homework assignment 11 - university of...

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 𝑟𝜋 = 2.5K, 𝑟𝑜 = 100K, 𝑔𝑚 = 40 mS, and 𝐶𝐿 = 1 nF?

(a) 65.25 kHz (b) 10 kHz (c) 1.59 kHz (d) 10.4 kHz

Answer: 𝐶𝐿 sees an equivalent resistance 𝑟𝑜 = 100K. (If one turns off 𝑉𝐼, 𝑔𝑚𝑣𝜋 = 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).

2. What is the time constant of the circuit?

Answer: The resistance the capacitor sees is 𝑅𝑇𝐻 = 10K||10K = 5K, so the time constant is 𝜏 = 𝑅𝑇𝐻𝐶𝐿 = (5 × 103)(1 × 10−6) = 5 ms.

3. Many BJT datasheets do not list 𝛽 explicitly, but list an equivalent h-parameter instead. What is this parameter?

Answer: 𝒉𝒇𝒆

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) 105⁄ = 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) –87.96 dB (b) 44 dB (c) -44 dB (d) 87.96 dB Answer: CMMR = 20 log10|𝐴𝑑 𝐴𝑐⁄ | = 20 log10|50 × 103 2⁄ | = 87.96 dB, so the answer is (d).

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6. For the following circuit, what is the numerical value for the two-port y-parameter 𝑦12?

Answer: Short port 1, then apply a voltage 𝑣2 and determine the current that flows into port 1, and apply the definition above. Then 𝑦12 = −1 300K⁄

7. For the following circuit, what is the numerical value for the two-port h-parameter ℎ21? The definition of the h-parameters are shown.(2 points)

𝑣1 = ℎ11𝑖1 + ℎ12𝑣2 𝑖2 = ℎ21𝑖1 + ℎ22𝑣2

Answer: To determine ℎ21we set 𝑣2 = 0 by shorting port 2, then use the second equation above to find ℎ21 = 𝑖2 𝑖1⁄ . However, 𝑖2 = −𝑖1, so that ℎ21 = −1.

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 𝑄1? (b) Write down one sentence/phrase that describes the purpose of the sub-circuit and constant current source.

Answers: (a) 𝑉𝐵𝐸 multiplier (b) Reduction of cross-over distortion

9. Explain what the difference is between the units “ms” and “mS”.

Answer: “ms” is millisecond and “mS” is milli-Siemens.

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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) 105⁄ = 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 (𝜇s) does it delay a 250-kHz sinusoidal signal? Answer: The amplifier’s phase is θ = − tan−1(f 1 × 106⁄ ) and at 250 kHz this is −14°. Further, the period of a 250-kHz sine wave is 4 µs and the delay is therefore:

Δ𝑡 =14

360× 4 𝜇s = 0.156 𝜇s

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) 105⁄ = 20 Hz

13. What is the impedance of a 0.1 𝜇F capacitor at 𝑓 = 1 kHz?

(a) ≈ −𝑗1.6 × 103 Ω

(b) 𝑗10 × 103 Ω (c) ≈ +𝑗1.6 × 103 Ω (d) −1.6 × 103 Ω (e) 10K

Answer: 𝑍𝐶 = −𝑗 (2𝜋𝑓𝐶)⁄ = −𝑗 (2𝜋 × 1 × 103 × 0.1 × 10−6) =⁄ − 𝑗1.592K. Thus, (a) is the answer.

14. A MOSFET has rated power of 50 W at an ambient temperature 𝑇𝐴 = 25oC and a maximum specified junction temperature of 105oC. What is the thermal resistance between the device case and the junction? Answer: 𝜃𝑑𝑒𝑣−𝑐𝑎𝑠𝑒 = 1.6 °C/W

15. A power MOSFET has rated power of 1,250 W at an ambient temperature 𝑇𝐴 = 25oC 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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16. What is the maximum theoretical efficiency for a class-B amplifier?

Answer: 78%

17. What is the purpose of 𝑅3 in the circuit below, and what should the value be to be effective?

Answer: This compensates for the op-amp’s input bias current. The value should be 𝑅1||𝑅2.

