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I was recently reading an article written about oscilloscope probing. And at one paragraph there is a comparison between active and passive probes as in the below quote:

"For high-frequency applications (greater than 600 MHz) that demand precision across a broad frequency range, active probes are the way to go. They cost more than passive probe and their input voltage is limited, but because of their significantly lower capacitive loading, they give you more accurate insight into fast signals."

What does "capacitive loading" means in this context? Why are passive scope probes more prone to it in high frequencies?

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For a passive device the probe, the cable and the scope input have capacitance seen at the input: -

enter image description here

Here's a circuit of an active probe: -

enter image description here

It uses a JFET (or MOSFET in the case above) to buffer the input. Here's a typical comparison between the probes: -

enter image description here

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  • \$\begingroup\$ That Rp and Cp is coupled only in 10x case? and if 1x case there is no Rp Cp? WhatRoughBeast says upstairs "For 1x probes, the coax cable from the test probe to the scope has significant capacitance, and there is not capacitor in the probe". \$\endgroup\$ – user16307 Dec 30 '15 at 17:09
  • \$\begingroup\$ @jjuserjr yes without the 10x bit on the end the capacitance can be 100pF (ish) as shown in my final diagram - note the 10x probe has an impedance that is 10x higher even at 100 MHz but it itself is rapidly becoming crappy at above 100 MHz whereas the active has a much bigger impedance and at over 1 GHz has the same impedance of a x1 at 10 MHz and a x10 at 100 MHz. \$\endgroup\$ – Andy aka Dec 30 '15 at 18:31
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It is not that the passive probe's capacitance increases at high frequencies, but that at high frequencies, the (typical) 10 pF input capacitance becomes a very significant load on the measured signal.

The impedance of 10 pF at 600 MHz is 1/(2.pi.600M.10p) = 27 ohm.

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  • \$\begingroup\$ why is there capacitor inside passive probes? \$\endgroup\$ – user16307 Dec 30 '15 at 5:18
  • \$\begingroup\$ @jjuserjr - For 1x probes, the coax cable from the test probe to the scope has significant capacitance, and there is not capacitor in the probe. For a compensated (10x) probe, the attenuation network both reduces the capacitance seen by the circuit under test and extends the frequency response at the scope. \$\endgroup\$ – WhatRoughBeast Dec 30 '15 at 5:23
  • \$\begingroup\$ so when we tune the probes with a screw driver, is that a simple variable resistor R coupled with C of the coax cable? but e still use coax in active probe meas. Why there is no capacitive coupling in that case? \$\endgroup\$ – user16307 Dec 30 '15 at 5:26
  • \$\begingroup\$ It's a variable capacitor in the base of the 10x probe, google "oscilloscope 10x probe schematic" to see the basic topology. The probe cable acts like a transmission line, with a fixed compensation network at the probe tip and an adjustable compensation network at the base of the probe, where it plugs into the oscilloscope. \$\endgroup\$ – MarkU Dec 30 '15 at 8:02
  • \$\begingroup\$ @WhatRoughBeast saya "For 1x probes, the coax cable from the test probe to the scope has significant capacitance, and there is not capacitor in the probe" what does that mean? there is not capacitor in 1x case?? \$\endgroup\$ – user16307 Dec 30 '15 at 11:47

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