# input resistance and input capacitance for scope probe

Discussion in 'Homework Help' started by Fraser_Integration, Nov 20, 2010.

1. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
5
Hi all.

I'm preparing for a lab regarding frequency response/stray reactances and the like in oscilloscope probes. If you look at the attached picture, this is the equivalent model they've given for us to consider.

One of the questions asks, "What is the effective input resistance and input capacitance for the probe?"

I'm really not sure how to go about answering that. It doesn't specify for any particular frequency so I don't think it will involve working out the impedance of C1 and C2. Oh by the way it asked you to work out C1 before using this relationship:

9*C1 = (C2 + C3) so I would say C1 is around 11pF.

Cheers.

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,301
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If the capacitors weren't there, R1 and R2 would form a 10:1 voltage divider.

Even with the capacitors present, at low frequencies, R1 and R2 still determine the voltage division ratio. At high frequencies, the capacitors determine the voltage division ratio. In order that there be no distortion in the waveform transmitted from Vin to Vout, the voltage division ratio for low frequencies and high frequencies must be the same.

For this to happen, R2/(R1+R2) = (C2+C3)/(C1+C2+C3)

Since the problem doesn't ask for the input impedance at any particular frequency, you should probably work it out for an arbitrary input frequency ω.

If the input impedance is expressed as an equivalent input resistance and capacitance,
you would have an expression like Zin = Rin + 1/(j ω C).

So work out the complicated expression involving R1, R2, C1, C2 and C3, separate the real and imaginary parts and equate them to the Zin expression parts just above.

Sgt.Incontro likes this.