18. Assume that your SPICE simulation software (such as Micro-Cap SPICE) do not have a photodiode “part”. Explain in 1–2 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 (𝜇𝑠) 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 × 106 100⁄ = 50 kHz. The phase at 10-kHz is

𝜃 = − tan−1 10 × 103

50 × 103 = −11.31°

The period of a 10-kHz sine wave is 100 𝜇s so that −11.31° corresponds to a delay of

Δ𝑡 =11.3360

× 100 𝜇s = 3.14 𝜇s

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Question 3 Determine the h-parameters for the circuit below. (16 points)

Solution

We obtain two h-parameters by setting 𝑣2 = 0, namely

ℎ11 =𝑣1𝑖1𝑣2=0

ℎ21 =𝑖2𝑖1𝑣2=0

With 𝑣2 = 0, the input voltage is equal to the voltage across the 20K resistor, and

𝑣1 = (20K)𝑖1 ⇒ ℎ11 =(20K)𝑖1

𝑖1= 20K

Use KCL to find 𝑖2:

𝑖1 + 50𝑖1 + 𝑖2 = 0 ⇒ 𝑖2 = −51𝑖1 and ℎ21 =−51𝑖1𝑖1

= −51 A/A

We obtain the other two h-parameters by setting 𝑖1 = 0, namely

ℎ12 =𝑣1𝑣2𝑖1=0

ℎ22 =𝑖2𝑣2𝑖1=0

With 𝑖1 = 0, the current through the current source is 0, so that

𝑖2 =𝑣2

200K ⇒ ℎ22 =

𝑖2𝑣2

=1

200K= 5 × 10−6 S

With 𝑖1 = 0, the current through the current source is 0, and there is no voltage drop across the 20K resistor, so 𝑣1 = 𝑣2, and

ℎ12 =𝑣1𝑣2𝑖1=0

= +1 V/V

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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 𝑦11 and 𝑦21 by setting 𝑣2 = 0, resulting in the circuit shown. KCL at node 𝑥 gives

−𝑖1 +𝑣𝑥2K

+𝑣𝑥4K

+ 2𝑖1 = 0 Further, 𝑣𝑥 = 𝑣1 − (8K)𝑖1 so that

−𝑖1 +𝑣1 − (8K)𝑖1

2K+𝑣1 − (8K)𝑖1

4K+ 2𝑖1 = 0

Solving yields 𝑦11 = 𝑖1 𝑣1⁄ = 0.15 mS. KCL at node 2 gives 𝑖2 + (2K)𝑖1 +

𝑣𝑥4K

= 0

𝑖2 + 2𝑖1 +𝑣1 − (8K)𝑖1

4K= 0

Solving yields 𝑦21 = 𝑖2 𝑣1 = −0.25 mS⁄ .

Determine parameters 𝑦21 and 𝑦22 by setting 𝑣1 = 0, resulting in the circuit shown. KCL at node 𝑥 gives

−𝑖1 +𝑣𝑥2K

+𝑣𝑥 − 𝑣2

4K+ 2𝑖1 = 0

Substituting 𝑣𝑥 = −(8K)𝑖1 gives

−𝑖1 − 4𝑖1 − 2𝑖1 −𝑣24K

+ 2𝑖1 = 0

Simplifying gives 𝑦12 = 𝑖1 𝑣2⁄ = −1 20K = 50 𝜇S⁄ .

KCL at the node at the bottom of the 2K resistor gives

𝑖2 =𝑣𝑥2K

− 𝑖1

Further, 𝑣𝑥 = −(8K)𝑖1 so that

𝑖2 =−(8K)𝑖1

2K− 𝑖1

Substituting 𝑖1 = 𝑦12𝑣2 and simplifying yields 𝑦21 = −5𝑦12 = 250 𝜇S.

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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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Question 6 You can assume that for all the transistors in the circuit below, 𝛽 is large. Show that 𝐼1 = 0.4 mA. Let 𝑣1 = 𝑣2 = 0 V and then determine VA, VB, VC, VD , 𝑉𝐸 , and 𝑣𝑂 . Assume

𝑉𝐵𝐸(𝑂𝑁) = 0.7 V for all the transistors. Further, note that 𝑅3 is small: for the purposes on this hand-analysis, ignore its effect. That is, assume 𝑅3 = 0. (14 Points)

Solution

Throughout we will assume 𝑉𝐵𝐸(𝑂𝑁) = 0.7 V for all the transistors. Further, since 𝛽 is large, we will ignore base currents and take 𝐼𝐸 = 𝐼𝐶 for all the transistors.

