My oscilloscope is rated: 1Mohm || 12pF. It's a 100 MHz oscilloscope. However, I don't get the point of the capacitance. If I set my probe on 10X (it's switchable), then it inserts 9Mohm in series. Now we've created an RC filter with -3dB break point of: ~1.473 kHz, and yet, I get higher bandwidth with 10X probes and I certainly don't get a 1.4 kHz bandwidth limiter! What am I missing?

Also, I was simulating the circuit on a circuit simulator. With no probe resistance a 10pF cap conducts 1A at 100 MHz, which would be massive loading compared to the 1 Mohm impedence.

  • \$\begingroup\$ I am not discouraging answers on this, but there is an excellent discussion of this in High Speed Digital Design(amazon.com/High-Speed-Digital-Design-Handbook/dp/0133957241/…). I would suggest you read through it, it gives you a much better idea of what scope bandwidth also means. \$\endgroup\$
    – Kortuk
    Commented Jan 10, 2011 at 16:37
  • \$\begingroup\$ This article explains what actually goes on behind the scenes with 1:10 probes, and why they work. dfad.com.au/links/… \$\endgroup\$ Commented Feb 10, 2021 at 12:11

2 Answers 2


Like pretty much all real circuits, oscilloscope inputs have a parasitic capacitance. No matter how small you made it by good design, it would still affect RF signal acquisition, except maybe for a defined 50 Ω connection and attenuation directly at the scope's input, for which case, with the numbers from your question -

$$f_{-3dB} = \frac{1}{2\pi \cdot R_{in,\ scope} \cdot C_{in,\ scope}} = \frac{1}{2\pi \cdot 50 \;\Omega \cdot 12 \;pF} = 256 \;MHz $$

Or even higher, if we would make the scope's input impedance Cin, scope smaller.

Usually, though, we don't want to load the circuit under test with a defined 50 Ω connection because most circuits under test will have any impedance but 50 Ω (like your signal generator's output would, because it is specifically designed for impedance-matched 50 Ω systems). So what can be done with a capacitance that can't be eliminated? It was chosen to use it in a clever way in the probe-and-scope combination. So clever, actually, that any unknown capacitance that may be caused by probe cables and other things in your connection can be compensated just like the scope's input capacitance, and all of them become don't-cares for most cases of practical measurement applications.

The 1:10 probe has an internal resistor of 9 MΩ and, in parallel, an internal capacitor of [1/9 * Cin, scope].

It is adjustable because the probe doesn't know the exact capacitance of the particular scope it is connected to.

With the capacitor in the probe properly adjusted, you have not only a resistive divider for the DC part of the signal (9 MΩ at the probe vs. 1 MΩ in the scope), but also a capacitive divider for the higher-frequency AC part of the signal (1.33 pF at the probe vs. 12 pF in the scope, using your numbers), and the combination works beautifully up to or beyond, say, 500 MHz.

Also, you get the advantage of inserting not 1 MΩ and 12 pF into your circuit when probing, but 9 MΩ + 1 MΩ = 10 MΩ and [the series equivalent of 12 pF and (12 pF / 9)] = 1.2 pF

enter image description here

Link to the source of the picture: Here.

What the picture in the link doesn't show and what we have neglected so far is the capacitance of the probe's cable, this would just add to the capacitance at the scope's input and can also be compensated for when turning the variable cap in the probe.

Using a 1:10 probe, the probe's small capacitance is in series with the scope's larger input capacitance. The total capacitance (approx. 1.2 pF) is in parallel to the point of your circuit that you are probing. Connecting the scope directly to the circuit, e.g. with just a straight BNC cable, you are indeed putting the entire input capacitance of the scope in parallel to what you are measuring - maybe loading your circuit under test so much that it will not work any more while being measured. At best, it might still work somehow, but the picture on your scope will show results far off the real waveforms in your circuit under test.

It would be possible to build scopes with a much smaller input capacitance - but then, there would be no way compensating the probe's cable capacitance with a small variable capacitor near the probe tip. After all, the 12 pF at the scope's input have been put there on purpose, to make the scope work well together with a good probe.

One last note: Using 1:100 probes, you load your circuit even less. In lack of an active probe with a really small capacitance at the tip, a 1:100 probe can be used in cases where even 1.2 pF would be too much load on your circuit - provided the signal is large enough that you still see something after the probe's 1:100 attenuation.

  • \$\begingroup\$ So the capacitor is in series with the input? \$\endgroup\$
    – Thomas O
    Commented Jan 7, 2011 at 20:55
  • \$\begingroup\$ Yes, the cap is in series with the input just like the resistor is. You could say that at the input, the two resistors divide the DC voltage, and the two capacitors divide the AC part (up to very high frequencies). Remember that a (1/10) resistive divider has the big resistance on the upper part, and a (1/10) capacitive divider has the small cap on top. \$\endgroup\$
    – zebonaut
    Commented Jan 7, 2011 at 20:59

Suppose you were checking a 100 MHz signal across a resonant parallel tuned circuit. The 12 pF would affect the accuracy of the measurement considerably, as it would be large compared with the capacitance across the coil, and would detune the circuit.

  • \$\begingroup\$ Yeah... I take that it's a bad thing, but how can it even measure at 100 MHz with such a capacitance? \$\endgroup\$
    – Thomas O
    Commented Jan 7, 2011 at 20:31
  • \$\begingroup\$ It's an extreme example, but it does show why the capacitance is important. It wouldn't cause a problem at a low impedance point. \$\endgroup\$ Commented Jan 7, 2011 at 20:33
  • \$\begingroup\$ Okay, but my simulator says a 10pF cap draws ±1 amp at 100Vp-p. My scope is rated 250Vp-p. Does this mean, a higher frequency could actually damage it(?) It can measure up to 5Vp-p, giving ±20V, although it can move ±16 divisions, so it could measure 100V maximum... I must be missing something?! \$\endgroup\$
    – Thomas O
    Commented Jan 7, 2011 at 20:37
  • \$\begingroup\$ @Thomas - Where is that 1A going to in your simulation? In the real world, it's going to a high-impedance analog input, not directly to ground. \$\endgroup\$ Commented Jan 7, 2011 at 20:55
  • 2
    \$\begingroup\$ Special secret trick for extreme low loading of the circuit under test (when you don't have an active FET probe): Use a 1:100 probe. They are designed primarily for high voltage probing, but they are also great when you want to insert an even lower capacitance into your circuit than you would using a 1:10 probe. \$\endgroup\$
    – zebonaut
    Commented Jan 7, 2011 at 21:21

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