Thermal Calculation of Semiconductor IC/Mosfet

 The article which I have wanted to write for so long time. Actually, this is the topic that I completed and got deleted completely before publishing it. That is one of the tragic stories. 

Ok, Back to the topic. I used to think that calculating thermal things was the most difficult thing in electronics because of the thermodynamics subject I studied during college. However, once I understand the logic behind the calculation, I feel that this is one of the easiest things in electronics.

Just have to remember two formulas, that is, everything else is that derivation.

The thermal resistance is given by

`R_{\theta_{JA}}=\frac{T_{J}-T_{Amp}}{P_{D}}` ----(1)

Also as per the resistance, the thermal resistance given by

`R_{\theta_{JA}}= R_{\theta_{JC}}+R_{\theta_{CS}}+R_{\theta_{SA}}`    ---(2)

Equating both equations we get,

`\frac{T_{J}-T_{A}}{P_{D}}=R_{\theta_{JC}}+R_{\theta_{CS}}+R_{\theta_{SA}}` ---(3)

The thermal resistance is expressed in °C/W.

That is, now I know the basic formula to calculate the junction temperature by using the above formula

But just by knowing the formula, obviously no use. Let's put that equation into action.

Let's take the example of one IC whose parameter is below.

 i. Maximum Junction Temperature , `T_{J}`=150 deg celsius

ii. Thermal Resistance Junction to case,  `R_{\theta_{JC}}`= 4 °C/W

iii. Thermal Resistance of Junction to Ambient, `R_{\theta_{JA}}`=60 °C/W

  iv. Asumming Thermal resistance of case to sink, `R_{\theta_{CS}}`=2 °C/W

v. Ambient Temperature= 30 °C

v. Power dissipation, `P_{D}` = 3W

Case 1: With no sink

From Eqn (1), Junction Temperature, `T_{J}=(R_{\theta_{JA}}\times P_{D})+ T_{Amp}`

=(60 x 3) + 30

`T_{J}`=210 °C

So, Without a heatsink, the temperature of the junction rises up to 210°C for the power dissipation of 3W.

Case 2: Using the heatsink at Maximum Junction temperature.

By using eqn (3),

                                                                `\frac{T_{J}-T_{A}}{P_{D}}=R_{\theta_{JC}}+R_{\theta_{CS}}+R_{\theta_{SA}}`

`{T_{J}-T_{A}}=(R_{\theta_{JC}}+R_{\theta_{CS}}+R_{\theta_{SA}})\times{P_{D}}`

                                                                `150-30=(4+2+R_{\theta_{SA}`) x 3

                                                                `R_{\theta_{SA}` = 34 °C/W

Heat sink with sink to ambient thermal resistance of 34 °C/W should be used to use the IC at the junction temperature of 150 °C.

Let's say I am using a heat sink with   `R_{\theta_{SA}`= 24 °C/W

Then, the Junction temperature of IC,  `T_{J}` = (4+2+ 24) x 3 + 30

`T_{J}`=126°C

The IC junction temperature rises to 126 °C.

There are some points to be noted.

1. The thermal resistance of the sink to the ambient depends on the selected heat sink. The thermal resistance should be calculated from the heat sink graph.
2. The thermal resistance of the case to the sink depends on the thermal conductor used to attach the case and sink. The pastes normally used are.
1. Thermal Grease-->0.2 °C/W
2. Mica-->0.4 °C/W
3. Silicon pad-->1.1 °C/W

By using the forced cooling method such as Fan, the heat can be further reduced. The thermal resistance while forced cooling will also be given in the heat sink graph based on the air velocity of the fan.

Few good references for understanding the thermal calculation

Power Electronics - Thermal Management and Heatsink Design-by  Power Electronics YouTube channel

Power Electronics - Thermal Considerations- by Power Electronics with Dr. K YouTube