (π -term in numerator) is 0 dB at 1 rad/s, and the constant term gives β72 dB and has a slope of 20 dB/decade. The pole at π = 20 is log10(20 1β ) = 1.3 decades higher so at the pole the Bode plot has value β72 + 20 Γ 1.3 = β46 dB β‘ 5 Γ 10β3.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Second transfer function
π(π ) =10(π + 10)(π + 100) =
(10)(10)(100) β
π 10 + 1
π 100 + 1
= π 10 + 1
π 100 + 1
The transfer function has a zero at π = 10 and a pole at π = 100. At π = 0, |π(π )| = 1 β‘ 0 dB
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 10
(a) Sketch the Bode magnitude plot of the following function. (6 points)
π(π ) =10π
(π + 10)(π + 500)
(b) What is the midband gain? (2 points) (c) Is there a dominant pole? If so, what is the approximate pole frequency? (2 points) (d) What is the low β3 dB frequency? (2 points)
Solution
Part (a)
π(π ) =10π
(π + 10)(π + 500) =10
(10)(500) βπ 1
π 10 + 1 π 500 + 1
= 0.002π 1
π 10 + 1 π 500 + 1
The transfer junction has zero at the origin, a pole at π = 10 and a pole at π = 500. The zero at the origin (π -term in numerator) is 0 dB at 1 rad/s, and the constant term gives 20 log10 0.002 =β54 dB, and has a slope of 20 dB/decade. The pole at π = 10 is 1 decade higher, so at this pole the Bode plot has value β54 + 20 = β34 dB β‘ 0.02
Part (b) Midband gain = 0.02
Part (c) π = 500 rad/s
Part (d) π = 10 rad/s
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 11 For the common-emitter amplifier below, the transistor parameters are π½ =100,ππ΅πΈ(ON) = 0.7 V, and ππ΄ = β.
ππΆπΆ = 12 π π πΆ = 1 kΞ© π 1 = 10 kΞ© π 2 = 1.5 kΞ© π πΈ = 0.1 kΞ© π π = 0.5 kΞ© πΆπΆ = 0.1 πF
(a) Determine πΌπΆπ , ππ and ππ (8 points) (b) Calculate the lower corner frequency (6 points) (c) Determine the midband voltage gain (6 points) (d) Sketch the Bode plot of the voltage gain magnitude (6 points)
Solution
Part (a) A dc analysis (not shown here) reveals that πΌπΆπ = 7.6 mA, and ππ = 40πΌπΆ =0.304 A Vβ . Thus, ππ = π½ ππβ = 0.329 kΞ©.
Part (b)
π π = (π 1βπ 2)β[ππ + (1 + π½)π πΈ] = 1.16 kΞ© Further
π = (π π + π π)πΆπΆ = 166 πs
β ππΏ =1
2ππ= 959 Hz
Part (c) The midband voltage gain from the base to the collector is
βπ πΆ
1ππ
+ π πΈ=
1 Γ 103
3.43 + 100= β9.68
A quick approximation, also acceptable here is π΄π£ β βπ πΆ π πΈβ = β10.
The source resistor π π forms a voltage divider with π π so that the overall midband voltage gain is
π΄π =π£ππ£π
= βπ π
π π + π π9.68 = β6.76
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Part (d)
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 12
For the three-stage amplifier above, make an estimate of the bandwidth and the rise time. (6 points)
Hint: note that the amplifiers do not load each other.
Solution
The circuit consists of three cascaded voltage amplifiers that do not load each other, and each amplifier has a single pole. The time-constants associated with the amplifiers are
π1 = π 1πΆ1 = 220 πs π2 = π 2πΆ2 = 330 πs π3 = π 3πΆ3 = 470 πs
The rise times of the amplifiers are
π‘π1 = 480 πs π‘π2 = 726 πs π‘π3 = 1.03 ms
An estimate of the rise time for the composite amplifiers is
π‘π = π‘π12 + π‘π22 + π‘π32 = 1.35 ms
An estimate for the β3 dB bandwidth of the composite amplifier is
BW = 0.35π‘π
= 260 Hz
A SPICE simulation gave π‘π = 1.47 ms and the β3 dB as 231 Hz respectively. The estimates are within about 10% of the SPICE simulation.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 13 For the circuit below π½ = 120,ππ΅πΈ(ππ) = 0.7 V, and ππ΄ = β
(a) Design a bias-stable circuit such that πΌπΆπ = 1 mA. (b) Determine the output resistance π π . (c) What is the lower 3 dB corner frequency?
