# Wheatstone bridge amplifier circuit

Discussion in 'Homework Help' started by jstrike21, Feb 25, 2010.

1. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0

I found Vout of just the wheatstone bridge part is:
Vin((10.01/20.01)-(10/20))
Using 5 as Vin i got Vout=0.00025V

Does the 0.00025 Volts go to the + terminal of both of the op amps?
I'm not sure how to solve from here.

I know the transfer function of the amplifier circuit:

is

would I use the 0.00025 as Vin2 and 0 as Vin1?

2. ### jlcstrat Active Member

Jun 19, 2009
58
3
I would think that sounds right. I have been wrong quite often though

3. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
thanks, can anyone verify this?

4. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
forgot to multiply by 5, Vout for the wheatstone bridge is .00125V

5. ### jlcstrat Active Member

Jun 19, 2009
58
3
The more I look at this, the more I think V2 should be 2.5, and V1 2.5012.

6. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
ya but still when you out those into the final equation you will subtract them and end up multiplying by .00125

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I would imagine in a real circuit implementation there would be a power supply common reference at some point - say at the lower point of the bridge. Otherwise one might have issues with lack of conducting paths for input bias currents - should they be non-zero.

8. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,222
Google -> instrumentation amplifier This will give you some tips. A instrumentation amplifier will amplify the difference between the two inputs.

9. ### jlcstrat Active Member

Jun 19, 2009
58
3
It's going to amplify the difference, but as far as the formula goes wouldn't you use 2.5 and 2.5012 for the V in 1 and 2? I thought that was the original question.