Diode Problems and Solutions PDF

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ECE 3410 – Homework 4 Problem 1. In each of the ideal-diode circuits shown below, vIN is a 1 kHz sinusoid with zero-to-peak amplitude 10 V. For each circuit, sketch the output waveform and state the values of the maximum and minimum output voltages. For this problem, use the ideal switch model of the diode.

D1

D2 v OUT

v IN

D1

D1 v OUT

v IN

1 kΩ

v OUT

1 kΩ

(A)

1 kΩ

(B)

D1

(C)

D1

v IN

v OUT D2

D1

v IN

v OUT D2

1 kΩ

(D)

v OUT D3

1 kΩ

v OUT

1 kΩ

(F)

1 kΩ

1 kΩ

1 kΩ

D2

v IN

(E)

v IN

D2

v IN

v OUT

v IN D1

D1

v OUT

v IN 1 kΩ

D2

D1

(G)

(H) (I)

Solution

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ECE 3410 – Homework 4

Solution (cont.) 1

0

10

0.5 5

−5

0

−10

0 −0.5

0

1

0

2

1

2

0

1

·10−3

(B)

(A)

(C)

10

10

10

5

0

5

0

0

−10 0

1

2

·10−3

·10−3

0

2

1

0

2

1

(F)

(E)

(D)

2 ·10−3

·10−3

·10−3

1

1

0.5

0.5

0

0

0

−2

−0.5

−0.5 −4 0

1

2

0

1

·10−3

·10−3

(G)

2 0

1

2 ·10−3

(H) (I)

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ECE 3410 – Homework 4

Problem 2. The circuit shown below uses a diode in the feedback path of the op amp. For each question, use only the ideal switch model to analyze the circuit’s behavior (i.e. ignore the diode’s 0.7V forward drop). Assume that R1 = 1kΩ, R2 = 4kΩ, and the op amp is ideal.

R2

R1 vIN

−

i1

vOUT +

(A) Suppose vIN = 2V. What direction is the current i1 passing through R1 ? Is the diode ON or OFF? What is vOUT?

Solution The current is passing forward (left to right), so the diode should be ON and we expect vOUT = 0 V since it is shorted to the virtual ground at the op amp’s inverting input terminal.

(B) Suppose vIN = −2V. What direction is the current i1 passing through R1 ? Is the diode ON or OFF? What is vOUT?

Solution In this case the current is passing backward (right to left), so the diode is reverse biased and should be OFF. With the diode deactivated, the circuit should behave like an ordinary inverting configuration with vOUT = − RR21 = 8 V.

Problem 3. The circuit shown below uses a diode in the feedback path of the op amp. For each question, use only the ideal switch model to analyze the circuit’s behavior (i.e. ignore the diode’s 0.7V forward drop). Assume that R = 1kΩ, and the op amp is ideal.

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ECE 3410 – Homework 4

D

vOUT

i

−

R

i

+

vIN

− +

(A) Suppose vIN = 2V. What direction is the current i passing through R? Is the diode ON or OFF? What is vOUT?

Solution With vIN positive, if the diode is OFF then the op amp will rail in the positive direction, so the diode will turn ON. Then the current i flows in the forward direction (consistent with the diode being ON). Since the feedback loop is closed, the op amp will regulate vOUT = vIN = 2 V.

(B) Suppose vIN = −2V. What direction is the current i passing through R1 ? Is the diode ON or OFF? What is vOUT?

Solution With vIN negative, if the diode is OFF then the op amp will rail in the negative direction, so the diode will stay OFF. In that case there is no current passing through R and vOUT = 0 V.

(C) Explain why vOUT can never be less than 0V.

Solution If vOUT is ever negative, then current will have to flow backwards through R, and then backwards through the diode. Since the diode will block any reverse current, this case is impossible.

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Problem 4. A half-wave rectifier circuit is shown below. You may assume that R = 1kΩ and the diode D has a 0.7V forward drop when its forward current equals 1mA. vOUT

vIN i

R

(A) Using the iterative analysis procedure, solve a precise solution for i and vOUT for the case when vIN = 2V. Show your results at each iteration, and state the number of iterations required to reach three significant digits of precision.

