I have an Texas Instrumentes OP07 and am trying to wire it as non-inverting amplifier. official wiring schematic non-inverting setup

I have made some photos.

On the "near" photo you can barely see the "orientation circle": near I was not 100% sure with dual power supply, so I made up this, see the "far" photo: far

I have connected oscilloscope 1MOhm probe to out and oscilloscope ground clip to circuit ground.

On the "far" photo you can see a blue breadboard wire connected to a white wire. This is my antenna which picks up some stray 50Hz fields easily measurable without amplifier, which I then expect to be amplified.

But I don't see any signal at my oscilloscope, only 4V from power supply. I have checked a lot for loose connectons, but didn't find any. Maybe I have done something wrong?


1 Answer 1


But I don't see any signal at my oscilloscope, only 4V from power supply

  • The minimum operational supply for the OP-07 is 6 volts
  • The guaranteed usable input range is between Vneg + 2 volts and Vpos - 2 volts
  • The guaranteed usable output voltage is Vneg +2.5 volts to Vpos - 2.5 volts

enter image description here

The fix should be something like a 10 MΩ resistor connected between IN+ and 0 volts (mid-rail). The IN+ pin produced up to 4 nA of leakage current on the OP-07 and if the open circuit impedance is 1 GΩ (a lot more of course), the input will rise or fall to +/- 4 volts. 10 MΩ guarantees that the input will remain within +/- 40 mV. It maybe less of course. And, remember that the gain of 11 will magnify this input offset by 11 at the output.

  • \$\begingroup\$ In dual supply mode the minimum operational supply is 3 volts. \$\endgroup\$ Commented Apr 27 at 11:23
  • \$\begingroup\$ Yes, half of 6 volts. \$\endgroup\$
    – Andy aka
    Commented Apr 27 at 11:25
  • \$\begingroup\$ What does "guaranteed usable input range" mean? The picked up 50Hz stray fields are much below 2V. \$\endgroup\$ Commented Apr 27 at 11:26
  • \$\begingroup\$ It's the range over which the input pins are guaranteed to work \$\endgroup\$
    – Andy aka
    Commented Apr 27 at 11:27
  • 1
    \$\begingroup\$ You cannot leave an input floating like that. Use a 10 Meg resistor to 0 volts at the very least. \$\endgroup\$
    – Andy aka
    Commented Apr 27 at 11:29

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