Re: Current source design (tricky?)

From: Larry Brasfield (donotspam_larry_brasfield_at_hotmail.com)
Date: 03/10/05 

Date: Thu, 10 Mar 2005 10:23:18 -0800

<fxalpha@gmail.com> wrote in message
 news:1110457920.604692.305640@g14g2000cwa.googlegroups.com...
> Hi,
Hi.
> thank you for taking the time to look through my requirements.
You are welcome.

> as for the calrifications:
>
> I'd say the maximum difference between Vcc and the load will be 3 to 4
> volts. the minimum should be as low as possible, say 0.5 Volt. on 2
> amps, that gives 8 watts of dissipation in the pass element. is it
> reasonble for to-220 and only a small heatsink?

That helps, but you are still looking at 18 W dissipation under
"normal" circumstances. (I rely on the 3 Amp output you
mentioned in your earlier post and battery charging voltage
of 16V.) During a load dump transient, that could briefly go
to 75W. (The transient is brief enough that you need not size
the heatsink for it, but it may mean that you want to keep the
junction cooler to provide some headroom for the thermal
transient.) Whether that can be managed with a small heatsink
is questionable. I would rely on the heatsink vendor's data***.

> as for the "reasonably constant current", I need about +/- 5%, as I
> specified later in the post.

So, the response to a voltage transient must stay within
that bound. This makes the design a little more interesting
and likely makes it a good idea to low pass filter the
incoming supply for both the FET and op-amp. You will
likely want to put a resistor in series with the integrator
feedback capacitor to improve the transient response
of the current feedback loop.

> you were right, since the load and Vcc are changing, the conditions
> should be more accurately called "nearly DC" I suppose. I approximate
> the load would only change at a max rate of 1-2% per second (because of
> heating). the voltage would also drop or rise slowly. the exception is
> that if this would be connected to a 12V source from a car, and the car
> starts, the voltage would rise from 12V to 13.8 or so pretty quickly
> (do you have an idea how fast, or how bad would transient spikes
> durning ignition would be?)

I was not trying to quibble about the meaning of "DC".
Ignition transients are pretty easy to filter out before
they hit the active part of your circuit. The load dump
transients change much faster than you appear to
suspect. You will need to research them.

> the idea of an opamp controlling a P mosfet sounds interesting. I
> suppose that for 0.5V minimum drop, I need a mosfet with Rds(on) of
> 0.25 Ohm. is that a reasonble requirement? (for a single digit price
> tag, that is). would the opamp work well if the drop on the sense
> resistor be only 100mV or so?

Yes. Maybe.

> if you can give me a recommendation for a single supply rail to rail
> opamp, as well as a mosfet the will do, I'll be most grateful. I don't
> do much analog, so some pointers would save me lots of time.

I will not claim these are the most cost effective parts,
but they will do the job:
  http://ec.irf.com/v6/en/US/adirect/ir?cmd=catProductDetailFrame&productID=IRF9Z24
  http://www.national.com/pf/LM/LM8261.html
This op-amp, suggested by Mr. Hill, is well suited for this application
due to its positive rail input range and capacitive load tolerance.

Just a few tips for your design:

Be sure to zener protect the op-amp supply.
Transients on an automobile "12V" rail can be
surprisingly large due to quickly changing loads,
especially removed loads, and the way alternators
are regulated. (This effect is often called "load
dump".) I would use a 24 or 27 Volt zener diode,
grounded at one end.

Unless you are willing to see an output current
spike sometimes, the MOSFET output stage
should get similar protection. I would use a
power rectifier from the low end of the current
sense resistor to the zener mentioned above.
You should satisfy yourself that the zener's
power rating is sufficient to absorb a load
dump transient. (I would have to research
this to know what that might be.)

Your op-amp circuit will be operating with
inputs at the positive (filtered, protected)
rail, and its output is referenced to that rail
by the feedback condition, so be sure not
to ground reference those parts of the
circuit that influence the controlled current.

> thanks again for the responses!
You're still welcome.

> Guy. 

-- 
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
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