```Geoff Raynak <graynak@u.washington.edu> writes:

>Sorry for the confusion.  I meant to say that I tried the circuits
>provided by H&H but they don't seem to be _precision_ in the sense that
>they cannot maintain the load against the resistance.  I am trying to
>sense geometry changes in a tube filled with saline.  Using a resistance
>method, i need a constant current source (very constant) to measure
>resistance-induced voltage changes without any current-induced voltage
>changes.

>thanks again
>geoff

A couple of thoughts come to mind:

Transistor based:

Cascode or modified Wilson source:

Gnd--R--+                |
|               \_/  Iin
+-----+          |
|     |          |
\    |         /
|---+--------|
/              \
|                |
|                |
|          +-----+
|          |     |
\         |    /
|--------+---|
/              \
|                |
|                |
-V               -V

All these attempted transistors are npn's (this is the modified Wilson
source).  It should keep compliance to within about 1V of the -V rail
which is well within your request.  You can set the output current
with the input resistor (approximate current is (V-2Vbe)/R.  You need
transistors with a VCEmax rating high enough to meet whatever voltages
you should expect to see.  This accuracy of this will require that the
temperature of the lower pair remain pretty constant (try touching one
of the lower transistors and watch the output current change...).  The
output resistance of this current source should be pretty large
O(beta*ro) where beta is the beta of the upper transistor on the right
and ro is the output resistance of the lower transistor on the right
(2N3904 at 1mA is about 100k, I think).

monolithic approach:
modified Howland Source

V2--R1-+---R1-----+
|          |
| |\       |
| | \     R2
+-|- \     |
|   >----+
+-|+ /     |
| | /     R2
| |/       |
|          |
V1--R1-+---R1-----+
|
|
\_/  Iout
|

This looks like a modified instrumentation amplifier.  There are 2
resistors in series going from the output to the input.  Your current
source output is actually between the 2 resistors.  With a difference
of voltage applied to the input, you will get an output of
Iout=(V1-V2)(1/R1 + 1/R2).  The output impedance depends on the CMRR
of the difference amplifier.  Burr-Brown (amongst others, I'm sure)
has an application note on current sources/receivers which includes
this circuit.  This circuit might be a bit overkill...  It is nice
though in the fact that you can get a nice bidirectional current
programmed by a voltage.  You can also boost the current by placing a
power buffer at the output of the opamp, if necessary.

There are plenty of other circuits...

Art Zirger.

```

Date: 13 Jun 96 01:01:05 GMT

Original Subject: Re: Constant Current Source (1-5 mA)