Reducing circuit to resistor and capacitor in parallel

Given the following circuit, in which the value of C1 is known to be 9.3 pF and in which the cable to the probe contributes an additional capacitance of 45 pF between the oscilloscope input and ground and an AC voltage is being applied at V1: I want to determine the input impedance as seen by V1, in the form of a single capacitor and resistor in parallel. How would one go about it?

I know that the impedance of a capacitor and resistor in parallel is given by:

$$Z = \frac{1}{1 + i \omega RC}$$

So I can find out the impedance of the C1 capacitor and 9 Mohm resistor, and of the 30pF capacitor and 1Mohm resistor and add them. How do I add, however, the impedance of the 45pF capacitor that is connected to the ground to the circuit? And, after that, how do I express the impedance of the whole circuit as a resistor and capacitor in parallel?

• Cable capacitance bit is easy it appears as a capacitor in parallel with the 30pF and you know how to add capacitors in parallel. ;) – RoyC Jan 1 '17 at 11:02

Let me redraw the circuit, and hopefully it will become clear: simulate this circuit – Schematic created using CircuitLab

Because we are interested in the impedance seen by V1, we want to reduce this entire circuit to a single impedance between V1 and ground (we don't care about V2). So:

• combine C2 and C3
• reduce C1/R1 to a single impedance (Z1)
• reduce R2/(C2+C3) to a single impedance (Z2)
• sum Z1 and Z2 to obtain the total impedance
• separate the total impedance into real (resistance) and imaginary (capacitance) to obtain the "resistor and capacitor in parallel" part.

Performing these calculations in the $s$-domain will make it much more straightforward.

Being picky: EEs use j rather than i for the square root of -1, because i is used for current.

The impedance of a resistor and capacitor in parallel is R/(jwRC+1): note the numerator. You can sanity check impedance expressions, and transfer functions in general, by making sure the values at DC and at very high frequency are correct. A parallel RC should look like R at DC, and a short circuit at high frequency.

It's a 10 to 1 probe compensation scheme. The capacitors are picked so that R1C1 = R2C2, making the overall attenuation a real number - frequency independent - in this case 1/10. The input impedance is the sum of the R1C1 and R2C2 impedances. (R1 + R2)/(jwRC +1), where RC is either R1C1 or R2C2. Resolving it into a single RC parallel equivalent takes some fiddling. We already know the numerator will be R1+R2. In the denominator we can replace R with R1+R2 and replace C with R1C1/(R1+R2), ending up with the same corner frequency. That makes our input equivalent 10 megohms in parallel with C1*0.9.

That's how it's all supposed to work. Your capacitor values are off somewhat. I suspect what you're showing as 9.3 pF is actually an adjustable compensation capacitor.