How do I calculate the size of the heat sink?

sky999

How do I calculate the size of the heat sink? [Copy link]

 

Assuming a MOS tube switches 24V/10A current, how do you calculate the size of the heat sink? The MOS tube model is IRF3205, thank you

luominxdy

Follow along and learn

qwqwqw2088

The power loss of the MOS tube is mainly the on-resistance Rds(on).

Total thermal resistance θJA=θJC+θCS+θSA

θJC Thermal resistance from junction to package

θCS Thermal resistance from package to heat sink

θSA Thermal resistance from heat sink to ambient

qwqwqw2088

in

θJC can be found in the datasheet, the typical value for IRF3205 is about 1.6°C/W

θCS and θSA need to be estimated or determined experimentally

qwqwqw2088

Power loss, at 10A current, power loss is calculated using the formula Pd = I square * Rds(on)

I is the current flowing through the MOS tube,
Rds(on) is the on-resistance of the MOS tube.

According to the datasheet of IRF3205, when Vgs=10V, Rds(on) is about 0.065Ω.

The power loss is: Pd = 10 square × 0.065 = 6.5W

qwqwqw2088

In actual operation, suppose we want the operating temperature of the MOS tube not to exceed 125°C

This is the maximum junction temperature of the IRF3205, and with an ambient temperature of 25°C, the maximum allowed temperature rise is 100°C.

ΔT=12525=100°C

We need to ensure that the total thermal resistance is low enough so that the temperature rise does not exceed 100°C for a power loss of 6.5W.

Total thermal resistance θJA = ΔT / Pd=100 / 6.5 ≈15.4°C/W

Choosing θJC = 1.6°C/W
, θCS + θSA must be less than about 13.8°C/W.
For simplicity, we assume that θCS is small or negligible, and focus on θSA.

The assumption is that θSA is between 10°C/W and 20°C/W.

A heat sink should be selected with a thermal performance θSA as close to 10°C/W as possible.

qwqwqw2088

The middle reply above, I don’t know which sensitive word is in it, I entered the review area and took a screenshot

There are many factors that affect the selection of heat sinks, such as the air flow in the circuit board and the natural convection cooling. For example, if a power supply has a fan inside, it is easier to choose a heat sink.

The calculation is only for reference, and in many cases the heat sink is based on empirical values.

tagetage

The easiest way is to refer to other devices and use the same size heat sink as the IRF3205.

sky999

qwqwqw2088 posted on 2024-7-28 09:31 I don’t know which sensitive words are in the middle reply above. I entered the review area and took a screenshot of the heat sink...

Thank you. I read your reply several times, but I still don't understand. Assuming the total thermal resistance is 15 degrees/W, what size heat sink should I choose? How do I calculate it?

qwqwqw2088

Based on the total thermal resistance and heat generation, the required heat dissipation area A can be calculated,
A = P / (θSA * ΔT), where P is the heat generation and ΔT is the temperature difference between the heat sink and the surrounding environment.

The specific heat dissipation area also needs to consider the number, height, and length of the fins.
This is the theoretical calculation. As mentioned on the 8th floor, the specific size of the heat sink should be the same as the heat dissipation surface of the IRF3205, or slightly larger.

okhxyyo

qwqwqw2088 posted on 2024-7-28 21:18 According to the total thermal resistance and heat generation, the required heat dissipation area A can be calculated, A = P / (θSA * ΔT), P is the heat generation, ΔT ...

It's too complicated. Great! QW has explained it in great detail.

sky999

qwqwqw2088 posted on 2024-7-28 21:18 According to the total thermal resistance and heat generation, the required heat dissipation area A can be calculated, A = P / (θSA * ΔT), P is the heat generation, ΔT ...

OK, thanks