I am trying to measure a DC signal in the mV range.

Below is the waveform I get after I connect the oscilloscope to its probe and without connecting it to anything:

enter image description here

I am getting a ripple like waveform of 50mV peak to peak. Why and from where am I getting this unwanted 50mV peak to peak ripple noise even when the scope and its probe is not connected to anything?

Why is there this noise and how to eliminate this noise ripple for proper and accurate measurements?

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    \$\begingroup\$ What you are seeing is the electric field in the air around you caused by AC power wiring in your building. I estimate that your country uses 50 Hz. \$\endgroup\$
    – Andy aka
    Oct 6, 2020 at 11:19
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    \$\begingroup\$ as I am measuring only DC signal with the oscilloscope, I could use the AC Coupling mode to remove this unwanted noise, ripple Uhm, no! AC coupling mode blocks the DC so that means if you measure a DC voltage with a small ripple: in AC coupling mode, you would only see the ripple. Looks like that ripple is 50 Hz, is the mains frequency in your country also 50 Hz perhaps? Also remember that a scope in put is high impedance so even the slightest coupling to mains wires will show up in the trace. \$\endgroup\$ Oct 6, 2020 at 11:20
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    \$\begingroup\$ also it might help: The signal ground of the scope has to be clipped/connected closest to the signal ground of the DC signal. I hope it helps \$\endgroup\$
    – Ruperto
    Oct 6, 2020 at 11:33
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    \$\begingroup\$ The coupling is capacitive from the wiring in your walls. It might be 1 or 2 pF. As you get closer it might be tens or hundreds of pF Work out the impedance at 50 Hz and see that it forms a potential divider with the unconnected probe input impedance. That's it basically. \$\endgroup\$
    – Andy aka
    Oct 6, 2020 at 11:56
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    \$\begingroup\$ If you connect your probe tip directly to probe ground, does this phantom signal go away? It should. After you note that effect, Google "Faraday's Law". The amount of signal you'll capture depends on loop area. THe reason for this may become more clear, and once you understand the reason, you're better able to avoid it! \$\endgroup\$
    – Kyle B
    Oct 6, 2020 at 13:23

2 Answers 2


Why and from where am I getting this unwanted 50mV peak to peak ripple

Start by thinking about 220 volts (if that is your AC supply running in your building). Because it runs in two wires where one is basically ground (0 volts), the net near-field voltage is 50% or 110 volts and that field disperses and reduces in value the further you are away from the wiring. But, we can also say that the tip on your scope probe is capacitively coupled to the 110 volts via capacitance but, how much capacitance.

I've not calculated this before so I'm also interested in what it might be.

It looks like your o-scope is showing about 40 mVp-p and it's definitely at 50 Hz: -

enter image description here

This is an RMS amplitude of about 14 mV.

This means that if your o-scope probe is 10 Mohm then it is receiving a current of 1.4 nA RMS. That current flows through the capacitance that exists between your wall wiring and your probe tip. So, work out the capacitive reactance: -

$$X_C = \dfrac{\text{voltage}}{\text{current}} = \dfrac{110 \text{ volts} - 14 \text{ mV}}{1.4 \text{nA}} = 78.57 \text{ Gohm}$$

How much capacitance is that: -

$$C = \dfrac{1}{2\pi f X_C} = 0.04\text{ pF}$$

So, if you have a voltage supply of 110 volts (and 50 Hz) feeding a 10 Mohm resistor via a 0.04 pF capacitor you would see 14 mV RMS (40 voltsp-p) across the 10 Mohm resistor.

If your scope input impedance is 1 Mohm then the current flowing into your probe tip is 14 nA. The knock-on effect is that the capacitance between wall wiring and probe tip will be ten times higher at 0.4 pF.

If you really wanted a more accurate answer you should model/factor the probe tip capacitance to ground - this is in parallel with your resistive impedance of 10 or 1 Mohm. It might be in the realm of 10 pF and right away you can imagine that this forms a serious extra attenuator with the previously predicted 0.4 pF of about 25:1. At this point, if I really wanted to know the capacitive value between wall-wiring and o-scope tip, I'd plug the circuit into a simulator and take the lazy route.

Short answer - it's a few pF.

Why is there this noise and how to eliminate this noise ripple for proper and accurate measurements?

Once you connect your probe to a real circuit node, the impedance drop massively and you won't see this effect.

  • \$\begingroup\$ OP could also connect the scope probe ground lead to the tip of the probe and hopefully see no signal pickup. If they do then there is likely to be a bad probe. \$\endgroup\$ Oct 6, 2020 at 12:40
  • \$\begingroup\$ I think the calculation is wrong. \$V_t = i\ Z\$ where \$Z=\sqrt{X_C^2+R^2}\$, not \$X_C+R\$. And thus, \$V_t = \sqrt{V_C^2 + V_R^2}\$, not \$V_C+V_R\$. The phasors, phasor angles... Or am I missing something? \$\endgroup\$ Oct 6, 2020 at 13:18
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    \$\begingroup\$ @RohatKılıç you are technically correct but, if X_C is massively greater than probe resistance (which it is), then the approximation I made is accurate to small sub-fractions of decimal places. \$\endgroup\$
    – Andy aka
    Oct 6, 2020 at 13:39
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    \$\begingroup\$ Thank you for the answer. But how is the near field voltage is taken as 50% of 220V? \$\endgroup\$
    – user220456
    Oct 6, 2020 at 14:53
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    \$\begingroup\$ @Newbie in the very, very near field you have a wire at 220 volts and a wire at 0 volts. As you move away from the very, very near field this averages out to a net e-field of 110 volts i.e. the not-very-close-near field begins to look like a single conductor carrying 110 volts. \$\endgroup\$
    – Andy aka
    Oct 6, 2020 at 15:10

As Andy computed, there is a voltage divider action between the wall power wiring, and the probe tip.

Place a piece of flat metal under your probe, large enough to coerce the electric field flux_lines to arrive orthogonally to the metal surface.

Ground this piece of metal to the scope chassis (scopes usually have bare-metal terminals on the front panel for this).

This allows the flat metal to maximally gather (almost all of) the electric field displacement current, and somewhat reduce the scope_probe displacement currents.

There are some natural_log coefficients involved here. Check for maths on wire/wire coupling, plate/wire coupling, plate/plate coupling. Though hyperbolic trig appears in the equations, you can convert these into equivalent natural_log, allowing easier causality reasoning versus the shapes/flatness/roundness.


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