Oscilloscope input

ZimmerJ

Joined Dec 9, 2020
58
Hello,

Simple question, but maybe not so simple to answer: Why is there a capacitor at the input of a oscilloscope? The one usually being around 20pF

I am leaning towards thinking that it has something to do with the amplifier that comes after this RC-network at the input, but i am not sure. I believe i am missing some basic circuitry-theory in regards to electronics in general.

In relation to the probe impedance, this capacitor seems to do nothing good really, except adding capacitance to cable-capacitance so that the bandwidth is decreased. To make my question even more clear, say that we use a 1x attenuation (meaning no attenuation), what happens if we remove the capacitor at the input? → In my mind, i only see the bandwidth increasing, because we are achieving less loading effect.

So unless it is meant for other levels of attenuation like 10x, 100x etc, which i don't expect because again, it just adds up to the total capacitance, i am left to believe that there is another purpose to why there is a capacitor at the input.

ZimmerJ

Joined Dec 9, 2020
58
Hm, i don't understand why it is needed for compensation?

What i see, is that the relationship between the different capacitances is suppose to be equal to the relationship between the resistances, 1MΩ/9MΩ, so why not skip the input capacitor and lower the variable capacitance on the probe? Well, since it is inverse, the variable capacitance should be higher.

It only says in the article that the probe must accommodate for that additional capacitance, not the reason why it is there, unless i am misunderstanding something.

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BobTPH

Joined Jun 5, 2013
4,244
If you could make the circuit without any capacitance, they would. But you cannot. There is an actual capacitor in the circuit, it is representing the stray capacitance.

Compensation. Since there is this stray capacitance in parallel with the input, we have a low pass filter. The compensation capacitor bypasses the input resistance to make a high pass filter that compensates for the low pass filter giving you a flatter response.

Bob

ZimmerJ

Joined Dec 9, 2020
58
R2 and C2 are not components.
They represent the model of the scope input.
That's what i started thinking, and i have been trying to find an answer to why there is capacitance at the input. Looking through block-schematics i can't really see it.

ZimmerJ

Joined Dec 9, 2020
58
If you could make the circuit without any capacitance, they would. But you cannot. There is an actual capacitor in the circuit, it is representing the stray capacitance.

Compensation. Since there is this stray capacitance in parallel with the input, we have a low pass filter. The compensation capacitor bypasses the input resistance to make a high pass filter that compensates for the low pass filter giving you a flatter response.

Bob
Okay, so there is an unintended capacitance going on? I read somewhere that you actually want to remove the DC-component before the ADC, but i don't remember exactly why. Hence why you place a capacitor at the input.

Deleted member 115935

Joined Dec 31, 1969
0
The capacitor at the input to an ADC is different,

The probe of the scope, connects via a length of coax cable,
coax cable, is one wire surrounded by another wire ( screen )
that makes a good capacitance,
coax cable is actualy specified with a capacitance per meter length,,,,,
and it varies as the temperature / time changes,
hence, to get a flat input, using a passive circuit, you need to cancel out that capacitance
the circuit, with the capacitor divder does that,

The capacitor on an ADC input,
if its there, ( some ADC are DC coupled )
is to move the centre of the input voltage to lign up with the centre of the ADC input range,
to remove any DC offset,

But thats a complete different reason,
( ADC dont have a Mohm input impedance _

MrChips

Joined Oct 2, 2009
24,990
You don't want to couple the input to the ADC via a capacitor if you want to measure DC.

sparky 1

Joined Nov 3, 2018
647
I found the 5 main mistakes video useful, since the controls can be based on the schematic above.
The small picture in the left hand corner helps to cope when your oscilloscope input is misbehaving.
He refers back to other videos depending on which of the 5 troubles you may have first, very systematic, sometimes it's just getting used to scope.

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KeepItSimpleStupid

Joined Mar 4, 2014
5,090
You want the divider to be purely resistive. Coax cable has a capacitance per foot The scope's input impeadace is intentionally 1M || 20 pf usually, so it can be nulled out.

Remember that the probe has an adjustable capacitor. It might be at the input connector or in the probe body.

You apply a square wave and adjust so the edges are clean. The Forrier Series of a perfect square wave requires infinate bandwidth. The capacitance of the probe is compensated when a square wave input looks like a square wave on the scope display.

ZimmerJ

Joined Dec 9, 2020
58
The capacitor at the input to an ADC is different,

The probe of the scope, connects via a length of coax cable,
coax cable, is one wire surrounded by another wire ( screen )
that makes a good capacitance,
coax cable is actualy specified with a capacitance per meter length,,,,,
and it varies as the temperature / time changes,
hence, to get a flat input, using a passive circuit, you need to cancel out that capacitance
the circuit, with the capacitor divder does that,

The capacitor on an ADC input,
if its there, ( some ADC are DC coupled )
is to move the centre of the input voltage to lign up with the centre of the ADC input range,
to remove any DC offset,

But thats a complete different reason,
( ADC dont have a Mohm input impedance _
Got it, the ADC capacitor is is not the same. Looking around, that stage comes after the RC network that i am asking about.

