# Regarding Passive Probes 1:1 and 10:1

I have a few questions regarding reduction of capacitive loading of oscilloscopes probes and cables in case of 10:1 probes compared to 1:1 probes. I am basing my research on these links:

a one meter direct (1X) coaxial probe will load a circuit with a capacitance of about 110 pF and a resistance of 1 MΩ.

Since the oscilloscope has 1MΩ||$C_{inp}$ (range up to 20pF), and coaxial cable will add 90pF/m, this will add capacitive load and reduce the bandwidth of measured signals.

Attenuator probes (10:1) will have 9MΩ||C (tunable) with the same circuit (again, probe capacitance is added in parallel and OSc input [1MΩ|20pF]). This attenuates the signal but increases the probe bandwidth to observe the high speed signals, if $$Rp \times Cp = R_{osc} \times (C_{osc} + C_{lumped\,cable\,capacitance} )$$

I didn't understand how the attenuator probe will present more bandwidth because 10:1 probe will also have the same cable capacitance (90pF/m), which is not reduced. How will this balancing increase the probe bandwidth and reduce loading on measured signal?

I've not been able to simulate and see the actual effect on measured signal spectrum. I have calculated that the impedances in parallel branches Z1 = R1||C1, Z2 = R2||C2||C3, and Z2/(Z1+Z3) should be 0.1.

I used Micro-CAP10 evaluation copy to simulate it, but I am getting weird results if I plot this quantity (Z2/Z1+Z2). I don't know what results CircuitLab would give.

Below is the equivalent schematic I used in Micro-CAP10. simulate this circuit – Schematic created using CircuitLab

Edit: The equivalent circuit is like 10MΩ shunted with ~14pF. So, the equivalent load on measured signal is reduced to 14pF. But it's not clear to me how one arrives to this equivalent circuit. Is there anyway to find the equivalent circuit of a more complex RC?