From: Tony Williams
Subject: Re: Limiting total wattage for a constant-current dummy load?
Date: Sun, 10 Nov 2002 08:46:39 +0000 (GMT)
References: <email@example.com> <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
NNTP-Posting-Date: Sun, 10 Nov 2002 08:53:39 +0000 (UTC)
User-Agent: Pluto/1.14i (RISC-OS/3.60)
In article <email@example.com>,
John Muchow wrote:
> Thanks for the explanations. I'm sure (ok, I'm hoping) that it will
> all be a lot clearer to me when I proto up a circuit.
/<--------------- V-wiper= (k * Vtop).
| Wiper position, k.
A pot is a simple analogue multiplier. If k is
the wiper position (varying from 0 to 1) then
V-wiper= (k * Vtop).
If the pot controls I-Fet over the range 0 to 10A,
then I-Fet= (k * 10) Amps. Therefore at any
current setting the wiper position, k = I-Fet/10.
If the pot is temporarily switched over so that
Vtop= V-Fet/10, then V-wiper2 = k * V-Fet/10.
OR; V-wiper2 = I-Fet/10 * V-Fet/10 = P-Fet/100.
V-wiper2 = >0.5V indicates that P-Fet will exceed 50W.
Any analogue multiplier will produce the same req'd
result, but the problem is finding one that works on
a single rail... in fact it might require generating
a negative rail (not too difficult though).
> Currently, I have a single +12V rail but would like to swap out the
> AC-to-DC power supply for a 9V battery (for portability reasons).
Two problems there.... 9V restricts your maximum
possible Vg-s on the MOSFET which means you may not
be able to get 10A, especially at low Vdrain-source.
Look at the data sheet for your MOSFET, at the output
curves. The second problem is fan power. At such a
high potential power dissipation you need a fan
with a serious airflow, which is possibly not battery