𝐼1 = (20 − 0.7) (48.25 kΩ) = 0.4 mA⁄ , which is mirrored as 𝐼𝑄

𝐼𝐵(𝑄3) is very small, so 𝐼𝐶1 = 𝐼𝑐2 = 0.2 mA

Thus, 𝑉𝐴 = 10 − 0.2 × 20 = 6 V

𝑉𝐵 = 6 − 1.4 = 4.6 V

Thus, 𝐼𝐸(𝑄4) = 4.6 11.5 = 0.4 mA⁄ , which is practically the same current that flows through 𝑅5

𝑉𝐶 = 10 − 0.4 × 5 = 8 V , and 𝑉𝐷 = 7.3 V

We are ignoring the effects of 𝑅3, so that IC(Q9) is the mirror of 𝐼1. That is 𝐼𝐶(𝑄5) = 𝐼𝐸(𝑄5) =0.4 mA.

𝐼𝐸(𝑄5) = 0.4 mA , so 𝑉𝐸 = 7.3 − (0.4 mA)(16.5 kΩ) = 0.7 V

Finally, 𝑣𝑂 = 𝑉𝐸 − 0.7 = 0 V

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Question 7 (diodes, load line) Consider the circuit below. Assume 𝑉𝑃𝑆 = 3.5 V, and 𝑅 =180 Ω. Also shown, are the LED’s voltage-current characteristics. Draw the circuit’s dc load line on the characteristics and find 𝐼𝐷 and 𝑉𝐷 (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: 𝐼 = 3.5 180⁄ = 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 𝐼𝐷 ≅6 mA and 𝑉𝐷 ≅ 2.8 V.

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Question 8

𝑅𝑖 = 30K 𝑅𝑃 = 10K 𝐶𝑆 = 10 𝜇F 𝐶𝑃 = 50 pF

(a) Determine the open-circuit constant associated with 𝐶𝑆, and short-circuit time constant associated with 𝐶𝑃 (4 points)

(b) Determine the corner frequencies and magnitude of the transfer function 𝑇(𝑠) = 𝑉𝑜(𝑠) 𝐼𝑖(𝑠)⁄ 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) 𝜏𝑆 = (𝑅𝑖 + 𝑅𝑃)𝐶𝑆 = 0.4 s 𝜏𝑃 = (𝑅𝑖‖𝑅𝑃)𝐶𝑃 = 0.375 𝜇s

Part (b)

𝑓𝐿 =1

2𝜋𝜏𝑆= 0.398 Hz

𝑓𝐻 =1

2𝜋𝜏𝑃= 424 kHz

At midband, 𝐶𝑆 → short, 𝐶𝑃 → open. Thus 𝑉𝑜 = 𝐼𝑖(𝑅𝑖‖𝑅𝑃). The transfer function is

𝑇(𝑠) =𝑉𝑜(𝑠)𝐼𝑖(𝑠)

= (𝑅𝑖‖𝑅𝑃) = 7.5 kΩ

Part (c)

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Question 9 Sketch the Bode magnitude plots of the following functions:

𝑇(𝑠) =−10𝑠

(𝑠 + 20)(𝑠 + 2000)

𝑇(𝑠) =10(𝑠 + 10)(𝑠 + 100)

Solution

First transfer function

𝑇(𝑠) =−10𝑠

(𝑠 + 20)(𝑠 + 2000) = −10

(2)(2000) ∙𝑠

𝑠20 + 1 𝑠2000 + 1

= 250 × 10−6𝑠

𝑠20 + 1 𝑠2000 + 1

The transfer function has zeros at 𝑠 = 0, 𝑠 = ∞ rad/s, and has poles at 𝑠 = 20, 𝑠 = 2,000 rad/s

The zero at the origin (𝑠-term in numerator) is 0 dB at 1 rad/s, and the constant term gives −72 dB and has a slope of 20 dB/decade. The pole at 𝑠 = 20 is log10(20 1⁄ ) = 1.3 decades higher so at the pole the Bode plot has value −72 + 20 × 1.3 = −46 dB ≡ 5 × 10−3.

