Re: DC voltage converter 0-110mV to 0-5V

On Sat, 15 Apr 2006 20:51:31 GMT, "Genome" <ilike_spam@xxxxxxxxxxx>
wrote:


"Jim Thompson" <To-Email-Use-The-Envelope-Icon@xxxxxxxxxxxxxxx> wrote in
message news:pqm2429aij3pbld3s4gpd61un3ajhfha9l@xxxxxxxxxx
On Sat, 15 Apr 2006 15:34:21 -0500, John Fields
<jfields@xxxxxxxxxxxxxxxxxxxxx> wrote:

On Sat, 15 Apr 2006 19:28:23 GMT, dscott@xxxxxxxxxxxxxxxx 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

And the input bias current?

---
The non-inverting input is going to be fed from a shunt, so that
smacks of way less than an ohm, and the inverting input is going to
be fed from what looks like a 5mA source, so even with something
horribly gross, like a 5µA bias current requirement, we're still
looking at only about a tenth of a percent of error in the output,
due to that, no matter what.

--
John Fields
Professional Circuit Designer
.

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