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Q&A Case temperature of MOSFET

Homework problem A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given t...

1 answer  ·  posted 2y ago by Carl‭  ·  edited 1y ago by Lorenzo Donati‭

#3: Post edited by user avatar Lorenzo Donati‭ · 2023-08-12T15:18:28Z (over 1 year ago)
Retagged.
#2: Post edited by user avatar Olin Lathrop‭ · 2022-02-25T20:11:01Z (over 2 years ago)
Always be clear about something being homework, when it is.
  • > A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given the thermal resistances: -
  • >
  • > Junction to case: $R_{th,jc}= 1.7 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > Isolation pad: $R_{th,cs}= 1.3 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > Heatsink to ambient air: $R_{th,sa}= 2.0 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > The ambient temperature around the MOSFET is 25°C. What is the temperature of the MOSFET case?
  • **My attempt**
  • The scenario must look something like this: -
  • ![Image alt text](https://electrical.codidact.com/uploads/rXQ96htJtGs3ob1fiMihNpyr)
  • Here I have represented the thermal resistances as resistors, the dissipated power as a current source, and the temperatures as potentials. From this I calculate the following: -
  • $$P_d = \frac{T_c-T_a}{R_{th,cs}+R_{th,sa}} \Leftrightarrow$$
  • $$T_c = P_d(R_{th,cs}+R_{th,sa})+T_a = (10\cdot 3.3)°C + 25°C = 58°C$$
  • So the case must be 58°C.
  • Is my thought process/calculations correct or have I misunderstood something? Moreover, are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?
  • **Homework problem**
  • > A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given the thermal resistances: -
  • >
  • > Junction to case: $R_{th,jc}= 1.7 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > Isolation pad: $R_{th,cs}= 1.3 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > Heatsink to ambient air: $R_{th,sa}= 2.0 \: \frac{°\text{C}}{\text{W}}$
  • >
  • > The ambient temperature around the MOSFET is 25°C. What is the temperature of the MOSFET case?
  • **My attempt**
  • The scenario must look something like this: -
  • ![Image alt text](https://electrical.codidact.com/uploads/rXQ96htJtGs3ob1fiMihNpyr)
  • Here I have represented the thermal resistances as resistors, the dissipated power as a current source, and the temperatures as potentials. From this I calculate the following: -
  • $$P_d = \frac{T_c-T_a}{R_{th,cs}+R_{th,sa}} \Leftrightarrow$$
  • $$T_c = P_d(R_{th,cs}+R_{th,sa})+T_a = (10\cdot 3.3)°C + 25°C = 58°C$$
  • So the case must be 58°C.
  • Is my thought process/calculations correct or have I misunderstood something? Moreover, are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?
#1: Initial revision by user avatar Carl‭ · 2022-02-25T17:54:04Z (over 2 years ago)
Case temperature of MOSFET
> A MOSFET has a power dissipation of $P_{d} = 10\text{W}$. The MOSFET is mounted on a heatsink. There is an isolation pad between the MOSFET case and the heatsink. We are given the thermal resistances: -
>
> Junction to case: $R_{th,jc}= 1.7 \: \frac{°\text{C}}{\text{W}}$
>
> Isolation pad: $R_{th,cs}= 1.3 \: \frac{°\text{C}}{\text{W}}$
>
> Heatsink to ambient air: $R_{th,sa}= 2.0 \: \frac{°\text{C}}{\text{W}}$
>
> The ambient temperature around the MOSFET is 25°C. What is the temperature of the MOSFET case?

**My attempt**

The scenario must look something like this: -

![Image alt text](https://electrical.codidact.com/uploads/rXQ96htJtGs3ob1fiMihNpyr)

Here I have represented the thermal resistances as resistors, the dissipated power as a current source, and the temperatures as potentials. From this I calculate the following: -

$$P_d = \frac{T_c-T_a}{R_{th,cs}+R_{th,sa}} \Leftrightarrow$$
$$T_c = P_d(R_{th,cs}+R_{th,sa})+T_a = (10\cdot 3.3)°C + 25°C = 58°C$$

So the case must be 58°C. 

Is my thought process/calculations correct or have I misunderstood something? Moreover, are calculations like these useful/used when designing circuits or are they only good for "back of the envelope" calculations?