Re: Simplest current regulator

On Jan 29, 5:07 am, whit3rd <whit...@xxxxxxxxx> wrote:
On Jan 28, 4:03 pm, Phil Endecott <spam_from_usenet_0...@xxxxxxxxxxxx>
wrote:

Of course I can use a zenner with a potential divider in an op-amp
circuit, but I need help to use it in something simpler.

Well, I'd think an LM10 and pass transistor is fairly
simple (half a square inch of circuit board), but there's
another option that hasn't been discussed yet: a current
mirror.

Just run a resistor from V+ to base of two transistors,
both emitters to GND, and connect collector 1 to base,
and LED from V+ to collector 2.

The trick is, the transistors have to be on a common heat sink
and the first has to have much smaller emitter area than the
second (the collector current is scaled by the emitter area,
all else being equal).    LED current is N*((V+) - Vbe))/R
and the 'N' represents the area ratio.

ICs use this kind of trick all the time, but buying transistors in
onesies it's hard to know what N is going to be.

You can buy dual transistors - Farnell list a bunch, of which the
PBSS4240Y looks good - and mimic the effect of emitter area by
arranging the Vbe for one transistor to be higher than that of the
other. At room temperature 86mV of extra bias gives you a 30:1 current
ratio.

You'd use a 300R resistor to draw some 12mA from the battery; 10mA of
that would go into the collector of the Vbe-setting transistor, 1mA
would go through an 82R resistor to the base of the Vbe-setting
transistor, and from there most of that 1mA would go on to the other
terminal of the battery through a 680R resistor across the Vbe of the
Vbe-setting transistor, while the remaining 1mA would provide the base
current for the other half of the dual transistor, which would be
sinking roughly 300mA from the battery via the white LED.

The junction of the drive transistor would be dissipating of the order
of 300mW, so it would be warmer than the junction of the Vbe-setting
setting transistor (which would be dissioating only about 8mW); in a
dual transistor both junctions should be on the same substrate and
physically close, so the difference ought to be small.

If I were building the circuit, I wouldn't start of with a 82R and
680R to bias the base of the Vbe-sensing transistor, but rather with a
200R pot in series with 560R, and use the wiper of the pot to drive
the base; you'd begin with the pot set to give you 0R and 760R, then
move the wiper down towards 82R and 678R while monitoring the current
drawn through the LED.

The circuit would be a bit crude, but it would work.

--
Bill Sloman, Nijmegen

.

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