Built a Wheatstone bridge with two 10k resistors, one 5k rheostat, and a thermistor. Adjusted rheostat so that voltage across bridge reads < 0.013v, close enough.

Powered LM324N, pin 4 to 6.07v, pin 11 to 0v (just one 6v battery supply)

When I connect pin 3 of the LM324 (Vin1+) the voltage of that leg of the Wheatstone bridge changes to 5.4v. I assumed I had blown the opamp previously, and tried a fresh one. Same result. So I simplified the circuit. The only connection to the Wheatstone bridge is a single wire that goes to pin 3 of the op amp. Nothing else is attached, and yet connecting that wire changes the voltage between the two resistors to 5.4v. When I disconnect, they go back to 3.03v as they should be.

Here's a picture of the breadboard in case I am just out of my mind. I can't believe power is leaking out of the input of an LM324, I must be doing something boneheaded but can't see it. Picture attached. The perspective isn't 100% ideal, but the wire from the Wheatstone bridge is connected to pin 3, and pin 4 is Vcc. enter image description here

This was my attempt to simplify the original problem. I built a differential amplifier to amplify the output of the Wheatstone bridge, and current was leaking out of both positive inputs back into the bridge. So I wanted to isolate the problem. I don't have a picture of the board implemented that way, but here's the schematic: enter image description here

R1 was 10k, Rgain was 1k, R2 and R3 were all 1k

  • 3
    \$\begingroup\$ Did you try to connect opamp in a voltage follower configuration? Because you left Vin- floating, Bad Things can happen, better to put an opamp in some proper configuration - either ground, Vin-, or connect it to positive rail, or make a voltage follower. \$\endgroup\$
    – miceuz
    Jan 15, 2013 at 21:54
  • \$\begingroup\$ What was your original configuration? Because as has been mentioned already, nothing good is going to happen with the inverting pin floating. \$\endgroup\$
    – Matt Young
    Jan 16, 2013 at 2:02
  • \$\begingroup\$ Ok, I edited the question to show the original op amp configuration. I will try to add the capacitor and feedback tomorrow. \$\endgroup\$
    – Dov
    Jan 18, 2013 at 3:42

3 Answers 3


Here is a 5 second answer to start. First, put a decoupling cap (0.1uF or something like that) between LM324 pins 4 and 11 (its just good practice). Just for testing connect LM324-1 to LM324-2 (unity gain), because an OpAmp is meant to be used with feedback.

Of course you know that you will need a differential amp sensing both sides of the bridge to get any advantage out of using the bridge.


Thanks for adding the schematic.

The output differential stage, as drawn is positive feedback. So, you will need to exchange pins 9 and 10 of the LM324 connection to have negative feedback.

Also, I didn't think about this much earlier, but the 4th channel would be best to be connected as a voltage follower with the (+) input connected to return. I know that each channel is supposed to be independent, but they are not quite really. The die will have parasitic structures that can affect all the channels. For example, parasitic SCRs are common in many monolithic ICs. So, it is always best to have all channels connected.


Connect all inputs to some proper configuration, for example - voltage follower. Inputs of other opamps in a package left floating can affect the performance of the whole package.

Also, connect opamp 1 in voltage follower.

Also, put a 100nF decoupling cap between pin 11 and pin 4.


The answer appears to have been none of the above. The op amp was blown. I replaced the op amp, and the circuit simply works. While placing a capacitor across Vcc+ and Vcc- is probably a good idea, as is connecting all negative inputs to outputs, it appears not to have any effect.


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