But the input capacitor is not needed to form the capacitance divider or am i missing something? Bob answered above that it's unavoidable stray capacitance.

ZimmerJ

Joined Dec 9, 2020
58
This is great, and in this they referred to another book with more focus on what i am looking for: https://www.davmar.org/TE/TekConcepts/TekVertAmpCircuits.pdf

On page 41 (47 in the browser), it says Fixed RC input and that "Maintaining a constant RC eliminates probe recompensation with attenuation changes". I am not sure what to make of this, but it sounds like there is intentional capacitance at the input, like keepitsimple answered aswell, and not only forms of stray capacitance.

ZimmerJ

Joined Dec 9, 2020
58
You want the divider to be purely resistive. Coax cable has a capacitance per foot The scope's input impeadace is intentionally 1M || 20 pf usually, so it can be nulled out.

Remember that the probe has an adjustable capacitor. It might be at the input connector or in the probe body.

You apply a square wave and adjust so the edges are clean. The Forrier Series of a perfect square wave requires infinate bandwidth. The capacitance of the probe is compensated when a square wave input looks like a square wave on the scope display.
But since you have a variable capacitor at the probe, why have a fixed capacitance at the input? This is what i cannot get inside my head.

ZimmerJ

Joined Dec 9, 2020
58
R2 and C2 are not components.
They represent the model of the scope input.
So far i am being told that stray capacitances are unavoidable at the input, i don't know the exact contributing factors to this but i can understand it. But articles also seem to press that the capacitance at the input is in fixed form, which tells me that there is probably a capacitor installed to form a constant RC network at the input. To me it does not make sense that a manufacturer can put i fixed value of a stray capacitance at the input, i mean the tolerances is what, 200 % then?

Maybe i am slow, i don't know. But i haven't quite found the answer yet.

tautech

Joined Oct 8, 2019
233
This is great, and in this they referred to another book with more focus on what i am looking for: https://www.davmar.org/TE/TekConcepts/TekVertAmpCircuits.pdf

On page 41 (47 in the browser), it says Fixed RC input and that "Maintaining a constant RC eliminates probe recompensation with attenuation changes". I am not sure what to make of this, but it sounds like there is intentional capacitance at the input, like keepitsimple answered aswell, and not only forms of stray capacitance.
This is necessary as every change in input attenuation (V/div) is in essence another circuit and therefore needs to be kept matched to the probe for accurate waveform reproduction and this issue is more prevalent with a 10x probe.
The modern DSO might only have 2 or 3 real attenuation steps when the smaller in-between steps are done in IC's where the attenuation steps are driven by SW.

ZimmerJ

Joined Dec 9, 2020
58
This is necessary as every change in input attenuation (V/div) is in essence another circuit and therefore needs to be kept matched to the probe for accurate waveform reproduction and this issue is more prevalent with a 10x probe.
The modern DSO might only have 2 or 3 real attenuation steps when the smaller in-between steps are done in IC's where the attenuation steps are driven by SW.
Okay, that makes sense. So let's sum it up a bit. Say we are going with 1x attenuation at the probe, regardless of the attenuation of the volt/div-circuitry, you maintain a "steady" compensation by having a fixed RC-network at the input?

If this is the case, i believe i am missing some theory here. Since there is already cable capacitance of the probe, i believe this is not to be expected a "steady" capacitance? Does this mean that applying a capacitor in parallel (input capacitance), counters this somehow?

Or is it that most probes have a certain level of capacitance, that needs to be increased with a parallel capacitor, in order to achieve good compensation?

MrChips

Joined Oct 2, 2009
24,990
Every oscilloscope without using a probe has a certain input impedance and frequency response.

You add a probe for one primary reason - to increase the input impedance of the measuring system.
The typical input impedance of the oscilloscope is 1MΩ.
You add a x10 probe to increase the measuring impedance to 10MΩ. The downside of this is you lose 20dB in signal amplitude. An upside is you extend the bandwidth of the measuring system, the probe and the oscilloscope.

ZimmerJ

Joined Dec 9, 2020
58
Every oscilloscope without using a probe has a certain input impedance and frequency response.

You add a probe for one primary reason - to increase the input impedance of the measuring system.
The typical input impedance of the oscilloscope is 1MΩ.
You add a x10 probe to increase the measuring impedance to 10MΩ. The downside of this is you lose 20dB in signal amplitude. An upside is you extend the bandwidth of the measuring system, the probe and the oscilloscope.
I appreciate the answer, but you are leaving out the capacitance value, which is what i am trying to figure out.

As you say, you add a 10x probe to increase impedance, and it accomplishes that for higher frequencies by including a capacitor that's ends up being in series to the cable capacitance. Serial capacitors sums up a lower capacitance, which in turn increases impedance. But this is the thing, because now there is also a capacitor at the input that ends up being parallel to the probe capacitance, and this means higher capacitance and lower impedance. So why apply input capacitance when it works against the whole purpose of attenuating the signal?

I am pretty sure it's not only stray capacitance so i have a strong feeling that the general 15-30 pF is there for a particular reason.