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Author Topic: transistor junction temperature  (Read 321 times)
AJ4SN
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Posts: 67




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« on: November 14, 2017, 01:33:19 PM »

I am working on a 200 watt solid state amplifier, and I want to manage the transistor heat properly.  The VRF151 transistor that I plan to use has a maximum junction temperature of 200 degrees centigrade. I'd like to run it at about 130 degrees centigrade for longer life. It has a thermal resistance rating of 0.6 degrees centigrade per watt. My question is the relationship between case temperature and junction temperature.

If I am running two transistors in push pull for a total output of 200 watts, and the amplifier efficiency is 50%, then each transistor has to dissipate 100 watts. The thermal "drop" across the case is: 100 watts x 0.6 C/W = 60 degrees C.

In order to keep the transistor junction temperature at 130 degrees, the transistor case temperature must not exceed 70 degrees (130 - 60 = 70). Is that correct? Does the ambient air temperature enter into the calculation?

Thanks for your help!

Stan
aj4sn
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AA4PB
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Posts: 14327




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« Reply #1 on: November 14, 2017, 02:06:40 PM »

Correct. The ambient temp comes into play when calculating the heat sink. The heat sink temperature rise above ambient will be rated in deg C per watt.

It's all like a series string of heat loss: junction - case - heat sink - ambient.
« Last Edit: November 14, 2017, 02:12:19 PM by AA4PB » Logged

Bob  AA4PB
Garrisonville, VA
N5EG
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Posts: 313


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« Reply #2 on: November 14, 2017, 02:13:01 PM »

Hi Stan - Yes, your calculations are in the right format. 
The thermal specification for the transistor is probably junction-to-case.
The computed temperature rise of 60C means the temperature rise of the
junction above the case of the transistor.

However there are some other sources of thermal resistance you also need to
include in your analysis:

1. Transistor case to heatsink thermal resistance.
2. Heatsink to ambient air thermal resistance. This thermal resistance will decrease
as you go from no forced air to forced air cooling.

Thermal resistance works like electrical resistance - you add the various
thermal resistances to compute the total thermal resistance.

As a hypothetical example:

Let's assume that the transistor-case-to-heatsink thermal resistance is 0.2 C/W, and that
the heatsink-to-forced-air thermal resistance is 0.5 C/W.    You'll need to find the actual
numbers for your exact hintsink, transistor mounting method, etc. These are just hypothetical.

The total thermal resistance from the transistor junction to ambient air would be:
0.6 C/W + 0.2 C/W + 0.5 C/W  or  1.3 C/W.

At 100 watts per transistor, then:

Tj = 100 * 1.3 = 130 C rise over the ambient air temperature.  If ambient air were
25C, then the heatsink temp would be  25C + 0.5 C/W * 100 W = 75 C
and the case of the transistor would be   75C + 0.2 C/W * 100 W = 95 C
while the junction would be  95C + 0.6 C/W * 100W = 155 C

-- Tom, N5EG





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AJ4SN
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Posts: 67




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« Reply #3 on: November 15, 2017, 05:54:49 PM »

Bob, Tom,

Thanks for taking time to reply. I appreciate it. Tom, your explanation makes it very clear. I think this is a fascinating subject. I'm going to try to learn more about it as I continue with this project.

73
Stan
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