# Keeping inputs to opamps

I'm trying to get the hang of op-amps, and I've run into a stop. I'm trying to implement a differential op-amp, where the two inputs are from a wheatstone bridge. Please see attached diagram

The problem I'm facing, is that the inputs $V_{left}$ and $V_{right}$ change, due to the impedance seen to the op-amp. I do not want these voltages to change, since that will cause the output of the op amp to be "wrong".

The immediate solution I see is to decrease the resistance of the bridge, and increase the resistance of the op amp. This only minimizes the error, and does not remove it. The op-amp is ment to output $V_{o}=2.2(V_{right}-V_{left})$. How do I deal with this? Thx in advance! :)

You can implement an instrumentation amplifier if you need high input impedance. This requires another 1-2 op-amps.

simulate this circuit – Schematic created using CircuitLab

The 3-opamp version is the best in many circumstances- you only need to change one resistor to change the gain (top right- change R10).

• Thx a lot! Just what I needed Jan 1, 2016 at 16:21

(Do you really intend R4 to be 10k ?)

As long as the opamp input resistors are matched (like you have), the 'error' is only an error in gain, and not in offset.

What this means is that when the bridge is balanced (e.g. if the opamp wasn't connected), when you connect the opamp, its output will be 0. Look at it like this: The right half of the bridge has a load of R6 plus R7; R6 is between the bridge and the opamp's '+' input.

When the opamp is working, it forces its '-' input to the same voltage as its '+' input. Therefore the left leg of the bridge also has a load resistance of R5 to the opamp '-' input. Notice that both R5 & R6 have the same voltage across them -- there is no offset error.

When the bridge is not balanced, there is a gain error -- this can be seen by analyzing the Thevenin equivalent of each half -- its output impedance if R1//R2 = 34k. This resistance effectively is in series with R5 (or R6), thus effectively increasing the value from 100k to 134k. If you reduced those resistors to 100k-34k = 66k, you would get the original gain you wanted.