Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Post History

71%
+3 −0
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 8mo ago by Lorenzo Donati‭

#3: Post edited by user avatar Lorenzo Donati‭ · 2023-08-12T15:18:28Z (8 months ago)
Retagged.
#2: Post edited by user avatar Olin Lathrop‭ · 2022-02-25T20:11:01Z (about 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 (about 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?