On Wed, Aug 13, 2014 at 9:59 PM, Paul Stoffregen <[email protected]> wrote:
> A "10X" passive probe attenuates the signal by a factor of 10, but it also > reduces the load it places on your circuit by a factor of 10. Only at DC. 10x probes will look like a large part of the scope input capacitance, say 50% or more. By the same token a 10x probe will have the full 100 MHz bandwidth, for a 100 MHz scope, whereas the 1x probe will be 15 MHz (rough numbers, you capacitance/bandwidth will vary, etc.). The capacitive divider of probe + scope doesn't match the 10x ratio of the DC divider, where the probe is basically a 9 meg series resistor. It's important that the compensation range of the probe covers a range that includes the input capacitance of the scope. If you really want low loading there are active (FET) probes, but they certainly aren't cheap. If you want to get a rough idea of how much having the ~15 pF of probe hooked into your circuit is affecting it simply hook up a second probe, comparing the waveform before and after. Of course for really sensitive things like xtal oscillators this might cause it to not work at all. More's the fun. :-) Regards, NealS
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