Understanding how oscilloscopes work

Thread Starter

umichfan1

Joined Jun 16, 2012
32
I am working through the labs in Hayes & Horowitz's "Student Manual for The Art of Electronics," and I am a little confused about their model of an oscilloscope and probe described in Class 3, pp. 62-63 of the 1989 edition (scan is attached). I have several questions:

1) Why do BNC cables have an intrinsic capacitance?
2) Given that BNC cables have an intrinsic capacitance, why is this capacitance modeled as being in parallel with the internal resistance of the scope itself? (It seems intuitive that the capacitor should be in series with the internal resistance, since you hook the BNC cable into the scope.)
3) What does the text mean when above Figure N3.4 when it says that "at the two frequency extremes one or the other dominates"? If you have a voltage divider composed of two resisters, it will have exactly the same output regardless of the frequency! I also do not understand why they choose the output to be V_in/10.
4) Finally, why must scope probes have 10x impedances? In the first paragraph, it says that doing so helps to prevent circuits from behaving in "strange ways," but I don't see why these "strange" effects would be present in the first place.

Thank you very much for the help.
 

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MrChips

Joined Oct 2, 2009
30,806
1) Why do BNC cables have an intrinsic capacitance?
Every piece of wire or conductor has capacitance. Capacitance is the ability to hold a charge. In the case of any paired conductor such as lamp cord, twisted cable or coaxial cable, the capacitance is increased because of the effect of having two conductors physically close to each other.

2) Given that BNC cables have an intrinsic capacitance, why is this capacitance modeled as being in parallel with the internal resistance of the scope itself? (It seems intuitive that the capacitor should be in series with the internal resistance, since you hook the BNC cable into the scope.)
The capacitance is between the two conductors, hence it is in parallel with the intrinsic resistance. If the capacitance were in series the probe would be of no use at DC or low frequencies.

3) What does the text mean when above Figure N3.4 when it says that "at the two frequency extremes one or the other dominates"? If you have a voltage divider composed of two resisters, it will have exactly the same output regardless of the frequency! I also do not understand why they choose the output to be V_in/10.
They demonstrate two types of voltage dividers, one with resistors and the other with capacitors. Both are set for 1:10 attenuation.
The resistor voltage divider works well for DC and low frequencies. The capacitor voltage divider works for high frequencies.

4) Finally, why must scope probes have 10x impedances? In the first paragraph, it says that doing so helps to prevent circuits from behaving in "strange ways," but I don't see why these "strange" effects would be present in the first place.
If you did not use an attenuation probe the input capacitance of the scope (120pF) and the input resistance (1MΩ) might have a detrimental effect on the circuit you are trying to monitor.

By adding the extra 9MΩ in the probe you are isolating (minimizing) the effects of the scope farther away from your circuit.

Probes are also available with 100x impedance for even better isolation.
 

Thread Starter

umichfan1

Joined Jun 16, 2012
32
Thank you very much for the reply--it was very helpful. I just have a few follow-up questions:

1) Because a voltage divider is used to attenuate the signal by 1/10, I assume the resulting voltage over the 1M resistor is then multiplied by 10 before it is output to the screen. Is this correct?

2) It seems to me that the voltage divider made up of resistors works well at *all* frequencies--so why not just use it?

Thanks!
 

MrChips

Joined Oct 2, 2009
30,806
1) Because a voltage divider is used to attenuate the signal by 1/10, I assume the resulting voltage over the 1M resistor is then multiplied by 10 before it is output to the screen. Is this correct?
The 10x probe attenuates the signal by a factor of 10. You now have to read the dial on your scope differently (on an old analog scope). So if your VOLTS/DIV setting is set to .2V/div you are really using it as if it were 2V/div.

On a digital scope, you have to change the input probe setting from 1x to 10x to get the correct readout.

2) It seems to me that the voltage divider made up of resistors works well at *all* frequencies--so why not just use it?
While this is true the coaxial cable and the scope input itself presents a capacitance to ground that attenuates the signal at high frequencies.
A trimmer capacitor is placed across the 9MΩ series resistor to compensate for the HF loss. When adjusted properly a frequency compensated 10x probe will extend the bandwidth beyond what you would have with just a 1x probe.
 
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