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Second transfer function

𝑇(𝑠) =10(𝑠 + 10)(𝑠 + 100) =

(10)(10)(100) ∙

𝑠10 + 1

𝑠100 + 1

= 𝑠10 + 1

𝑠100 + 1

The transfer function has a zero at 𝑠 = 10 and a pole at 𝑠 = 100. At 𝑠 = 0, |𝑇(𝑠)| = 1 ≡ 0 dB

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Question 10

(a) Sketch the Bode magnitude plot of the following function. (6 points)

𝑇(𝑠) =10𝑠

(𝑠 + 10)(𝑠 + 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 −3 dB frequency? (2 points)

Solution

Part (a)

𝑇(𝑠) =10𝑠

(𝑠 + 10)(𝑠 + 500) =10

(10)(500) ∙𝑠1

𝑠10 + 1 𝑠500 + 1

= 0.002𝑠1

𝑠10 + 1 𝑠500 + 1

The transfer junction has zero at the origin, a pole at 𝑠 = 10 and a pole at 𝑠 = 500. The zero at the origin (𝑠-term in numerator) is 0 dB at 1 rad/s, and the constant term gives 20 log10 0.002 =−54 dB, and has a slope of 20 dB/decade. The pole at 𝑠 = 10 is 1 decade higher, so at this pole the Bode plot has value −54 + 20 = −34 dB ≡ 0.02

Part (b) Midband gain = 0.02

Part (c) 𝜔 = 500 rad/s

Part (d) 𝜔 = 10 rad/s

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Question 11 For the common-emitter amplifier below, the transistor parameters are 𝛽 =100,𝑉𝐵𝐸(ON) = 0.7 V, and 𝑉𝐴 = ∞.

𝑉𝐶𝐶 = 12 𝑉 𝑅𝐶 = 1 kΩ 𝑅1 = 10 kΩ 𝑅2 = 1.5 kΩ 𝑅𝐸 = 0.1 kΩ 𝑅𝑆 = 0.5 kΩ 𝐶𝐶 = 0.1 𝜇F

(a) Determine 𝐼𝐶𝑄 , 𝑔𝑚 and 𝑟𝜋 (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 𝐼𝐶𝑄 = 7.6 mA, and 𝑔𝑚 = 40𝐼𝐶 =0.304 A V⁄ . Thus, 𝑟𝜋 = 𝛽 𝑔𝑚⁄ = 0.329 kΩ.

Part (b)

𝑅𝑖 = (𝑅1‖𝑅2)‖[𝑟𝜋 + (1 + 𝛽)𝑅𝐸] = 1.16 kΩ Further

𝜏 = (𝑅𝑖 + 𝑅𝑆)𝐶𝐶 = 166 𝜇s

⇒ 𝑓𝐿 =1

2𝜋𝜏= 959 Hz

Part (c) The midband voltage gain from the base to the collector is

−𝑅𝐶

1𝑔𝑚

+ 𝑅𝐸=

1 × 103

3.43 + 100= −9.68

A quick approximation, also acceptable here is 𝐴𝑣 ≅ −𝑅𝐶 𝑅𝐸⁄ = −10.

The source resistor 𝑅𝑆 forms a voltage divider with 𝑅𝑖 so that the overall midband voltage gain is

𝐴𝑉 =𝑣𝑜𝑣𝑖

= −𝑅𝑖

𝑅𝑖 + 𝑅𝑆9.68 = −6.76

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Part (d)

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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

𝜏1 = 𝑅1𝐶1 = 220 𝜇s 𝜏2 = 𝑅2𝐶2 = 330 𝜇s 𝜏3 = 𝑅3𝐶3 = 470 𝜇s

The rise times of the amplifiers are

𝑡𝑟1 = 480 𝜇s 𝑡𝑟2 = 726 𝜇s 𝑡𝑟3 = 1.03 ms

An estimate of the rise time for the composite amplifiers is

𝑡𝑟 = 𝑡𝑟12 + 𝑡𝑟22 + 𝑡𝑟32 = 1.35 ms

An estimate for the −3 dB bandwidth of the composite amplifier is

BW = 0.35𝑡𝑟

= 260 Hz

A SPICE simulation gave 𝑡𝑟 = 1.47 ms and the −3 dB as 231 Hz respectively. The estimates are within about 10% of the SPICE simulation.

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Question 13 For the circuit below 𝛽 = 120,𝑉𝐵𝐸(𝑂𝑁) = 0.7 V, and 𝑉𝐴 = ∞

(a) Design a bias-stable circuit such that 𝐼𝐶𝑄 = 1 mA. (b) Determine the output resistance 𝑅𝑜 . (c) What is the lower 3 dB corner frequency?

