[sdiy] Linear Current Source Math Question

[Scott Bernardi]

The best way to understand these is by using the ideal opamp
assumptions: In an opamp employing negative feedback,
1. the output will go whereever it needs to to force the negative input
equal to the positive input
2. the current drawn by the opamp inputs is negligible (zero).

Applying this to your circuit, Vin is applied to the + input, so
feedback forces the - input to be Vin also. One end of your resistor is
therefore at Vin, and the other side is at Vcc, therefore a current of
(Vcc-Vin)/R flows through it. Since the opamp input current is
negligible, this current flows into the pnp emitter, and approximately
the same out the collector.
Someone mentioned output compliance: The minimum Vce on the pnp
transistor should be about a diode drop (.6v - .7v) for good current
source operation. Since the pnp emitter is at Vin, then the pnp
collector should be no higher than Vin - .7v.
Another source of error comes from the fact that the opamp input current
is not zero. Use a JFET type, and unless you are trying to make a sub uA
current source, that error should be negligible.
There are useful variations of this circuit. For example, ground the +
input, and then make the one end of the current setting resistor be Vin
rather than Vcc. You can show that the output current will be Vin/R.
This is a great circuit for linearly driving OTA's (like for a VCA).
Vin must not go below ground. One way to protect from that is to add a
diode across the BE junction in the reverse direction (i.e., anode to
pnp base, cathode to pnp emitter).
See http://home.attbi.com/~sbernardi/elec/og2/og2_dual_linear_vca.gif
for example.

Scott Gravenhorst 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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