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Old 28th December 2004, 02:18 PM   #1
DarkOne is offline DarkOne  Slovakia
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Default Thermal compensation

Hello,
does anybody have some clever idea how to implement thermal compensation to attached circuit? Maybe I'm too tired from work or something, I cannot think about it anymore.
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Old 29th December 2004, 02:39 PM   #2
DarkOne is offline DarkOne  Slovakia
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Nobody has some clever idea?

OK, I think that I can run without compensation with 8 pairs of output devices and 0R47 source resistance. Bias current with cold heatsink(0.1K/W) would be at about 2A and after warmup maybe 4-5A (not bad). According to graphs in IRFP240 datasheet Vgs value changes 0.6V in a range from 25degrees to 150degrees. My heatsink might have temperature about 50degrees so worst case is change in Vgs maybe 0.2V which gives current increase about 0.4A per output pair (8x0.4=3.4A). After warmup bias current will reach 5.4A
If this will not work, I can increase source resistance to 1R

Isn't it nice?
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Old 29th December 2004, 03:14 PM   #3
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The way I see it, the bias is determined by the standing current in the ltp's. You may want to try thermal coupling from the heatsink to D1/M6 and/or D2/M5, but I don't know what the thermal gain is, or if the phase is correct. Did you think about that?

Or put diodes in series with the load resistors in the ltp's and couple those thermally to the heatsinks.

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Old 29th December 2004, 03:22 PM   #4
DarkOne is offline DarkOne  Slovakia
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If I attach D1/M6 and D2/M5 to heatsink the CCS will produce more current, voltage across drain resistors will rise, then current in output stage will rise and my fuses will blow.
Diodes are not bad, but then I have to use about 6 diodes to compensate Vgs.
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Old 30th December 2004, 05:28 AM   #5
PRR is offline PRR  United States
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> how to implement thermal compensation to attached circuit?

Well, what is the idle current in the output stage? Something like I(M5) times R10 times M9's Vg/I curve. Therefore I(M5) must track M9's needs. So replace D2 with an IRFP9240's Gate-Source voltage, biased at the desired output current, thermally bonded to an output device. (Sort of like a diode-strapped BJT, only with a MOSFET.) Then replace M5 and its R14 with a precision voltage-to-current converter.

I omitted a "/2" in the equation, and the fact that you need to duplicate (or mirror) the reference to the other rail, but these tidbits should be obvious.
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Old 30th December 2004, 08:55 AM   #6
DarkOne is offline DarkOne  Slovakia
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Thanks everybody.

Version of PRR seem to me too complicated and I want to go as simple as possible. So I will not use any compensation and I'll see.

NP wrote in one thread about temp compensation that he gives more source resistance and sleep well at night. I think I'm gonna sleep well.
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