Solutions for Homework 3: Peak Detector Circuit Analysis, Exercises of Electronics

The solutions to Problem 3.28 and related problems in the homework set, focusing on the analysis of a peak detector circuit. The solutions involve calculating the steady-state value of the output voltage for different capacitor ratios and determining the circuit simplification and related parameters.

Typology: Exercises

2019/2020

Uploaded on 01/17/2020

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Solutions for HW3
Divija Gogineni
September 25, 2019
1. A peak detector is shown on page 27 of note set set2.pdf. Assuming ideal diodes with
VD,ON = 0, determine the steady-state value of the output if the input is switched between
±A, for C2= 2C1, and C2=C1/2.
For C2= 2C1: The following terms contribute to VOUT (k)
Charge sharing from C1to C2:C1A
C1+C2
=A
3
Contribution from input voltage: C1A
C1+C2
=A
3
Contribution from VOUT (k1):C2VOU T (k1)
C1+C2
=2
3VOU T (k1)
Thus VOUT (k) = A
3+A
3+C2VOU T (k1)
C1+C2
Since VOU T (0) = 0, we have VOUT (k ) = 2A
3+2A
3.2
3+2A
3.(2
3)2+· · · = 2A
Similar proof for C2=1
2C1for which
VOU T (k) = 2A
3+2A
3+1
3VOU T (k1)
1
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Solutions for HW

Divija Gogineni

September 25, 2019

  1. A peak detector is shown on page 27 of note set set2.pdf. Assuming ideal diodes with VD,ON = 0, determine the steady-state value of the output if the input is switched between ±A, for C 2 = 2C 1 , and C 2 = C 1 / 2.

For C 2 = 2C 1 : The following terms contribute to VOU T (k)

  • Charge sharing from C 1 to C 2 :

C 1 A

C 1 + C 2

A

  • Contribution from input voltage:

C 1 A

C 1 + C 2

A

- Contribution from VOU T (k − 1): C 2 VOU T (k − 1) C 1 + C 2

VOU T (k − 1)

Thus VOU T (k) =

A

A

C 2 VOU T (k − 1) C 1 + C 2

- Since VOU T (0) = 0, we have VOU T (k → ∞) =

2 A

2 A

2 A

)^2 + · · · = 2A

  • Similar proof for C 2 =

C 1 for which

VOU T (k) =

2 A

2 A

VOU T (k − 1)

  • Problem 3.

Figure2: 3.28(c)

Figure4: 3.28(d)

First, find rd:

rd = V IDT = (^265) mAmV = 5.2Ω

Since il = +1mA,id = − 1 mA

Therefore, change inVout,

ie, Vout = (− 1 mA)(3 ∗ 5 .2) = − 15. 6 mV