[sdiy] Linear Current Source Math Question

[Scott Bernardi]

In normal mode you don't really have to worry about limiting the base current
because the current is limited at the emitter. One exception to that is that
if you saturate the transistor, then current will flow from the opamp output
to the now forward biased BC junction.
You're right about the difference between emitter and collector currents. Ic
= Ie*(Beta/(Beta+1)), so for a Beta of 100 that is a 1% error. That's running
into the tolerance range for your current setting resistor, so often it's
ignored (especially since Beta is usually 200 or more for discrete
transistors).
There are other useful uses for transistors in feedback loops of opamps. For
example, If you want to make a voltage source with boosted current output
(that is, more than the 5mA-10mA an opamp will give you.  In this case, for a
positive reference, use an npn transistor with collector to Vcc, emitter to -
input, and base to opamp output. Input voltage goes to the opamp + input,
output voltage is at the emitter.  This circuit should have current
protection added, as shorting the Vout to ground will fry the transistor.
Jim Patchell's temp compensated VCO uses these for voltage references (see
http://www.silcom.com/~patchell/synthmodules/vco.html).


Scott Gravenhorst wrote:

> Thanks to all who helped me with this.  I think I finally understand how
> this cicuit works.  If I may:
>
> Ie is really the sum of Ic and Ib, something not mentioned in A of E,
> they probably thought this was obvious.  But because of the transistor's
> current gain (beta, I believe), Ib will be a small fraction of Ie and Ic
> will be the bulk.  This is what causes the small error in output current
> and is why using a darlington further minimizes this error due to the
> current gain being much larger, the product of the gains of the two
> transistors.
>
> Further thinking here has me note that this curcuit requires that the
> collector be connected to it's load, otherwise, the opamp will try to
> push all of the current through the b-e junction of the transistor.  Most
> of the times I've seen this circuit, it has no resistor to limit base
> current at all.  In the FatMan, there's a 47 ohm in series with the base
> and 1K in series with the emitter, but the total being 1047 means a
> possible base current of about 11.5 milliamps.  So then I looked at the
> data sheet I have for a 2N3906, and I see no maximum base current
> parameter listed.  ??  There must be *some* limit, all diodes have a
> maximum forward current...
>
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