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Resistance minimizes capacitance?

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Thank you very much.

The ratio between the overall capacitance in 1x probe (=90pf) and the overall capacitance in the 10x probe (=9pf) is only 10.
So you're saying that this 10's ratio is what causes the diode in the 10X circuit to stay ON during discharging?
Meaning, if the scope's input capacitance in the 1x circuit would be 9pF and not 90pF, then the diode in this circuit would also be ON all the time?
That's correct.
I hope you have noticed that the signal amplitude at the 10X output is only 1/10 that of the 1X output. In this example, since the cap never fully discharges in the 1X probe, that ratio doesn't hold, but if we reduced the frequency, it would.
You can't get something for nothing.:D
I'm thinking I should have simulated a simple RC circuit instead of using a diode. This is more complication than you need. at this point.
 
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That's correct.
I hope you have noticed that the signal amplitude at the 10X output is only 1/10 that of the 1X output. In this example, since the cap never fully discharges in the 1X probe, that ratio doesn't hold, but if we reduced the frequency, it would.
You can't get something for nothing.:D
I'm thinking I should have simulated a simple RC circuit instead of using a diode. This is more complication than you need. at this point.

I dont know how can I thank you enough.
Take this advice, watch 'The reader', seen it today and its a great movie (I bet the book is even more awesome than the movie).

I read now the datasheet of the 1N4148, and its VF is pretty large, at least ~0.6V, Is it possible that the capacitor discharges that faster than the source declines, to keep the diode's voltage above 0.6V?
 
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