---
title: "Diode leakage current at 60°C"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-leakage-current-60c"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-leakage-current-60c
site: BCT Engineering
---

# Diode leakage current at 60°C

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2074 Chaitra, 4 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-leakage-current-60c
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

The leakage current of a silicon diode is Is = 10⁻⁹ A at 25°C, and the emission coefficient is η = 1.6. The operating junction temperature is Tj = 60°C. Determine (i) the leakage current Is and (ii) the diode current ID at VD = 0.8 V.

## Short answer

IS(60°C)=10⁻⁹×2^(3.5)≈11.3 nA. With VT=28.71 mV, ηVT=45.93 mV, ID=11.3 nA×exp(0.8/0.04593)≈0.41 A.

## Answer

The reverse saturation (leakage) current roughly **doubles for every 10 °C rise** in temperature, and the thermal voltage VT = kT/q also increases with temperature.

#### Given
IS(25 °C) = 10⁻⁹ A, η = 1.6, Tj = 60 °C (333 K), VD = 0.8 V.

#### (i) Leakage current at 60 °C
```text
IS(T2) = IS(T1) × 2^[(T2 - T1)/10]
       = 10⁻⁹ × 2^[(60 - 25)/10] = 10⁻⁹ × 2^3.5
       = 10⁻⁹ × 11.31 = 1.13×10⁻⁸ A ≈ 11.3 nA
```

#### (ii) Diode current at VD = 0.8 V
```text
VT = kT/q = 8.617×10⁻⁵ × 333.15 = 28.71 mV
η VT = 1.6 × 28.71 mV = 45.93 mV

ID = IS [exp(VD/(η VT)) - 1]
   = 11.31×10⁻⁹ × [exp(0.8/0.04593) - 1]
   = 11.31×10⁻⁹ × [exp(17.42) - 1]
   = 11.31×10⁻⁹ × 3.66×10⁷
   ≈ 0.41 A
```

**Answer: (i) IS ≈ 11.3 nA; (ii) ID ≈ 0.41 A (about 414 mA).**

*Note:* such a large current only shows that 0.8 V is a high forward voltage for this diode at 60 °C; in practice an external resistor limits the current.

## Check yourself

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

1. How does leakage current change with temperature?
   A. Halves every 10 degC
   B. Triples every degC
   C. Doubles for about every 10 degC rise
   D. Is constant
   Answer: C. IS roughly doubles per 10 degC. (Key point: Leakage vs temperature)

2. What is IS at 60 degC from 1 nA at 25 degC?
   A. About 35 nA
   B. About 11.3 nA
   C. About 1.1 nA
   D. About 3.5 nA
   Answer: B. IS = 1 nA x 2^3.5 = 11.3 nA. (Key point: Leakage at 60 degC)

3. How is VT affected by temperature, and what is eta VT at 333 K, eta = 1.6?
   A. VT falls; about 25 mV
   B. VT is constant; 40 mV
   C. VT rises with T; about 45.9 mV
   D. VT rises; 28.7 mV
   Answer: C. VT = kT/q = 28.7 mV, so eta VT = 45.9 mV. (Key point: Thermal voltage)

4. Which formula gives ID at VD = 0.8 V?
   A. ID = IS x VD
   B. ID = IS ln(VD)
   C. ID = IS[exp(VD/(eta VT)) - 1]
   D. ID = VD/(eta VT)
   Answer: C. Using IS = 11.3 nA and exp(17.43), ID is about 0.41 A. (Key point: Diode current)

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