Back feeding when connecting one side of circuit to differential op-amp

simulate this circuit – Schematic created using CircuitLab

I am building four 30v voltage sensing circuits to measure battery+/battery-, load+/load-, battery+/load- and battery-/load+ voltages to verify contactor open/closing sequence.

1. V1 is 30v battery
2. Load will be DC motor switched by 2 contactors, now im just using a open circuit to do testing.
3. The monitoring circuit can use battery negative as circuit GND, but I prefer not because I dont want connect low voltage side to high voltage side.
4. The circuit will be powered independently of battery because of above reason.
5. Schematic above only shows battery+/battery- and load+/load- sensing circuits.

When sw1 and sw2 are both open or closed, I got expected results (1.99v at outputs with 0.066 gain). However, when sw1 is open sw2 is clsoed (or sw2 is open s1 is closed), I see a weird 0.66v at output 1 while output 2 is 2v. What would be the reason?

• your circuit doesn't make much sense – I presume you meant to build something different? Jul 27, 2017 at 18:17
• it is a conntactor voltage feedback circuit, thats why I need to measure both voltage before and after sw1 and sw2 Jul 27, 2017 at 18:19
• Why would you get 1.99 volts when both switches are open? Is there something missing from your circuit? What power rails are on the op amps? Jul 27, 2017 at 18:27
• @yxing: I tried to help on your previous question. Why don't you explain what you are actually trying to do instead of asking us to fix a flawed design. What is the 30 V supply and why can we not use one terminal of it for GND? Jul 27, 2017 at 19:02
• Again, you are explaining how you think it should work. We want you to explain what you are trying to do. Edit your question to explain the project requirements something along the lines of: (1) I am building a 30 V 50 Ah battery voltage monitor. (2) I require no-load and on-load measurements. (3) Load will be 1 Ω switched by contactor. (4) The monitoring circuit can / can not (explain why) use battery negative as circuit GND. (5) The response time should be < 50 ms. (6) The circuit will be powered independently of battery. (7) Required output is 0 to 2.54 V for 0 - 40 V input. (8) My circuit. Jul 27, 2017 at 19:25

Really, that circuit doesn't make sense.

Leave SW1 open and remove all the components that don't carry any current anymore.

You'll notice that OA4 needs to make the voltage between its two inputs 0V (golden rule of opamp with inverting feedback), and since the positive input is pulled to ground (no current in our out of inputs, hence no current through R27, hence no voltage over R27), the output of OA4 needs to become 0V to. It tries to, but I'd guess it has a minimum output voltage of GND+0.66V.

Also, notice that you're feeding -30 V into that OA4. It's voltage supply range is 15 V – that's a recipe for disaster.

So, I don't know how to save that schematic. Design it from scratch, anew?

• please see update, 30v is seprate from circuit GND, input voltage of both OP amp is 5v Jul 27, 2017 at 18:45
• How can you claim the -30V is independent from GND? The positive end of your 30V voltage source is connected to GND via R5--R7. So, there's definitely a negative voltage at the left terminals of R4 and R24. And you don't have dual-supply for your Opamp, so this simply can't work in any way. Your circuit is broken by principle. Jul 27, 2017 at 18:51

When sw1 and sw2 are both open or closed, I got expected results (1.99v at outputs with 0.066 gain). However, when sw1 is open sw2 is clsoed (or sw2 is open s1 is closed), I see a weird 0.66v at output 1 while output 2 is 2v. What would be the reason?

Use a multi meter to measure between your load+ and load- when either one of the SW is close. The 0.66V is actually the 10V potential difference across the load.

Another possibility is you're using an opamp that is not rail to rail, the negative terminal is picking up a high potential floating voltage; thus making the opamp pushing out the lowest output voltage possible.