---
title: "Dynamic resistance of diode, numerical"
subject: "Electronics Devices and Circuit"
chapter: 1
key: "semester_3/electronics_devices_and_circuit:diode-dynamic-resistance-numeric"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/diode-dynamic-resistance-numeric
site: BCT Engineering
---

# Dynamic resistance of diode, numerical

Electronics Devices and Circuit · Chapter 1: Diodes

- Asked in: 2070 Chaitra, 5 marks
- Video: [Example on AC resistance or Dynamic Resistance](https://www.youtube.com/watch?v=EBjo6m-8o-A) — Salaar Khan
- 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-dynamic-resistance-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-1

## Question

Find the value of dynamic resistance if voltage in the diode is 650mV and IRS is 10pA = (10×10⁻¹² A) (Given n = 2 and VT = 25 mV).

## Short answer

From the Shockley equation, diode current ID = 10 pA × (e^(650/50) − 1) = 4.42 µA. Dynamic resistance rd = nVT / ID = 50 mV / 4.42 µA ≈ 11.3 kΩ.

## Answer

The dynamic resistance is rd = n VT / ID, so first find the diode current from the Shockley equation.

#### Given
VD = 650 mV, IS = 10 pA = 10×10⁻¹² A, n = 2, VT = 25 mV (n VT = 50 mV).

#### Step 1: diode current
```text
ID = IS [exp(VD/(n VT)) - 1]
   = 10×10⁻¹² × [exp(650/50) - 1]
   = 10×10⁻¹² × [exp(13) - 1]
   = 10×10⁻¹² × 4.424×10⁵
   = 4.42×10⁻⁶ A = 4.42 µA
```

#### Step 2: dynamic resistance
```text
rd = n VT / ID = 0.05 V / 4.42×10⁻⁶ A ≈ 1.13×10⁴ Ω
```

**Answer: rd ≈ 11.3 kΩ.**

## Check yourself

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

1. Which formula gives rd?
   A. rd = VD / ID
   B. rd = n VT x ID
   C. rd = n VT / ID
   D. rd = ID / (n VT)
   Answer: C. rd = n VT/ID, so ID must be found first. (Key point: Formula)

2. What is ID at 650 mV with IS = 10 pA, n VT = 50 mV?
   A. About 13 mA
   B. About 4.42 uA
   C. About 0.13 uA
   D. About 44 mA
   Answer: B. exp(13) is about 4.42e5; ID = 10 pA x 4.42e5 = 4.42 uA. (Key point: Current)

3. Why is the Shockley equation used first?
   A. To find the ac signal
   B. To find the dc bias current needed by rd
   C. To find temperature
   D. To find breakdown
   Answer: B. rd depends on the dc current. (Key point: Why use Shockley)

4. What is rd?
   A. About 50 ohm
   B. About 113 ohm
   C. About 1.13 ohm
   D. About 11.3 kohm
   Answer: D. rd = 0.05/4.42e-6 = 11.3 kohm. (Key point: Result)

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