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When I touch the positive end of the oscilloscope probe without touching the ground, I see the following waveform on the screen. The Vpp is around 45 volts as shown in the picture below:

enter image description here

When I touch the ground end of the probe while touching the positive tip, the Vpp drops to 3.6V as shown in the image below.

What is going on?

enter image description here

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    \$\begingroup\$ you made yourself into an antenna. \$\endgroup\$ Jul 12, 2021 at 7:40
  • \$\begingroup\$ Is the oscilloscope getting properly earthed via the mains plug and socket? Or is it connected to an unearthed socket? \$\endgroup\$
    – Justme
    Jul 12, 2021 at 8:21

2 Answers 2

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Your ungrounded body is picking up power line noise, which is shown in the first trace.

When you touch the scope ground at the same time, you’re grounding your body, which conducts away most of the coupling. The remainder that isn’t coupled away is what’s shown in the second trace. The closer you place the probe tip to where you touch the ground, the smaller that residual voltage will be.

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Notice the period of the signal in the first example. it is more or less 20ms which translated to frequency is 50Hz. Your ungrounded body has become "an antenna" for mains voltage. When you ground yourself it becomes less noticeable.

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    \$\begingroup\$ More or less 50 Hz? Actually it's exactly 60 Hz, the other common mains frequency. \$\endgroup\$
    – DamienD
    Jul 12, 2021 at 8:06
  • \$\begingroup\$ As I said more or less 20ms. which translates to more or less 50Hz. Frequency=1/Period. My mistake is not taking a closer look at the picture provided and realizing that the period is actually much closer to 4/5 of the time division selected (20ms) which would be more or less 16ms which translates to more or less 60Hz. Hope this clarifies any confusion. \$\endgroup\$ Jul 12, 2021 at 8:23
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    \$\begingroup\$ No, not an antenna but one plate of a capacitor. \$\endgroup\$
    – Andy aka
    Jul 12, 2021 at 10:01
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    \$\begingroup\$ Yes! I guess what I mean is, if you live in a 50 Hz country, and you pick up a sine wave at, say, 52 Hz, then there's something quite mysterious going on. The grid is quite tightly regulated. \$\endgroup\$
    – DamienD
    Jul 12, 2021 at 10:05
  • \$\begingroup\$ It's exactly 60 Hz. Five 20ms periods of 20ms is six periods of the noise pickup. 20ms * 5/6 = 16.6667ms, or 1/60Hz. \$\endgroup\$ Jul 12, 2021 at 16:47

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