Solution

Part (a) The Thevenin equivalent circuit for the base-emitter bias circuit is

𝑅𝑇𝐻 = 𝑅1||𝑅2

𝑉𝑇𝐻 = 12𝑅2

𝑅1 + 𝑅2

𝑉𝐵𝐸 = 0.7 V 𝐼𝐶𝑄 = 1 mA 𝛽 = 120

𝐼𝐵 must be 𝐼𝐶𝑄 𝛽 = 8.33 𝜇A⁄ , 𝑉𝑅𝐸 ≅ 𝐼𝐶𝑄𝑅𝐸 = 4 V, so that 𝑉𝐵 must be 4.7 V to meet the design specifications. The figure below summarizes the requirements.

𝑅𝑇𝐻 = 𝑅1||𝑅2 =𝑅1𝑅2𝑅1 + 𝑅2

𝑉𝑇𝐻 = 12𝑅2

𝑅1 + 𝑅2

There are many combinations of 𝑅1,𝑅2 that will meet these requirements. “Design” 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 − 200 kΩ. Choose 𝑅𝑇𝐻 = 47 kΩ. Then

𝑉𝑇𝐻 = 4.7 + (8.33 × 10−6)(47 × 103) = 5.1 V

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𝑅𝑇𝐻 =𝑅1𝑅2𝑅1 + 𝑅2

= 47 × 103

5.1 = 12𝑅2

𝑅1 + 𝑅2

Solving yields 𝑅1 = 110.6 kΩ and 𝑅2 = 81.74 kΩ. The closest standard values are 𝑅1 =110 kΩ and 𝑅2 = 82 kΩ.

Part (b) Using BJT impedance scaling

𝑅𝑜 = 𝑅𝐸 𝑟𝜋

1 + 𝛽

𝑟𝜋 =𝛽𝑔𝑚

=120

40𝐼𝐶𝑄= 3 kΩ

⇒ 𝑅𝑜 ≅ 25 Ω

Part (c) Use the time-constant technique to determine the lower corner frequency. The time constant associated with the coupling capacitor is

𝜏 = 𝐶𝐶2(𝑅𝑜 + 𝑅𝐿) = 2 × 10−6(25 + 4 × 103) = 8.1 ms

The lower corner frequency is then

𝑓𝐿 =1

2𝜋𝜏= 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 𝛽 replaced with 𝐕𝐀𝐅 = 1G and 𝐁𝐅 = 120.

Parameter Calculated/Design SPICE 𝐼𝐶𝑄 1 mA 1.02 mA 𝑉𝐵𝐸 0.7 V 0.642 V 𝐼𝐵𝑄 8.33 𝜇A 12.5 µA 𝑉𝑅𝐸 4 V 3.89 V 𝑓𝐿 19.8 Hz 18.8 Hz

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Question 14 For the amplifier below, 𝛽 = 100, and a dc analysis reveals that 𝐼𝐶𝑄 = 0.838 mA. Estimate the 3-dB bandwidth. Assume that 𝐶𝑐2 = 0.1 𝜇F. (15 points)

Answer

For the transistor 𝑟𝜋 = 𝛽 𝑔𝑚 = 100 40𝐼𝑐 ≈ 3K⁄⁄ . Using BJT impedance scaling, we estimate the resistance looking into the emitter as

𝑅𝑜 =𝑅𝑆||𝑅𝐵 + 𝑟𝜋

(1 + 𝛽)≈ 35 Ω

The time constant associated with 𝐶𝐶2 is

𝜏 = 𝐶𝑐2(𝑅𝑜||𝑅𝐸 + 𝑅𝐿)

Since 𝑅𝑜 is so much smaller than 𝑅𝐸 and 𝑅𝐿 we can write

𝜏 ≈ 𝐶𝐶2𝑅𝐿 = 1 ms

The 3-dB bandwidth is 𝐵 = 1 2𝜋𝜏 = 159 Hz⁄ .

