midterm 1 exam name: max: 90 points - university of...

55:041 Electronic Circuits. The University of Iowa. Fall 2012.

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Midterm 1 Exam

Name: _____________________ Max: 90 Points

Question 1 (Short Takes)

1. For the circuit shown, with , and with .

Estimate . Ignore the effects of bias current the inverting input. (3 points)

Solution

so with , the output is

simply . Thus

2. A linear power supply consists of a transformer with 6.3-V secondary, full-wave bridge

rectifier, and smoothing capacitor . The load resistance is . The measured ripple voltage

is 100 mV. The load resistance is doubled ( ). What is the new ripple voltage?

(2 points)

(a) 200 mV

(b) 100 mV

(c) 50 mV

(d) 150 mV

Answer: Option (c)

3. A linear power supply consists of a transformer with 6.3-V secondary, full-wave bridge

rectifier, and smoothing capacitor . The load resistance is . What is the output voltage?

Take into account. (2 points)

(a) 6.3 V

(b) 8.9 V

(c) 7.5 V

(d) 4.9 V

Answer: The peak secondary output voltage is

(√ )( ). For a full-wave bridge, the dc output voltage is

then (√ )( ) ( ) . Thus, option (c).


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4. An engineer measures the bandwidth of the circuit below by sweeping across a range of

frequencies and noting the output amplitude on an oscilloscope response. The engineer uses

a standard probe rather than a probe. What is the measured bandwidth?

(2 points)

(a) About the same as the true bandwidth

(b) About 2 times larger than the true bandwidth

(c) About 2 times smaller than the true bandwidth

(d) About √ times larger than the true bandwidth

Answer: A standard, scope probe has an internal resistance of . This about the same

as the 910K. Thus, the resistance that sees, will be about 2 times smaller once the probe is

connected. This will increase the bandwidth by a factor 2. Thus, option (b).

5. The op-amp circuit shown has a serious flaw. What is it? (1 point)

Answer: There is no dc path to bias the non-

inverting input.

6. An engineer measures the (step response) rise time of an amplifier as . Estimate

the 3 dB bandwidth of the amplifier. (2 points)

Answer:

7. True or false: The turn-on voltage of blue LEDs is larger than the turn-on voltage of green

LEDs. (1 point)


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8. Consider a forward-biased silicon diode with Next, is increased 10 mA.

Circle the true statement below. (2 points)

(a) The diffusion capacitance increases and junction capacitance decreases

(b) The diffusion capacitance decreases and junction capacitance increases

(c) Only the diffusion capacitance increases

(d) Only the junction capacitance increases

Answer: will increase significantly, and will decrease slightly. Thus, option (a)

9. True or false: a silicon diode is biased so that at 25 oC. VD changes with 2 mV/

oC,

so that at 125 oC, will be 0.7 + 100×0.002 = 0.9 V (2 points)

Answer: False. decreases with increasing temperature

10. True or false: a diode, forward biased at ID = 1 mA, has a small-signal or incremental

resistance of about . (2 points)

Answer: False, because ⁄ ⁄

11. Which of the following depicts the correct current direction? Circle one. (1 point)

Answer: (a)

12. In the context of diodes, the term “PIV” means: (1 point)

Answer: Peak Inverse Voltage


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13. A current source has an output resistance and drives a load.

What current flows through the load? (2 points)

Answer: [ ( )⁄ ] A

14. True or false: in the circuit below, even though the diode equation is nonlinear, the

photocurrent is essentially linear with photon flux density. (1 point)

Answer: True

15. 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 and the gain-bandwidth product is 2 MHz.

Thus, the open-loop bandwidth is ( ) ⁄

16. Consider a frequency . How many decades higher is the frequency ?

(2 points)

Answer: Each decade means a frequency higher. Thus, we have to find in . Substituting values gives ( ⁄ ) decades.

17. Consider a first-order RC low-pass filter with 3-dB frequency . What is the phase

shift in degrees at 75 Hz? (3 points)

Answer: The phase shift at 25 Hz is and increases at / decade. 75 Hz is

( ) ⁄ decades higher than 25 Hz. Thus, the phase shift is . A more accurate calculation gives the phase shift as ( ⁄ ) .


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18. The following circuit has a time-constant of . What is the attenuation (in dB) at a

frequency of (4 points)

Answer. This is a 1st order low-pass network with a corner frequency of

( ) ⁄ . The attenuation is 20 dB/decade above and 1.6 kHz is 1 decade

higher than 159.2 kHz. Thus, the network will attenuate at 20 dB at 1.6 kHz. An alternate

calculation is (√ ( ⁄ ) ) .

19. Consider the Bode plot of a 1st order RC network. What is the attenuation of the network at

? Provide your answer in dB. (4 points)

Answer: 60 Hz is ( ⁄ ) decades higher than the 2.5 Hz corner frequency.

The attenuation increases by 20 dB per decade, so that at 60 Hz | ⁄ | (in dB) is dB. The attenuation is 31.1 dB.

