V G wrote:
> I didn't know the gain worked like that. I assumed 10 mA into the
> base =3D 1000 mA pulled by the collector.

It does.  The "gain" of a BJT (more specifically Hfe) is the collector
current divide by the base current.  If the base current is 1, then the
collector current is GAIN, and the emitter current is GAIN+1.  In your case=
,
the collector current is the battery current, but what you measure with a
low side current sense is the emitter current.  Note that the ratio of GAIN
to GAIN+1 approches 1 as the gain gets large.

> Also, if I use a Darlington pair, should I use a small transistor as
> the first transistor and the large transistor for the second? Or
> should both of them be the same large transistor?

Stop and think.  Even a big fat brick-outhouse NPN power transistor can do =
a
gain of 10.  By definition, its base current will therefore be 1/10 the
battery current.  The second transistor only handle this base current, so
doesn't need to be anywhere near that beefy.

As I said before, use a 2N3055 for the main transistor.  These come in TO-3
package and can dissipate lots of power with a decent heat sink.  You can
then probably use a TO-220 resistor in free air for the base drive.


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