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enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

I am new to hardware experiments, although I have acceptable theoretical knowledge. My setup is:

  • a signal generator
  • a breadboard with an op-amp (not powered)
  • an oscilloscope

The output of the generator has a splitter, one cable goes directly to the oscilloscope plotting the input (a sine wave). The second cable leaving the splitter is a crocodile clip cable. The oscilloscope has a probe.

There is this phenomenon I don't understand. When neither the probe nor the crocodile cable are attached to anything the sine wave shows correctly on the oscilloscope (I am just talking about the direct line from the generator, not the probe). When I connect the ground of the oscilloscope probe to the Vcc- of the op-amp, and the red crocodile cable to the IN+ of the op-amp, the sine wave is distorted (see photo attached).

I am sure there must be a simple answer, maybe something to do with the device impedance etc. but I just can't wrap my head around it. This prevents me from properly measuring the op-amp's output for an input sine wave.

The following schematic shows the fully connected setup. The input sine wave is still distorted, however the op amp works fine (don't have a photo, but it's constant high value when input is positive and constant low when negative)

schematic

simulate this circuit

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    \$\begingroup\$ Where are the attached photos? What is the period and amplitude of the sine wave shown on the oscilloscope? What are the settings of the signal generator? Influence of selecting different waveforms? \$\endgroup\$
    – Uwe
    Feb 4, 2023 at 17:55
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    \$\begingroup\$ Are you really connecting this to an unpowered op-amp? In that case you might be powering the op amp via the input (+ scope ground), hence the distortion. (This would depend on the op-amp, but don't assume any behaviour for powered-down active components unless stated: all kinds of fun things can happen (as different parts of the chip cross conduction thresholds, draw too much power and turn off again, causing oscillation, etc). You've effectively created a severe brown-out situation. \$\endgroup\$
    – 2e0byo
    Feb 4, 2023 at 18:15
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    \$\begingroup\$ Please edit your question by including a schematic of the circuit. If you absolutely don't know how to make a schematic then a picture of the board or a muntzing diagram, along with a statement that you can't make schematics, is -- OK. And -- what @2e0byo said about trying to inject a signal into a powered-down op-amp. \$\endgroup\$
    – TimWescott
    Feb 4, 2023 at 18:38
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    \$\begingroup\$ Lacking a drawing showing the exact setup, let me venture a guess that the OP has neglected the fact that the ground of the oscilloscope probe, and the ground of the signal generator output, are connected via the mains ground. \$\endgroup\$
    – sh-
    Feb 4, 2023 at 19:14
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    \$\begingroup\$ I don't believe the period is 200 ms, it is 20 ms and the frequency is 50 Hz. It is not a generator signal, it is a pick up of the mains voltage from the power grid. \$\endgroup\$
    – Uwe
    Feb 4, 2023 at 19:57

2 Answers 2

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Signal generators like that have a finite output impedance -- usually 50 ohms.

You're putting a signal into an unpowered amplifier. Unless a chip is specially designed for that purpose, applying a voltage to an unpowered chip will make it conduct. Probably what is happening is that the positive-going swing is powering the op-amp through various parasitic diodes, while the negative-going swing is just turning a diode on -- hence the asymmetrical response.

In general, you need to observe the device's absolute maximum ratings, which almost invariably call for not applying input voltages outside of the bounds of the positive and negative supplies.

In specific, in this case, you should build an op-amp circuit of some sort (the chip by itself is not a useful amplifier; just the core of one). Then you should power it up and see how things work.

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  • \$\begingroup\$ Just to quote what I said in one of the comments (there were quite a lot) "Additionally, when I fully hook up the op amp the output is as expected, piecewise constant, with the high value when the waveform is positive." When fully operational the amp does its job however the input still looks the same (distorted). Trying to change parameters doesn't work. The only thing that fixes it is adding a bias to the input, putting it in positive territory. So something is happening when it goes negative... \$\endgroup\$ Feb 5, 2023 at 22:32
  • \$\begingroup\$ That picture shows just two wires going to the op-amp -- that's not a fully operational amp. Perhaps you could show a picture of the actual setup then. \$\endgroup\$
    – TimWescott
    Feb 5, 2023 at 22:47
  • \$\begingroup\$ With a closeup of how the opamp is connected on the board -- if you'd had an actual circuit in that shot I wouldn't have been able to make out the details. \$\endgroup\$
    – TimWescott
    Feb 5, 2023 at 23:04
  • \$\begingroup\$ Is there a way I can add two schematics? I don't have a photo with the whole circuit and I don't want to confuse by suddenly changing the diagram \$\endgroup\$ Feb 5, 2023 at 23:36
  • \$\begingroup\$ I think so -- just put your cursor somewhere and push the "schematic" button. \$\endgroup\$
    – TimWescott
    Feb 6, 2023 at 2:10
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It turns out it was indeed a grounding issue. The closest answer came from @2e0by0. As the schematic shows, I was connecting the Vcc- pin to ground. Now I am using two power sources in series, so I have Vcc -Vcc and ground. The input is fed into the amp using ground as reference. The signal distortion is not present anymore.

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