---
title: "Diode current at 100°C"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-current-temperature-100c"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-current-temperature-100c
site: BCT Engineering
---

# Diode current at 100°C

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2081 Baisakh, 6 marks; 2078 Bhadra, 5 marks; 2071 Chaitra, 4 marks (asked 3 times)
- Video: [Diode Current Equation (Numerical 1)](https://www.youtube.com/watch?v=_jDpVxps0u4) — EnggClasses
- 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-temperature-100c
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

A diode conducts 1mA at 20°C. If it is operated at 100°C, What will be its current? Given data are η = 1.6 and negative temperature coefficient value = -2.2mV/°C.

## Short answer

The voltage at 1 mA drops by 176 mV from 20°C to 100°C. At constant applied voltage, the current rises exponentially: I2 = 1 mA × exp(176 mV / (1.6 × 32.15 mV)) ≈ 30.6 mA, using VT = kT/q at 100°C.

## Answer

**Idea:** at a fixed current the diode voltage falls by about 2.2 mV for each °C rise. If the voltage across the diode is held constant, this fall appears as an exponential rise in current.

#### Given
I1 = 1 mA at T1 = 20 °C, T2 = 100 °C, η = 1.6, temperature coefficient = -2.2 mV/°C.

#### Step 1: change in voltage for the same current
```text
ΔV = (-2.2 mV/°C)(100 - 20) = -176 mV
```
So at 100 °C the voltage needed for 1 mA is 176 mV lower. At the original voltage the diode is 176 mV "more forward biased" than needed for 1 mA.

#### Step 2: thermal voltage at 100 °C
```text
VT = kT/q = 8.617×10⁻⁵ × 373.15 = 32.15 mV
η VT = 1.6 × 32.15 = 51.4 mV
```

#### Step 3: current at constant voltage
```text
I2 = I1 · exp(|ΔV| / (η VT))
   = 1 mA × exp(176 / 51.4)
   = 1 mA × exp(3.42) ≈ 30.6 mA
```

**Answer: I ≈ 31 mA** (about 30 times larger) at 100 °C for the same applied voltage.

*Assumption:* VT is taken at 100 °C. Using VT = 25 mV throughout would give ≈ 81 mA, so state the VT you use.

## Check yourself

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

1. What is the effect of a temperature rise on the forward voltage needed for a fixed current?
   A. It rises by about 2.2 mV/degC
   B. It falls by about 2.2 mV/degC
   C. It doubles every 10 degC
   D. It stays constant
   Answer: B. The temperature coefficient is negative, so the required voltage falls. (Key point: Temperature effect on voltage)

2. What is the voltage change DeltaV between 20 and 100 degC for -2.2 mV/degC?
   A. -2.2 mV
   B. -27.5 mV
   C. -176 mV
   D. -220 mV
   Answer: C. DeltaV = -2.2 mV x 80 = -176 mV. (Key point: Step 1 value)

3. What is eta VT at 100 degC (373 K) with eta = 1.6?
   A. About 40 mV
   B. About 25 mV
   C. About 51.4 mV
   D. About 32 mV
   Answer: C. VT = 8.617e-5 x 373.15 = 32.15 mV, and 1.6 x 32.15 = 51.4 mV. (Key point: Thermal voltage)

4. Which formula gives the new current at the same applied voltage?
   A. I2 = I1 exp(|DeltaV|/(eta VT))
   B. I2 = I1 / exp(|DeltaV|/(eta VT))
   C. I2 = I1 x DeltaV
   D. I2 = I1 + |DeltaV|/(eta VT)
   Answer: A. The diode is 176 mV more forward biased, so I2 = 1 mA x exp(176/51.4), about 31 mA. (Key point: Current at constant voltage)

## Close variants set in other exams

- [Diode current at 100°C (eta 1.8)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adiode-current-temperature-eta1p8) — Electronics Devices and Circuit, chapter 1, asked 4×

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