I would like to state the problem I am dealing with which I cannot explain.

The set up I have is as follows:

  • 2 x resistors 1 kΩ,
  • 1 x breadboard,
  • Wires,
  • 1 x Voltcraft DSO-6202FM oscilloscope which has 2 channels and dual channel arbitrary function generator.

I have a breadboard where I built a resistor voltage divider of two 1k Ohm resistors. At one end of the first resistor I connected a probe coming from the function generator. The signal from the function generator is a square wave at 200 Hz, start phase 0 degrees, a high level 500 mV, low level 0 mV. A probe is used to display the output signal at the measuring point and is connected to channel 1 CH1 of the oscilloscope.

First observation:

When I connect the output probe from the function generator to the probe connected to CH1 directly, I see a square wave, without needing to connect the two ground clips of both probes.

Q. How is this possible? I thought I need to connect the ground clips together to have a common ground and then be able to see the waveform from the function generator!

Second observation:

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As shown in the diagram above, the output of the function generator is connected to the one end of the first 1 kΩ resistor. The probe of CH1 is connected to the output resistor of the voltage divider. The second end of the output resistor is NOT connected to a common ground. A signal is recorded but the value of which is the same as in the first observation. If I connect the ground clip of the function generator, the ground clip of CH1 and the second end of the output resistor together, the recorded waveform on CH1 of the oscilloscope is constant 0 volt.

Q. What is the explanation of this phenomenon that when I connect all the ground clips and the resistor of the voltage divider together to form ground I get a 0 volt signal on the oscilloscope?

Your wise comments and suggestions are welcomed with high regards.

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  • 3
    \$\begingroup\$ Can you post a picture of your setup? The zero volt thing is probably a wiring mistake. \$\endgroup\$
    – JRE
    Commented Nov 20, 2021 at 20:26

2 Answers 2


If you're using a 10x oscilloscope probe on the output of the function generator, there is a 9 Megohm resistor in series with the probe. When the floating end of the resistor divider is grounded, this will form a resistor divider that will severely attenuate your signal.


simulate this circuit – Schematic created using CircuitLab

As for the grounding bit, since the signal generator is part of your oscilloscope, they share grounds internally to the instrument.

  • \$\begingroup\$ Thank you very much qrk. That is the solution. Great. Now I get a clear signal from the function generator. I have no problem with the voltage divider where the signal is half the input. \$\endgroup\$
    – Jeddison
    Commented Nov 20, 2021 at 21:00
  • \$\begingroup\$ @Jeddison, if qrk's answer is the solution, please can you look at accepting his answer. You can choose to upvote it too, if you wish. Accepting answers concludes the question and acknowledges the work of the community members here. Thanks. \$\endgroup\$
    – TonyM
    Commented Nov 21, 2021 at 20:34

Both your signal generator and oscilloscope are powered via mains input.

They both have three-prong earthed mains plugs.

The earthing goes to the metal chassis and common ground on the devices.

So basically, the oscilloscope BNC connector ground, and signal generator BNC connector ground are already connected together via mains wiring, so they are at the same potential.


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