# Trying to learn about devices but fails.

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#### alphacat

##### New Member
Trying to learn about devices but fail.

Hey,
I always find myself fails in understanding about how devices work and I dont know why it happens all the time.

For example, I decided to understand how probes work, and read in Wikipedia that in 10x probes, there's a 9Mohm resistor that is connected in series with the probe tip in order to decrease the loading that the cable capacitance would impose on the DUT (Device Under Test).

How can resistance decrease capacitance?

I might be just a student who finished only two years in EE but we're never taught that resistance can decrease capacitance.

This kind of things always stop me from learning about devices that I use in the everyday without knowing how they work.

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1. The 9Mohm resistor is said to decrease the loading capacitance because it requires a parallel capacitor to reach 10x attenuation also for AC signals?

2. How do they choose that parallel capacitance to reach constant 10x attentuation for all frequencies?

Thanks.

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#### flat5

##### Member
You won't get a response until I answer you and then someone will not be able to resist telling me I'm wrong.

The resistor isolates the probe from the load. This is true for dc and ac. I mean the resistor also isolates the capacitive loading of the probe.

The parallel cap provides a high frequency boost by somewhat bypassing the resistor. Whereas the capacitance of the probe lowers the high freq. response, the attenuating resistor bypass cap., in effect, boosts the high freq. response. The idea is to balance the two and get a flat response. Most 'scopes provide a square wave test signal that you can use to adjust the probe trimmer for the best looking waveform.
Apparently it is a rather broadband effect.

#### MikeMl

##### Well-Known Member
Suppose you have a transistor circuit you wish to measure. The circuit generates a 0.2us pulse, repeated every 1us. The circuit has an impedance of 1KΩ, meaning that you will "load" the circuit if you connect a measurement device which has an input impedance of much less than 10KΩ.

You have a standard 10MHz O'Scope, which has an input impedance of 1megΩ and an input capacitance of 20pF. So Far so good. You need to get the signal from the circuit to the scope, so you grab a 2m piece of RG58 coax cable with clip leads on one end and a BNC plug on the other. You observe a waveform.

Now you repeat the measurement using a 10X scope probe with a 2m cable; and behold, even allowing for the one tenth amplitude attenuation, the signal doesn't look much like the previous measurement.

Why are they different?

Look at the attached simulation. See if the light bulb goes off??

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#### crutschow

##### Well-Known Member
The 9Mohm probe resistance increases the scope input resistance by a factor of 10:1. The probe parallel capacitance is adjusted to give a similar 10:1 decrease in the input capacitance (increase in capacitive impedance). Thus if the probe capacitance was adjusted to 9pF for optimum square-wave response, that means the stray probe cable and oscilloscope input capacitance are a total of 81pF.

If the two capacitances are constant with frequency, then you will have a voltage divider that is 10:1 over frequency. You have a resistive divider in parallel with a capacitive divider with the same attenuation ratio.

That make sense?

#### MrAl

##### Well-Known Member
Hey,
I always find myself fails in understanding about how devices work and I dont know why it happens all the time.

How can resistance decrease capacitance?

I might be just a student who finished only two years in EE but we're never taught that resistance can decrease capacitance.

Thanks.

Hello,

That is because resistance does not decrease capacitance,
but it can sometimes help to reduce the 'apparent' capacitance, and in

When you reduce the capacitance in a circuit you change the value
from say 100pf to 10pf, but you can reduce the loading effect the
capacitor has by connecting a resistor between the circuit under test
and the capacitor.

So you see, it's not actually lowering the capacitance, it is lowering
the apparent capacitance. Thus, when a book says, "it decreases
the capacitance", it could really mean that it just reduces the *effects*
of that capacitance, not the actual value itself.

To see how this is relevant, imagine connecting a 100pf cap across two
nodes in some circuit and the effect that cap might have on that circuit.
Next, imagine connecting that same cap in series with a 1 megohm resistor
first before connecting the combination across the same two nodes, then
think about the different effect the addition of the extra resistor has.

EE is all about ratios and what is more significant than something else.

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