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Question 15 Consider the amplifier below. In the circuit, 𝐼𝑃ℎ represents a photodetector. Ignore the MOFET’s 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 𝜏 = 𝐶𝐶1𝑅𝑝ℎ + 𝑅1||𝑅2 = (2.2 × 10−9)(60 × 103) =132 𝜇𝑠. The 3-dB frequency is the 𝐵 = 1 2𝜋𝜏 = 1.21 kHz.⁄

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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

𝑃𝐷,𝑚𝑎𝑥 = 𝑉𝐶𝑄𝐼𝐶𝑄 = 25 W

and

𝐼𝐶𝑄 =2512

= 2.08 A, and 𝑅𝐿 =12𝐼𝐶𝑄

= 5.76 Ω

𝑅𝐵 =24 − 0.72.08/80

= 896 Ω

Note: load power is 𝑃𝐷,𝑚𝑎𝑥 = 𝑉𝑃/√2 × 𝐼𝑃/√2 = 12/√2 × 2.08/√2 = 12.48 W and the supply power is 2.08×24 = 49.92 W, and 𝜂 = 12.48/49.92 = 0.25, as expected.

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Question 17 Q3, Q4, RP2, RP4 below form a short circuit protection network for the amplifier. If RP3 = RP4 = 0.15 Ω, 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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Question 18 (BJT impedance scaling) In the amplifier below, assume 𝑉𝐵𝐸(𝑂𝑁) = 0.6 V for all the transistors. (a) Show that the collector current of 𝑄3 is approximately 1.6 𝜇A. (2 points) (b) Estimate the output resistance. (8 points) (c) Estimate the voltage gain 𝑉𝑜 𝑉𝑠⁄ . (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 𝑄3 is about (10 mA) (𝛽1𝛽2)⁄ = 1.6 𝜇A. 𝑄3′𝑠 base current is 0.1 𝜇A.

Part (b) We can view the three transistors as a single BJT with beta 𝛽𝑐 = 100 × 63 × 16 =100,800. For this composite transistor 𝑔𝑚𝑐 = 40𝐼𝑐 = (40)(10 mA) = 0.4 S, and 𝑟𝜋𝑐 = 𝛽𝑐 𝑔𝑚𝑐 = 100,800 0.4⁄ = 252K⁄ .

Use BJT scaling to estimate 𝑅𝑂:

𝑅𝑂 =10K + 𝑟𝜋𝑐

1 + 𝛽𝑐=

10K + 252K100,801

= 2. 5 Ω

Part (b) The arrangement is that of a voltage follower, so the gain is essentially 1.

Part (c) The base current of 𝑄3 is 0.1 𝜇A so the base voltage of 𝑄1 is −1 mV ≅ 0. The dc output voltage is then −3𝑉𝐵𝐸 = −3 × 0.6 = −1.8 V

𝛽 = 63

𝛽 = 100

𝛽 = 16

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Question 19 (Frequency Response) The transistor in the amplifier shown has 𝛽 = 350 and 𝑉𝐵𝐸(𝑂𝑁) = 0.65 V. Ignore the BJT capacitances. (a) Make reasonable assumptions and show that 𝐼𝐶𝑄 ≈

1 mA (3 points) (b) Show that 𝑅𝑖 ≈ 13.7K (5 points) (c) Estimate the lower 3-dB frequency if 𝐶𝐶 = 1 𝜇F

(3 points)

Solution

Part (a) Since 𝛽 is large, ignore 𝐼𝐵𝑄 so that 𝑉𝐵 = (9)(27 (100 + 27)⁄ ) = 1.9V. Since 𝑉𝐵𝐸(𝑂𝑁) = 0.65 V, then 𝑉𝑅𝐸 = 1.9 − 0.64 = 1.25 V. Consequently, 𝐼𝐶𝑄 ≅ 𝐼𝐸 = 1.25 1.3K = 0.962 mA ≅ 1 mA⁄ .

Part (b) 𝑟𝜋 = 𝛽 𝑔𝑚 = 350 40𝐼𝐶𝑄 =⁄⁄ 8.75K. Using BJT scaling,

𝑅𝑖 = 65K18K𝑟𝜋 + (1 + 𝛽)(1.3K) = 13.68K

Part (c)

𝑓3𝑑𝐵 =1

2𝜋𝑅𝑖𝐶𝑐=

1(2𝜋)(13.68K)(1 × 10−6) = 11.64 Hz

For comparison, SPICE gives 𝑟𝜋 = 8.15K, 𝑅𝑖 = 13.59K, 𝑓3𝑑𝐵 = 11.8 Hz

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homework assignment 11 - university of...

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