I just need to select a Heat sink for Lm317/337, i have Texas Instrument T220 devices and i am a little confuse about; the Junction-to-ambient thermal resistance is 23.3 °C/W, the Junction-to-case (top) thermal resistance 16.2°C/W.With the same T220 (Case 221AB) device with On conductor the Junction-to-ambient thermal resistance is 65 °C/W and the Junction-to-case (top) thermal resistance 5°C/W.
What values should i use for heatsink calculation ??
What values should i use for heatsink calculation ??
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Yes that is right and ?? why that huge difference??Junction to ambient is for free air (no sink).
Yes that is right and ?? why that huge difference??
I would trust the mfr specs, and use those for your device. Small differences in the construction
can make large differences in the thermal resistance. As can mounting torque, esp with TO-220.
i means that when it produces 1 watt of heat it will get 5degrees hotter than the heatsink,if you disipate 10 watts ,it will get 50 degrees hotter than the heatsink.You need to calculate the watts lost (amps drawn X voltage dropped),and then find a heatsink that has a low enough c/w rating that it won't get over 90 degrees and then add the internal temperature to that
A Rtja of 23.3°C/W for a TO220 sounds like a fairy tale: it means the naked TO220 could dissipate over 5W....
It probably has something to do with the so called new thermal metrics
You should study them thoroughly, to find where the catch hides, and use the proper figure for your calculations
It probably has something to do with the so called new thermal metrics
You should study them thoroughly, to find where the catch hides, and use the proper figure for your calculations
For 1A and 11V delta (Vin-Vout) i need 10℃/W with TI Mfr specs and 1℃/W with On Semi Mfr specs...10 Time Morei means that when it produces 1 watt of heat it will get 5degrees hotter than the heatsink,if you disipate 10 watts ,it will get 50 degrees hotter than the heatsink.You need to calculate the watts lost (amps drawn X voltage dropped),and then find a heatsink that has a low enough c/w rating that it won't get over 90 degrees and then add the internal temperature to that
Yes for sure, something do not song right.......look to good to be right, i will test with a 10℃/W heatsink...A Rtja of 23.3°C/W for a TO220 sounds like a fairy tale: it means the naked TO220 could dissipate over 5W....
It probably has something to do with the so called new thermal metrics
You should study them thoroughly, to find where the catch hides, and use the proper figure for your calculations
You should study them thoroughly, to find where the catch hides, and use the proper figure for your calculations
When you download the SPRA953 document and read it through, you find this :
In fact, in still-air JEDEC-defined R θJA measurements, almost 70%–95% of the power generated by the chip is dissipated from the test board, not from the surfaces of the package.
Anyone fancy downloading the JEDEC specs to find out what the test board looks like? My guess is it has an awful lot of copper surface area.
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