From: Tony Williams
Newsgroups: sci.electronics.basics,sci.electronics.design,sci.electronics.misc
Subject: Re: Limiting total wattage for a constant-current dummy load?
Date: Sun, 10 Nov 2002 15:07:00 +0000 (GMT)
Organization: None
Message-ID: <4b933f88aetonyw@ledelec.demon.co.uk>
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In article <4b931cb6d0tonyw@ledelec.demon.co.uk>,
Tony Williams wrote:
> Vtop
> +
> |
> \
> /<--------------- V-wiper= (k * Vtop).
> \ /|\
> | Wiper position, k.
> | \|/
> --+----------------0v.
I've got a germ of an idea, but I'm stuck with
it... can't yet proceed beyond a certain point.
All we need to know is the wiper position, k,
of the current setpoint pot, and multiply that
by the V-Fet. The only way to do that so far
is to use a dual-pot, or multiplex a single.
I2= (G*VFet) +-->--+ +--<--+ I1= Iref
| <--R--> |
V2<--+---/\/\/\/--+-->V1
/|\
(1-k) | k
|
-------------+---------
If the pot has value R and the wiper is grounded then
we get two linked variable resistors, one of value
k*R and the other of value (1-k)*R.
Feed a fixed reference current (Iref) into the right
hand side, and V1 = (k * Iref * R), which is the variable
control voltage we need for the current servo opamp.
Feed a current proportional to VFet, (G*VFet), into the
left hand side. V2 = ( (1-k)*(G*VFet)*R ). This can
be expanded to....
V2 = (G*VFet)*R - k*(G*VFet)*R
~~~~~~~~~~ ~~~~~~~~~~~~
/|\ /|\
Variable offset Proportional to PFet.
That right hand term is the power in the MOSFET,
but it is accompanied by that variable offset.
I can't think of any easy ways (on a single supply
rail) to factor out the variable offset.
Any bright ideas?
--
Tony Williams.