---
title: "Diode circuit current, eta 1.74"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-current-eta-1p74"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-current-eta-1p74
site: BCT Engineering
---

# Diode circuit current, eta 1.74

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2080 Baisakh, 5 marks; 2072 Kartik, 6 marks (asked 2 times)
- Video: [Diode Current Equation || Lecture 10 Electronic Devices & Circuit || Explained in Urdu / Hindi](https://www.youtube.com/watch?v=uwg1sZuMPYQ) — Tuition Point
- 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-eta-1p74
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

In the given circuit the diode used has its η = 1.74 and it conducts 1mA at forward bias voltage of 0.7V. Find the current flow in the circuit.
[Figure: VDD = 5 V in series with a 500 Ω resistor and the diode]

## Short answer

Iterate ID = (5–VD)/500 and VD = 0.7 + 0.0435 ln(ID/1mA). The solution converges to VD ≈ 0.793 V and ID ≈ 8.41 mA.

## Answer

Use the diode equation to relate current and voltage, then solve the circuit by iteration, starting from the 0.7 V point.

#### Given
VDD = 5 V, R = 500 Ω, η = 1.74, ID = 1 mA at VD = 0.7 V, VT = 25 mV (so η VT = 43.5 mV).

#### Diode equation (two points)
Since I2/I1 = exp[(V2 - V1)/(η VT)]:
```text
VD = 0.7 + η VT ln(ID / 1 mA) = 0.7 + 0.0435 ln(ID/1 mA)
```
Circuit (KVL): ID = (5 - VD)/500.

#### Iteration
| Step | VD guess (V) | ID = (5 - VD)/500 (mA) | New VD = 0.7 + 0.0435 ln(ID) |
|---|---|---|---|
| 1 | 0.700 | 8.600 | 0.7937 |
| 2 | 0.7937 | 8.413 | 0.7926 |
| 3 | 0.7926 | 8.415 | 0.7927 |

The values have converged: VD ≈ 0.793 V.

```text
ID = (5 - 0.793)/500 = 8.41 mA
```

**Answer: ID ≈ 8.4 mA** (VD ≈ 0.79 V).

A quick check with the constant-drop model (0.7 V) gives 8.6 mA, within 2.5 %, so the exponential effect is small.

## Check yourself

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

1. Which relation links the diode voltage and current given 1 mA at 0.7 V?
   A. VD = 0.7 + ID R
   B. VD = 0.7 + eta VT ln(ID/1 mA)
   C. VD = 0.7 x ID
   D. VD = 0.7 - eta VT ln(ID)
   Answer: B. Since I2/I1 = exp[(V2 - V1)/(eta VT)], the voltage follows the logarithm of the current ratio. (Key point: Diode equation from two points)

2. Which equation from KVL is used with the diode equation?
   A. ID = 5 x 500
   B. ID = 5/(500 + VD)
   C. ID = (5 - VD)/500
   D. ID = VD/500
   Answer: C. KVL: VDD = ID R + VD. (Key point: Circuit equation)

3. Why is iteration used?
   A. Because the diode is reverse biased
   B. The two equations (exponential and linear) cannot be solved directly in closed form
   C. Because VT is unknown
   D. Because R is unknown
   Answer: B. The log/exponential diode law with the linear KVL is solved by repeatedly updating VD and ID. (Key point: Method)

4. What is the converged current?
   A. About 18 mA
   B. About 8.4 mA
   C. About 8.6 mA
   D. About 4.2 mA
   Answer: B. VD converges to 0.793 V, giving ID = (5 - 0.793)/500 = 8.41 mA. (Key point: Result)

## Close variants set in other exams

- [Diode circuit current, eta 1.74, 10 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adiode-current-eta-1p74-10v) — Electronics Devices and Circuit, chapter 1, asked 4×
- [Diode circuit current, eta 1.74, 6 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adiode-current-eta-1p74-6v) — Electronics Devices and Circuit, chapter 1, asked 4×

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