---
title: "Diode current and voltage, VDD 5 V, R 1 kΩ"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-current-vd-eta2"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-current-vd-eta2
site: BCT Engineering
---

# Diode current and voltage, VDD 5 V, R 1 kΩ

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2082 Shrawan, 6 marks
- 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-current-vd-eta2
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

Determine the current ID and diode voltage VD with VDD = 5 V and R = 1 kΩ. Assume that the current through the diode is 1 mA at a voltage of 0.7 V. Assume η = 2.
[Figure: series circuit of source VDD, resistor R and a diode, with current ID and diode voltage VD]

## Short answer

Iterative solution of diode law VD=0.7+0.05 ln(ID/1mA) and KVL ID=(5−VD)/1k yields VD≈0.772 V, ID≈4.23 mA.

## Answer

Combine the diode's exponential law with KVL and solve by iteration.

#### Given
VDD = 5 V, R = 1 kΩ, ID = 1 mA at 0.7 V, η = 2, VT = 25 mV (η VT = 50 mV).

#### Equations
```text
Diode:  VD = 0.7 + η VT ln(ID/1 mA) = 0.7 + 0.05 ln(ID/1 mA)
KVL  :  ID = (VDD - VD)/R = (5 - VD)/1 kΩ
```

#### Iteration
| Step | VD (V) | ID (mA) = 5 - VD | New VD |
|---|---|---|---|
| 1 | 0.70 | 4.300 | 0.7729 |
| 2 | 0.7729 | 4.227 | 0.7721 |
| 3 | 0.7721 | 4.228 | 0.7721 |

Converged: VD = 0.772 V, ID = (5 - 0.772)/1 kΩ = 4.228 mA.

**Answer: ID ≈ 4.23 mA, VD ≈ 0.772 V.**

## Check yourself

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

1. What is the diode voltage relation used?
   A. VD = 0.7 x ID
   B. VD = 0.7 + 0.05 ln(ID/1 mA)
   C. VD = 0.7 + 0.025 ln(ID)
   D. VD = 0.7 + 2 ID
   Answer: B. eta VT = 2 x 25 mV = 50 mV, so VD = 0.7 + 0.05 ln(ID/1 mA). (Key point: Diode law)

2. With R = 1 kohm, ID in mA equals what?
   A. VD - 5
   B. 5 - VD
   C. 5 + VD
   D. 5/VD
   Answer: B. ID = (5 - VD)/1 kohm gives ID in mA = 5 - VD. (Key point: KVL)

3. Starting at VD = 0.7 V, what is the first new VD?
   A. 0.6 V
   B. 0.85 V
   C. 0.7 V
   D. 0.7729 V
   Answer: D. ID = 4.3 mA, so VD = 0.7 + 0.05 ln(4.3) = 0.7729 V. (Key point: First iteration)

4. What is the converged result?
   A. ID = 5 mA, VD = 0
   B. ID = 4.23 mA, VD = 0.772 V
   C. ID = 4.30 mA, VD = 0.7 V
   D. ID = 3.2 mA, VD = 0.85 V
   Answer: B. The iteration converges to VD = 0.772 V and ID = 4.228 mA. (Key point: Final answer)

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