---
title: "Need of heat sink; thermal Ohm's law"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:heat-sink-thermal-resistance"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/heat-sink-thermal-resistance
site: BCT Engineering
---

# Need of heat sink; thermal Ohm's law

Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers

- Asked in: 2068 Baisakh, 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/heat-sink-thermal-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

Why heat sink is necessary in power transistor? Explain with the help of thermal Ohm's law or thermal resistance method.

## Short answer

Power transistors dissipate large heat; without a heat sink the junction temperature can exceed limits causing damage. A heat sink reduces thermal resistance from junction to ambient, allowing safe operation at higher power. Thermal Ohm's law: TJ - TA = PD × (θJC + θCS + θSA).

## Answer

#### Why a heat sink is needed
A power transistor dissipates a large amount of power as heat at its collector junction (PD ≈ VCE · IC). If this heat is not removed, the junction temperature rises. Above the maximum rated junction temperature (about 150 to 200 °C for silicon), the transistor can be permanently damaged. Also, higher temperature increases leakage current ICBO, which increases collector current and heating further (**thermal runaway**).

The case of the transistor has a small area, so it cannot lose heat quickly to the air. A **heat sink** (a finned aluminium block attached to the case) provides a large surface area, and so lowers the thermal resistance between the junction and the ambient air. This keeps the junction temperature within the safe limit and allows the transistor to handle a higher power.

#### Thermal Ohm's law
Heat flow is analogous to current flow in a resistor:

| Thermal | Electrical |
|---|---|
| Temperature difference ΔT (°C) | Voltage V |
| Power dissipated PD (W) | Current I |
| Thermal resistance θ (°C/W) | Resistance R |

TJ − TA = PD · θJA

The path from junction to ambient has three thermal resistances in series:

θJA = θJC + θCS + θSA

```text
 Junction --[θJC]-- Case --[θCS]-- Sink --[θSA]-- Ambient
   TJ                TC               TS              TA
```

- θJC: junction to case (device data)
- θCS: case to heat sink (insulator or grease)
- θSA: heat sink to ambient (decreases with larger sink)

**Example:** TJ(max) = 150 °C, TA = 25 °C, θJC = 1.5, θCS = 0.5 °C/W, PD = 20 W. Required total θJA = (150 − 25)/20 = 6.25 °C/W, so the sink must have θSA ≤ 6.25 − 2 = 4.25 °C/W. Without the sink, θJA is typically 40 to 60 °C/W and the allowed power would be only about 2 to 3 W.

## Check yourself

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

1. Why is a heat sink needed on a power transistor?
   A. To increase the gain
   B. To provide a large surface area, lowering thermal resistance and keeping the junction below its maximum temperature
   C. To insulate the transistor
   D. To raise the leakage current
   Answer: B. The small case cannot lose heat quickly; excess junction temperature damages the device and can cause thermal runaway. (Key point: Need for heat sink)

2. Which is the thermal Ohm's law?
   A. TJ + TA = PD / θJA
   B. TJ - TA = PD / θJA
   C. TJ - TA = PD x θJA
   D. PD = TJ x θJA
   Answer: C. Temperature difference is analogous to voltage, power to current and θ to resistance: TJ - TA = PD θJA. (Key point: Thermal Ohm's law)

3. What is the total thermal resistance from junction to ambient?
   A. θJA = θJC x θCS x θSA
   B. θJA = θJC + θCS + θSA
   C. θJA = θJC
   D. θJA = θSA - θJC
   Answer: B. Junction-case, case-sink and sink-ambient resistances are in series and add. (Key point: Series thermal resistances)

4. How does thermal runaway develop?
   A. Higher temperature lowers ICBO, cooling the device
   B. Higher temperature raises leakage ICBO, raising IC and heating further
   C. The heat sink adds heat
   D. Lower PD raises temperature
   Answer: B. Rising temperature increases ICBO and collector current, which raises dissipation and temperature in a positive feedback loop. (Key point: Thermal runaway)

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