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'Constant Current source'
1997\07\02@122617 by

Hi!

Anyone know how to build a constant current source using Op-amps. I know
how to do it with transistors, but I want to try it with op-amps.

Thanks.
--
eric van es
cape town, south-africa
http://www.nis.za/~vanes/
>Anyone know how to build a constant current source using Op-amps. I know
>how to do it with transistors, but I want to try it with op-amps.

Hi Eric,

Try this, from a National Semiconductor Ap Note:

R3
-----/\/\/\----
|                |
|                |
R1     |    | \         |
E_in ---/\/\/\-------| - \       |      I = E_in * R3
|    \------          -----------
----|+   /      |           R1 * R5
|    | /         |
|                |          R3 = R4 + R5
|                \          R1 = R2
|                /   R5
|                \
|                /
|       R4       |
|-----/\/\/\-----|
|                |  <---Constant current = I
|                |
\              -----
/             |     |
/             |     |
|              -----
|                |
GND              GND

--
Bob Fehrenbach    Wauwatosa, WI     bfehrenbexecpc.com
On Wed, 2 Jul 1997 18:24:28 +0200 EJA van Es <vanesilink.nis.za> writes:
>Hi!
>
>Anyone know how to build a constant current source using Op-amps. I
>know
>how to do it with transistors, but I want to try it with op-amps.
>

If you can put up with a floating load, just put the load between
the output of an op-amp and the inverting input, then put a resistor from
the inverting input to ground.  Apply a voltage to the non-inverting
input.  Since the inputs will "follow each other", the inverting input
will be at about the same voltage, causing I=V/R through the resistor to
ground.  Since an ideal op amp has no input current, this current also
You can also ground the noninverting input and apply a voltage to
the end of the resistor that was grounded.  The "happy op amp" will keep
the two inputs together, keeping the inverting input at ground.  The
current through the resistor and the load will be I=V/R, since the
voltage at one end of the resistor is zero.  This is the basis of an
inverting integrator.  The controlled current flows through the capacitor
(the load).  Since the inverting input is at ground, the op amp output
voltage is the same as the capacitor voltage.
Another circuit with a grounded load is described below (my ascii
art isn't very good).  Imagine two equal value resistors in series.
Connect the load from the junction of the two resistors to ground.
Connect the reference voltage to one end of the series resistors.
Connect the junction of the resistors (and the load) to the input of a
noninverting amp with a gain of 2.  Connect the output of the amp to the
other end of the two resistor network.  Clever circuit!  Try a few ohm's