---
title: "Diode ac amplitude with 60 Hz superimposed"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-ac-ripple-amplitude-10v"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-ac-ripple-amplitude-10v
site: BCT Engineering
---

# Diode ac amplitude with 60 Hz superimposed

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2079 Bhadra, 5 marks; 2073 Chaitra, 5 marks (asked 2 times)
- Video: [Electronics I: Diodes: Consider the circuit shown in Fi. 4.14(a) for the case in which R=10k...](https://www.youtube.com/watch?v=xrDfrymGmVg) — Computer Engineering life
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-ac-ripple-amplitude-10v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

In the circuit given below the DC power supply Vcc = 10 V is superimposed with 60 Hz sinusoid of 1 V peak to peak amplitude. Calculate the amplitude of the sine wave signal appearing across the diode for the case R1 = 10 KΩ. Assume the constant voltage drop of 0.7 V in the diode.
[Figure: supply Vcc in series with resistor R1 and a diode]

## Short answer

DC bias: ID = (10–0.7)/10k = 0.93 mA. Small-signal resistance rd = 25 mV/0.93 mA ≈ 26.9 Ω. The AC voltage across the diode (peak-to-peak) = 1 V × 26.9/(10000+26.9) ≈ 2.7 mV.

## Answer

The 60 Hz signal is small, so we find the dc bias current first and then use the diode's small-signal resistance rd.

#### Given
Vcc = 10 V dc; ac = 1 V peak-to-peak (0.5 V peak); R1 = 10 kΩ; VD = 0.7 V (constant drop); VT = 25 mV; η = 1 is assumed, since it is not given.

#### Dc bias
```text
ID = (10 - 0.7)/10 kΩ = 0.93 mA
```

#### Dynamic resistance
```text
rd = VT / ID = 25 mV / 0.93 mA = 26.9 Ω
```

#### Ac signal across the diode (voltage divider between R1 and rd)
```text
vd = vs × rd / (R1 + rd)
   = 1 V(p-p) × 26.9 / (10 000 + 26.9)
   = 2.68 mV  (peak-to-peak)
```

**Answer:** the ac signal across the diode is ≈ 2.7 mV peak-to-peak (≈ 1.34 mV peak). If η = 2 is taken, rd doubles to 53.8 Ω and the signal doubles to ≈ 5.4 mV peak-to-peak. Almost none of the ripple appears across the diode because rd ≪ R1; this is why a forward-biased diode can act as a nearly constant-voltage source.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The ac is 1 V peak-to-peak. What is its peak value?
   A. 0.7 V
   B. 2 V
   C. 1 V
   D. 0.5 V
   Answer: D. Peak = peak-to-peak/2 = 0.5 V. (Key point: Peak vs peak-to-peak)

2. What is the dc diode current for Vcc = 10 V, R1 = 10 kohm, VD = 0.7 V?
   A. 0.07 mA
   B. 0.93 mA
   C. 9.3 mA
   D. 1.07 mA
   Answer: B. ID = (10 - 0.7)/10 kohm = 0.93 mA. (Key point: DC bias)

3. What is rd with eta = 1 and VT = 25 mV?
   A. 25 kohm
   B. 269 ohm
   C. 2.69 ohm
   D. 26.9 ohm
   Answer: D. rd = 25 mV/0.93 mA = 26.9 ohm. (Key point: Dynamic resistance)

4. Why is the ripple across the diode so small (about 2.7 mV p-p)?
   A. rd is much larger than R1
   B. The diode blocks ac
   C. rd is much smaller than R1 so almost all of the ac drops across R1
   D. Vcc cancels the ac
   Answer: C. The voltage divider R1/rd puts nearly all the ac across R1, so the diode acts almost like a constant voltage source. (Key point: Result and meaning)

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