Solution We start with the initial guess that vD = 0.7 V and iD = 1 mA. We then iterate these steps: i. Solve vOUT = vIN − vD and update iD = vOUT/R. ii. Solve vD = 0.7 V + 26 mV × ln (iD /1 mA) . Then the table of iterations is as follows: vD iD vOUT ∆vD ∆iD ∆vOUT 0.70000 1.00000 1.30000 0.00000 0.00000 0.00000 0.70682 1.30000 1.30000 0.00682 0.30000 0.00000 0.70668 1.29318 1.29318 -0.00014 - 0.00682 -0.00682 0.70669 1.29332 1.29332 0.00000 0.00014 0.00014 After three iterations (four rows including the initial guess) we can verify there is no change in the three most significant digits.

(B) Using the small-signal model, solve an approximate solution for i and vOUT for the case when vIN = 2V. How close is the small-signal result compared to the iterative method, and compared to the constant voltage drop method?

Solution Using the linearized method, we insert a 26 Ω resistor to account for the error in our initial guess. Then we have this equivalent circuit:

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ECE 3410 – Homework 4

Solution (cont.) 0.7 V

26 Ω

− + vOUT

vIN

1 mA R

NOT FINISHED YET!!!

(C) Repeat the iterative and small-signal solutions for the case when vIN = 0.5V. How do they compare in this case?

Problem 5. A 0.7V regulator circuit is shown below. You may assume that R = 1kΩ and the diode has a 0.7V forward drop at 1mA. R vout

vin i

(A) Suppose vIN has a sinusoidal “ripple” such that its maximum and minimum values are vIN(max) = 3.5V vIN(min) = 3.0V Using the small-signal procedure, estimate the maximum and minimum values that will appear at vOUT. Calculate your answer with mV precision. (B) Calculate the line regulation, defined as the ratio LR =

vOUT(max) − vOUT (min) vIN(max) − vIN (min)

Give your result in mV/ V.

Problem 6. The peak rectifier circuit shown below is used to convert a 1kHz AC signal into a DC signal. The input signal has a zero-to-peak amplitude of 15V. The input signal also Utah State University

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ECE 3410 – Homework 4

has a frequency f = 1kHz and period T = 1ms. The resistor and capacitor values are C = 10µF and R = 10kΩ, respectively. The diode has a 0.7V forward drop when its forward current is 1mA. vOUT

vIN R

C

(A) Assuming that vOUT charges all the way up to +15V during the positive cycle of vIN, calculate the ripple amplitude Vr that happens during the period when the diode is OFF. Note that, since RC ≫ T , if the diode switches OFF at some time t0 then we may approximate vOUT as vOUT ≈ 15V −

t − t0 RC

(B) Calculate the ripple amplitude when f is increased to 10kHz.

Problem 7. The half-wave precision rectifier circuit suffers from three problems: First, when the diode is off, the feedback loop becomes an open-circuit and the op amp is driven to a rail voltage. Second, when the diode turns on again, the op amp must slew back from the rail, which delays its response. Third, only half of the waveform is rectified. The circuit shown below rectifies (ehem) these problems. Study the circuit and answer the following questions. +

vIN

vX

D

1

−

2

D

RL +

D

4

vOUT

3

D

vY

−

R

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(A) First, let vIN = 0 V and suppose has a (very large) open-loop gain A but is otherwise ideal. Using the ideal switch model, predict the voltages that will appear at vX , vY and vOUT. (B) Now let vIN = 100 mV. Using the ideal switch model for the diodes, predict which diodes will be on and which will be off, and report the state (ON/OFF) for each diode. Is the op amp’s negative feedback loop closed? Predict the value at vY , vX and vOUT. (C) Now let vIN = −100 mV. Again using the ideal switch model, predict which diodes will be on and which will be off, and report the state for each diode. Is the op amp’s negative feedback loop closed? Predict the value at vY , vX and vOUT. (D) If you use the diode’s full exponential model instead of the ideal switch model, how will your results change for the last two sub-problems? You may state your answer symbolically, in terms of the diodes’ unknown forward voltage drop, vD .

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