On Sat, 15 Apr 2006 15:34:21 -0500, John Fields
On Sat, 15 Apr 2006 19:28:23 GMT,
[email protected] wrote:
I need to make a "voltage transducer" that would convert 0-100 mV
(from a DC current shunt) to 0-5V range. The output will be fed into
some sensor, so, input impedance will be quite high and there is no
need for high power. All that is needed is low power voltage
multiplication. Thanks to all you guru's
---
Since:
5V
------ = 50,
0.1V
you'll need a gain of 50, and you don't need an inversion, so you
can do this:
Vin>------|+\
| >--+-->Vout
+--|-/ |
| |
+--[R2]--+
|
[R1]
|
GND>---+----------->GND
An easy way to work out the resistors is to decide how much current
you want in the divider and then choose R1 to drop the maximum
voltage on the + input with Vout at its maximum output also.
In your case, if you wanted to have 1mA in the divider, then with
0.1V across R1 it would have to have a value of:
E 0.1V
R = --- = -------- = 100 ohms
I 0.001A
and R2 would have a value of:
E 4.9V
R = --- = -------- = 4900 ohms
I 0.001A
Looking at the divider like this:
E1
|
[R2]
| 0.1V
+----E2
|
[R1]
|
GND
and choosing 1% resistors for R1 and R2, the opamp output voltage
with 0.1V on the + input will be:
E2 (R1+R2) 0.1V * 4970R
E1 ------------ = -------------- = 4.97V
R1 100R
So, you've got a 30 millivolt error.
If you diddle around with the resistors you can probably do better
than that, but the easy way out is just to use a decent multi-turn
trimpot, put 100mV on the + input, and crank away until you get
precisely 5.00V on the output.
BTW, the classical formula for the circuit is:
R1 + R2
Vout = -------- Vin
R1
± 50X the offset voltage of the OpAmp
...Jim Thompson