[sdiy] Linear Current Source Math Question

[Tim Ressel]

Scott,

It's almost that easy, yes. The voltage drop across R
due to the current flowing is held equal to the input
voltage. If this is a current sourse, then the input
voltage is referenced to Vcc, because R goes there. In
a current sink, it is all ground referenced.

There are a few caveats. Compliance range has to do
with the maximum voltage the collector can do. In
other words, if the load resistance times the current
equals more than Vcc minus about a volt and a half,
then the source quits working.

The accuracy is quite good. The opamp cancels the
nasty things the transistor does. The better the
opamp, the better the accuracy, of course. And the
tolerance (and tempco for larger currents) of the
resistor R affects accuracy.

--Tim

--- Scott Gravenhorst <music.maker at gte.net> wrote:
> I've been reading in chapter 4 of "The Art of
> Electronics"
> about linear current sources that use a bipolar
> transistor
> (b-e junction) in the feedback loop of an op-amp. 
> It states
> that the emitter current is (Vcc-Vin)/R and that
> Iout is the
> same as Ie.  Is it really that simple that one needs
> only
> calculate the current in the feedback loop and that
> current
> will be the output current (Ic)?
> 
>
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