Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I thought 10X probes are only used for impedance matching at high freq. circuits.

But I read in a text that they are also used in high impedance circuits to prevent loading. But isn't a scope's input impedance already huge? I mean what is high impedance in this case and how can 10X help the situation?

share|improve this question
up vote 8 down vote accepted

A typical scope input impedance is 1Mohm + 20pF. This is not huge, even with a 100k source I get 10% error DC and 20pF is only 8k impedance at just 1MHz. A 10x porbe makes the input 10meg + 2pF, much better. It also allows for tuning of rise times by trimming a small capacitor in series with the R in the probe.

share|improve this answer
thanks. just one thing.. imagine i connect the probe to a high impedance and set the probe to 10X to reduce loading. are you saying the input impedance of the scope becomes almost 10 times more? does that mean the 10X probe itself has 9 times more resistance than the scope's input resistance? – user16307 Jan 28 at 18:42
Yes, exactly, the probe forms a potential divider with the scope input, 9/10 of the voltage is dropped across the probe. – user1582568 Jan 28 at 18:44
just last question.. is probe's resistance in series or shunt with scope's input resistance? – user16307 Jan 28 at 18:46
In series, it is essentially a resistor and capacitor (9 x the value of the scope input) in parallel with each other, and the that parallel combination is in series with the scope input. – user1582568 Jan 28 at 18:49
very clear thanks – user16307 Jan 28 at 18:49

You can refer to the data sheet of the probe. Below is a typical set of specifications. Consider the 200MHz probe.

enter image description here

The input impedance will be 10M\$\Omega\$ || 15pF. That's a complex number in general but it will be dominated by the capacitance at high frequencies so we can pretty much ignore the real (resistive) part. At 200MHz, x10 the capacitance has an impedance of about 53\$\Omega\$, which is quite a bit of loading.

In the x1 position, it's 47pF plus the 'scope input impedance, in parallel with 1M, so more like 12\$\Omega\$, which, of course, is much worse.

You could also consider the frequency at which the capacitive reactance equals the resistance- so the magnitude of total impedance is about 7M. That's at about 1kHz- hardly high frequency but that's where the capacitive loading begins to exceed the resistive loading.

If you require very low capacitance loading then there are active probes that have only a couple pF input capacitance, but they tend to be quite expensive.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.