Solution
Part (a) The Thevenin equivalent circuit for the base-emitter bias circuit is
π ππ» = π 1||π 2
πππ» = 12π 2
π 1 + π 2
ππ΅πΈ = 0.7 V πΌπΆπ = 1 mA π½ = 120
πΌπ΅ must be πΌπΆπ π½ = 8.33 πAβ , ππ πΈ β πΌπΆππ πΈ = 4 V, so that ππ΅ must be 4.7 V to meet the design specifications. The figure below summarizes the requirements.
π ππ» = π 1||π 2 =π 1π 2π 1 + π 2
πππ» = 12π 2
π 1 + π 2
There are many combinations of π 1,π 2 that will meet these requirements. βDesignβ implies choices by the designer, so different engineers will design different circuits to meet the specifications. Small values for the bias the resistors will work, but will result in low input resistance. Generally, BJT bias resistor values are 20 β 200 kΞ©. Choose π ππ» = 47 kΞ©. Then
πππ» = 4.7 + (8.33 Γ 10β6)(47 Γ 103) = 5.1 V
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
π ππ» =π 1π 2π 1 + π 2
= 47 Γ 103
5.1 = 12π 2
π 1 + π 2
Solving yields π 1 = 110.6 kΞ© and π 2 = 81.74 kΞ©. The closest standard values are π 1 =110 kΞ© and π 2 = 82 kΞ©.
Part (b) Using BJT impedance scaling
π π = π πΈ ππ
1 + π½
ππ =π½ππ
=120
40πΌπΆπ= 3 kΞ©
β π π β 25 Ξ©
Part (c) Use the time-constant technique to determine the lower corner frequency. The time constant associated with the coupling capacitor is
π = πΆπΆ2(π π + π πΏ) = 2 Γ 10β6(25 + 4 Γ 103) = 8.1 ms
The lower corner frequency is then
ππΏ =1
2ππ= 19.8 Hz
The table below compares the design values with a SPICE simulation that started with the 2N2222 BJT with the Early voltage and π½ replaced with πππ = 1G and ππ = 120.
Parameter Calculated/Design SPICE πΌπΆπ 1 mA 1.02 mA ππ΅πΈ 0.7 V 0.642 V πΌπ΅π 8.33 πA 12.5 Β΅A ππ πΈ 4 V 3.89 V ππΏ 19.8 Hz 18.8 Hz
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 14 For the amplifier below, π½ = 100, and a dc analysis reveals that πΌπΆπ = 0.838 mA. Estimate the 3-dB bandwidth. Assume that πΆπ2 = 0.1 πF. (15 points)
Answer
For the transistor ππ = π½ ππ = 100 40πΌπ β 3Kββ . Using BJT impedance scaling, we estimate the resistance looking into the emitter as
π π =π π||π π΅ + ππ
(1 + π½)β 35 Ξ©
The time constant associated with πΆπΆ2 is
π = πΆπ2(π π||π πΈ + π πΏ)
Since π π is so much smaller than π πΈ and π πΏ we can write
π β πΆπΆ2π πΏ = 1 ms
The 3-dB bandwidth is π΅ = 1 2ππ = 159 Hzβ .
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 15 Consider the amplifier below. In the circuit, πΌπβ represents a photodetector. Ignore the MOFETβs capacitances. (a) Classify the type of frequency response (high-pass or low-pass) the circuit has. (2 points) (b) Determine the 3-dB frequency. (5 points)
Answer
Part (a) The amplifier has a high-pass response.
