From: Bill Bass
Subject: Re: Limiting total wattage for a constant-current dummy load?
Date: Sat, 09 Nov 2002 21:42:03 -0600
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John Muchow wrote:
> I've got a 1.2-50V (approx.), 0 -10A constant current dummy load
> circuit (MAX480 op-amp, ICL8069 reference, and IRF540 for the load)
> working beautifully, but I have a problem I can't seem to find a
> solution for.
> I'm limited to about 50 watts for the load right now with the single
> MOSFET device and heat sink I'm using now. But, I can use the load
> with a current "setting" and load voltage that can significantly
> exceed the 50 watt limit if I'm not careful to figure out the wattage
> in advance.
> How can I allow both extremes of current/voltage (10A at up to 5V, or
> 1A(at up to 50V) to flow in this circuit but still limit the overall
> wattage dissipated? I can easily control the current to limit it to
> 10A through the IRF540, but I'm stumped as to how to use the voltage
> to determine the wattage the load would have to dissipate *before* I
> start the darn thing up.
> I'd love have an LED come on that told me "power is too high" if I
> selected a current level that would exceed 50 watts (or any arbitrary
> value) with the voltage present at the MOSFET load.
> Can anyone point me in the right direction? I couldn't find any info
> on the IR, National, Zetex, TI, Fairchild, or Linear Tech. web sites.
I've looked through the other responses and it suddenly came to mind that
if what you want to do is limit watts, you can do the same by limiting
Tie a thermistor to the heat sink and use it to sense excessive heat (or
think is too hot) and let it fold back the current driver. If you
tie it close to your power FET, it should respond quickly enough.
Just trying to take the easy way out.