---
title: "Diode dc voltage and ac amplitude"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-dc-ac-superimposed"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-dc-ac-superimposed
site: BCT Engineering
---

# Diode dc voltage and ac amplitude

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2080 Bhadra, 4 marks; 2076 Ashwin, 5 marks (asked 2 times)
- 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-dc-ac-superimposed
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

Consider the circuit below for which R = 10kΩ. The power supply V+ has a dc value of 10V on which is superimposed a 60 Hz sinusoid of 1-V peak amplitude. Calculate both the dc voltage of the diode and the amplitude of the sine wave signal appearing across it. Assume the diode to have a 0.7V drop at 1 mA current and n=2.
[Figure: supply V+ in series with resistor R and a diode to ground]

## Short answer

DC bias: ID = (10–0.7)/10k = 0.93 mA, VD ≈ 0.7 V. Dynamic resistance rd = 2×25 mV/0.93 mA ≈ 53.7 Ω. AC peak amplitude across diode = 1 V × 53.7/(10000+53.7) ≈ 5.3 mV.

## Answer

Treat the problem in two parts: the dc analysis gives the bias current and the diode voltage; the small-signal (ac) analysis gives the signal across the diode using its dynamic resistance.

#### Given
V+ = 10 V dc + 1 V peak, 60 Hz; R = 10 kΩ; diode: 0.7 V at 1 mA; η = n = 2; VT = 25 mV.

#### Dc analysis
The diode voltage at current I: VD = 0.7 + n VT ln(I / 1 mA) = 0.7 + 0.05 ln(I/1 mA).

Assume VD = 0.7 V first:
```text
I = (10 - 0.7)/10 kΩ = 0.93 mA
```
Improve: VD = 0.7 + 0.05 ln(0.93) = 0.7 - 0.0036 = 0.6964 V, giving I = (10 - 0.6964)/10 kΩ = 0.9304 mA (no further change).

**DC voltage across the diode, VD ≈ 0.696 V (≈ 0.7 V).**

#### Ac analysis
Dynamic resistance:
```text
rd = n VT / ID = 2 × 25 mV / 0.93 mA = 53.7 Ω
```
Ac equivalent: the source (1 V peak) in series with R and rd. By voltage division:
```text
vd(peak) = 1 V × rd / (R + rd)
         = 1 × 53.7 / (10000 + 53.7)
         = 5.34 mV
```

**Answer:** dc voltage across the diode ≈ 0.7 V; ac amplitude across the diode ≈ 5.3 mV peak (10.7 mV peak-to-peak).

The small-signal approximation is valid because 5.3 mV is much smaller than n VT = 50 mV.

## Check yourself

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

1. How is the problem solved?
   A. Only dc analysis
   B. Only ac analysis
   C. Fourier analysis
   D. Dc analysis for bias, then small-signal analysis with rd for the signal
   Answer: D. Dc analysis gives the bias current and VD; ac analysis uses rd = eta VT/ID. (Key point: Two-part method)

2. With the diode taken as 0.7 V, what is the dc current?
   A. 0.7/10 kohm = 0.07 mA
   B. 10/10 kohm = 1 mA
   C. (10 - 0.7)/10 kohm = 0.93 mA
   D. (10 + 0.7)/10 kohm = 1.07 mA
   Answer: C. I = (10 - 0.7)V/10 kohm = 0.93 mA. (Key point: DC bias current)

3. What is rd with n = 2, VT = 25 mV, ID = 0.93 mA?
   A. 26.9 ohm
   B. 5.37 kohm
   C. 107 ohm
   D. 53.7 ohm
   Answer: D. rd = 2 x 25 mV/0.93 mA = 53.7 ohm. (Key point: Dynamic resistance)

4. How is the ac amplitude across the diode found?
   A. vd = 1 V x rd/(R + rd), about 5.3 mV
   B. vd = 1 V x rd
   C. vd = 1 V x R/(R + rd)
   D. vd = 1 V - 0.7 V
   Answer: A. The ac source divides between R and rd, so vd is about 5.34 mV peak, much smaller than the 1 V source. (Key point: Ac voltage division)

## Related in Electronics Devices and Circuit

- [Diode ac amplitude with 60 Hz superimposed](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adiode-ac-ripple-amplitude-10v) — Electronics Devices and Circuit, chapter 1, asked 2×