Part (b) The circuit time constant is π = πΆπΆ1π πβ + π 1||π 2 = (2.2 Γ 10β9)(60 Γ 103) =132 ππ . The 3-dB frequency is the π΅ = 1 2ππ = 1.21 kHz.β
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 16
The maximum transistor power is PD,max = 25 W. Design the circuit (i.e., determine RL and RB) such that maximum power is delivered to the load. (8-points)
Solution
We ignore transistor saturation. This is reasonable, since the saturation voltage is ~ 0.2V, while the supply voltage is 24 V. This is a class-A amplifier, and for maximum power, bias it such that VCQ = VCC/2 = 12 V. This allows maximum voltage swing and maximum power to the load. We need to find an RL to maximize power, but not exceed the 25 W power rating of the transistor. Drawing the load line is not required, but helps, as I pointed out many times in class.
The quiescent point is where the transistor dissipates the most power, so that
ππ·,πππ₯ = ππΆππΌπΆπ = 25 W
and
πΌπΆπ =2512
= 2.08 A, and π πΏ =12πΌπΆπ
= 5.76 Ξ©
π π΅ =24 β 0.72.08/80
= 896 Ξ©
Note: load power is ππ·,πππ₯ = ππ/β2 Γ πΌπ/β2 = 12/β2 Γ 2.08/β2 = 12.48 W and the supply power is 2.08Γ24 = 49.92 W, and π = 12.48/49.92 = 0.25, as expected.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 17 Q3, Q4, RP2, RP4 below form a short circuit protection network for the amplifier. If RP3 = RP4 = 0.15 Ξ©, what is the maximum io in case there is a short. That is, when RL = 0? (3 points)
Solution
Assume Q3, Q4 turn on when VBE = 0.5 V (0.7 V, 0.65 V, etc. would also be acceptable). This will occur when io = 0.5/0.15 = 3.3 A, which will starve Q1, Q2 from further base current. Thus, the maximum io is 3.3 A.
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 18 (BJT impedance scaling) In the amplifier below, assume ππ΅πΈ(ππ) = 0.6 V for all the transistors. (a) Show that the collector current of π3 is approximately 1.6 πA. (2 points) (b) Estimate the output resistance. (8 points) (c) Estimate the voltage gain ππ ππ β . (2 points) (d) Find the quiescent, dc voltage at the output. (3 points)
Hint: view the transistors as a composite transistor and use BJT impeadance scaling.
Solution
Part (a) The collector current of π3 is about (10 mA) (π½1π½2)β = 1.6 πA. π3β²π base current is 0.1 πA.
Part (b) We can view the three transistors as a single BJT with beta π½π = 100 Γ 63 Γ 16 =100,800. For this composite transistor πππ = 40πΌπ = (40)(10 mA) = 0.4 S, and πππ = π½π πππ = 100,800 0.4β = 252Kβ .
Use BJT scaling to estimate π π:
π π =10K + πππ
1 + π½π=
10K + 252K100,801
= 2. 5 Ξ©
Part (b) The arrangement is that of a voltage follower, so the gain is essentially 1.
Part (c) The base current of π3 is 0.1 πA so the base voltage of π1 is β1 mV β 0. The dc output voltage is then β3ππ΅πΈ = β3 Γ 0.6 = β1.8 V
π½ = 63
π½ = 100
π½ = 16
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55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Question 19 (Frequency Response) The transistor in the amplifier shown has π½ = 350 and ππ΅πΈ(ππ) = 0.65 V. Ignore the BJT capacitances. (a) Make reasonable assumptions and show that πΌπΆπ β
1 mA (3 points) (b) Show that π π β 13.7K (5 points) (c) Estimate the lower 3-dB frequency if πΆπΆ = 1 πF
(3 points)
Solution
Part (a) Since π½ is large, ignore πΌπ΅π so that ππ΅ = (9)(27 (100 + 27)β ) = 1.9V. Since ππ΅πΈ(ππ) = 0.65 V, then ππ πΈ = 1.9 β 0.64 = 1.25 V. Consequently, πΌπΆπ β πΌπΈ = 1.25 1.3K = 0.962 mA β 1 mAβ .
Part (b) ππ = π½ ππ = 350 40πΌπΆπ =ββ 8.75K. Using BJT scaling,
π π = 65K18Kππ + (1 + π½)(1.3K) = 13.68K
Part (c)
π3ππ΅ =1
2ππ ππΆπ=
1(2π)(13.68K)(1 Γ 10β6) = 11.64 Hz
For comparison, SPICE gives ππ = 8.15K, π π = 13.59K, π3ππ΅ = 11.8 Hz
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