Wheatstone Bridge Interface

Dear all,

I hv some questions here abt the Wheatstone bridge and hope to seek for ur help. I m simulating a wheatstone bridge by using 5 resistors which of 4 resistors r the same resistance value n another 1 is smaller.

**broken link removed**

Let's say my R1, R2, R3 and R4 are 1KΩ and I hv one more resistor across the S+ and S- which is 2.8KΩ. If I supply a DC voltage of +5V across E+ and E-, the voltage across S+ and E- will be +5/2 = +2.5V same for the S- and E-.

If now I supply a signal of 25µV to S+ without reference to S-, wat would be the output voltage across S+ and S-? Let say the supplied signal of 25µV to S+ comes out directly from an IC in which it is taking the E- and its GND.

I really hv no idea with the operation of wheatstone bridge in this configuration as the 4 core resistors are of fixed value which differs from the normal way implementing a wheatstone bridge.

Thanks for any expert here to enlighten me in advance.

Best regards,
Eric

Hi Ericmar,

I'm not very clear about S+ being driven by an IC.
Assuming an opamp output stage drives S+ with 25uV with respect to E-, it will have enough output drive to hold the value of S+ at 25uV.

Now the question is, what happens to S-..
Just apply superposition theorem for voltage at S-
One is 5V - this creates around 2.12V at S-.(Imagine 2.8k and 1k in parallel and a 1k in series)
The other is 25uV. Very negligible 12.5uV.

So the voltage at S- is 2.12V.
v(S+,S-) = 25uV-2.12V.

Hope this helped!

Giri