From: "Boris Nogoodnik"
Subject: Re: Limiting total wattage for a constant-current dummy load?
Organization: KGB, Ltd.
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Date: Sat, 09 Nov 2002 09:51:52 GMT
NNTP-Posting-Date: Sat, 09 Nov 2002 04:51:52 EST
"John Muchow" wrote in message
> I've got a 1.2-50V (approx.), 0 -10A constant current dummy
> circuit (MAX480 op-amp, ICL8069 reference, and IRF540 for the
> working beautifully, but I have a problem I can't seem to find
> solution for.
> I'm limited to about 50 watts for the load right now with the
> MOSFET device and heat sink I'm using now. But, I can use the
> with a current "setting" and load voltage that can
> exceed the 50 watt limit if I'm not careful to figure out the
> in advance.
> How can I allow both extremes of current/voltage (10A at up to
> 1A(at up to 50V) to flow in this circuit but still limit the
> wattage dissipated? I can easily control the current to limit
> 10A through the IRF540, but I'm stumped as to how to use the
> to determine the wattage the load would have to dissipate
> start the darn thing up.
> I'd love have an LED come on that told me "power is too high"
> selected a current level that would exceed 50 watts (or any
> value) with the voltage present at the MOSFET load.
> Can anyone point me in the right direction? I couldn't find
> on the IR, National, Zetex, TI, Fairchild, or Linear Tech. web
Perhaps you need to use an analogue multiplier to multiply
current and voltage and then feed the output into contol
cuircuit. Look at AD835AN. Just an idea.