An alternate calculation is (√ ( ⁄ ) ) .

20. In the circuit, , and compensates for the op-amp’s input bias current.

What should its value be to be effective? (2 points)

(a) 10K

(b) 15K

(c) 6K

(d) 25K

(e) Need

Answer: Choose ‖ , so (c) is the answer.


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21. Shown are the characteristics for a FET. What type of FET is this (circle one)? (1 point)

a) Enhancement PMOS

b) Enhancement NMOS

c) Depletion PMOS

d) Depletion NMOS

Answer: Option (a)

22. True or false: the threshold voltage ( ) for a particular MOSFET is, as is the case

with the cut-in voltage ( ) of diodes, well-defined and not subject to large variation between

samples of the same part number. (1 point)

Answer: False

23. A MOSFET is biased such that and . To get a change in

what is the required change in ? (3 points)

Answer

⁄ ( ) ( )⁄⁄

24. Using the MOSFET equations and one or two sentences, explain why the MOSFET below is

always operating in the saturation region. (2 points)

Answer:

No current flows into the gate so that no current flows through

and consequently there is no voltage drop across and .

( ) becomes ( ) . Since

it follows that ( ) and the FET is always

operating in saturation.


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25. The abbreviation/term “ESR” is often encountered in data sheets of capacitors. What does

ESR stand for? (1 point)

Answer: Equivalent Series Resistance

26. What is in the following circuit if = 1.2 V, , and ?

(3 points)

Answer: The current through is 1.2/240 = 5 mA, which also flows through . Thus,

the output voltage is 1.2 + 0.005× 820 = 5.3 V

27. The output voltage of a three-terminal voltage regulator is 3.3 V @ 5 mA load, and 3.25 V

@ 3 A load. What is the regulator’s output resistance? (2 points)

(a)

(b)

(c)

(d)

Answer: ⁄⁄ , so (a)

28. The output voltage of a three-terminal voltage regulator is 5 V @ 5 mA load, and 4.96 V @

1.5 A load. What is the regulator’s load regulation? (2 points)

Answer:

( ) ( )

( )


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29. The figure below depicts a MOS capacitor. The thickness of the oxide insulator is on the

order of (circle one). (1 point)

( ) ( ) ( ) (d) 0.5 nm

Answer: Option (c)

30. Consider two MOSFETs A and B that are identical in all respects except that A’s channel is

twice as wide as channel B’s channel: . Under identical bias what is the

relationship between the drain current of A and B? (2 points)

(a)

(b)

(c)

(d) √

Answer: Option (b)

31. Write down the dc load line equation for the MOSFET in the circuit below. (2 points)

Answer:


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Question 2 What is the voltage gain ⁄ of the circuit below if and

? (5 points)

Solution

Replace the 10K and with a single 9.091K resistor. KCL at the output node, assuming

currents flow away from the node, gives

However, from the diagram , so that

Thus

( )


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Question 3 In the circuit below, , the op-amps are ideal, except they both have an offset

voltage .

(a) Find magnitude of (3 points)

(b) Find worst-case magnitude for (4 points)

Solution

The equivalent circuit with and the

offset voltages indicated, is shown.

Part (a) With respect to its offset voltage, the first amplifier is a noninverting amplifier with gain

so that | | .

Part (b) | | is then amplified by the second amplifier by a factor 5. With respect to its offset

voltage, the gain of the second amplifier is 6, so that the worst-case | | (using superposition) is

| |


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Question 4 Suppose the amplifier shown has a differential-

mode gain of 2,500 and a CMMR of 80 dB. What is the

output voltage if , and ?

(6 points)

Solution

The differential-mode gain is 2,500 and the common-mode gain is 80 dB less, which corresponds

to a factor ⁄ . Thus, the common-mode gain is (⁄ ) .

The output voltage is

( )( ) ( )( )

Question 5 The parameters of the transistor in the circuit are ⁄ , and . Determine for the circuit when

. (6 points)

Solution

For

( )

( )( )

The solution is the proper solution since would imply that the

FET is off, because this is less than .


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Question 6 The op-amp in the circuit below is ideal, except for a finite open-loop gain

. Determine the closed-loop gain ⁄ and be sure to provide your answer to four

decimal places. (6 points)

Solution

KVL around the loop shown below gives

Where is the voltage across . However, no current flows into the op-amp, so and

the KVL equation becomes

Now , which means ⁄ , so the KVL equation becomes

Solving for ⁄ yields ( ) ⁄


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Question 7 The circuit below uses an NMOS transistor to implement a current source. For the

transistor, and ⁄ . What is the required value op so that

What is the compliance voltage? (The compliance voltage is the smallest voltage

across a current source while still keeping it a current source.)

(6 points)

Solution

( )

( )( )

( )

√

To function as a current source the transistor must be in saturation, or ( ). Now,

( ) . Thus, the compliance voltage is 1.414 V.

midterm 1 exam name: max: 90 points - university of...

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