# Why does overshooting occur when probe's adjustable capacitor is under-value

I can't wrap my head around how overshooting (the spike) happens when the 10X probe capacitor trimmer (Cp) is under-tuned (i.e. adjustable capacitor's value is not high enough). There are countless of tutorials online but none I found explains how the shape takes its form. I've been staring at the simulation and played with the values, but can't figure intuitively as in where these spikes and dips come from, other than knowing I'm compensating the capacitor divider to achieve the same ratio as the 9:1 voltage divider so I measure the right voltage when the signal increases in frequency.

Mathematically, the RC rise or fall time is $$\ V = V_{0}(1 - e^{\frac{-1}{RC}}) \$$ so no where in time should $$\V\$$ "shoots" above or "dips" below $$\V_{0}\$$. The scope input voltage is essentially the voltage drop across $$\Cp\$$ - the smaller the capacitance, the earlier the capacitor hits the target voltage (i.e. a smaller rounded corner); vice versa, a larger capacitor takes longer to settle to its target voltage (i.e. a bigger rounded corner). But both my oscilloscope and the simulation contradicts my reasoning, and I'm not sure where I'm thinking this wrong.

• Haven't thought about it, but what happens with the charge? If the compensation cap is too small will the same amount of charge be transferred and thus (same charge on smaller cap) lead to overshoot / undershoot? – Arsenal Sep 3 '20 at 14:04
• @Arsenal a bigger cap keeps a larger amount of charges, and hence takes longer for charges to build up to a target voltage or delay to 0. So either way I should only see a "rounded corner" around the square wave, where the cap size is proportional to the size of the "rounded corner". The spikes and dips happen at the transition between 0V and 3V of the square wave, where you have the fastest rate of change in voltage (or $dV/dt$) .... then I get the spikes and dips. But I can't fill up the dots in between – KMC Sep 3 '20 at 14:26
• Look up transmission lines. What happens if you're pumping water into buckets of a certain size and someone suddenly sticks in a smaller bucket? It overflows (overshoots)because you were pumping water expecting a larger bucket. It takes time to react to the change in conditions. – DKNguyen Sep 3 '20 at 14:54
• @KMC Water height in the bucket typically represents voltage. A bucket overflowing is the equivalent of a capacitor charging above the regular voltage since capacitors can't actually overflow. You are also wrong about the current/water being higher. The current/water flow is at the same level as it was before (at least until it has time to react), but it is at a level higher than required to achieve the target line voltage. – DKNguyen Sep 3 '20 at 16:43
• The charge sent down the line was established to fill up a sequence of capacitors of a certain size, then suddenly encounter a smaller capacitor but the same level of charge is already set to be flowing down the line and has inertia (inductance) behind it , ergo, that charge flows into the cap and raises the voltage higher than the capacitors before it. – DKNguyen Sep 3 '20 at 16:45

It isn't the rise time or fall time of the signal edges that has the exponential effect; it happens once the probe has let the edge through - that's when the RC time constant takes place.

If your scope probe is over-compensating for the scope input capacitance, the ratio of probe capacitor to scope input capacitance might be 1.1 higher than the resistive divider gain so inevitably you would get the "edge" of a transient being passed through with a "gain" of 1.1 however, once the "edge" has finished, the resistive part of the divider erodes the "gain" of 1.1 back down to 1 and that is where the exponential decay takes place.

A typical probe capacitor setup

Image source.

I'm talking about the capacitor referred to above as "LF Compensation Adjustment" i.e. that capacitor is 1.1x higher than for the required compensation.

Idealized capacitive voltage divider

So, if you were to have a scope that used a purely capacitive voltage divider (with a "gain" of 1.1"), the waveform would be more like these additions in yellow: -

• What is 1.1? Over-compensating means tuning adjustable capacitor to smaller then desired value. Because reactance is inversely proportional, the voltage drop across this "smaller" capacitor becomes larger at the exact moment after the edge (or the extra "gain" you termed here?). Once square wave voltage stabilizes on 3V or 0V after the transient, the resistors dominate impedance over capacitors and exponentially approach to 0.3V or 0V (1:9 ratio). That sort of make sense to me, so shouldn't the green curve meets yellow at the righter edge, and shoots above or dips under at the lefter edge? – KMC Sep 3 '20 at 15:04
• My point is if input square wave is 3V / 0V. A properly compensated reading should be 0.3V / 0V. If compensated capacitance is smaller (over-compensated?), the voltage drop at transient should be higher than 0.3V (spikes) or below 0V (dips) before resistor dominates and stabilize the reading to 0.3 / 0V. – KMC Sep 3 '20 at 15:10
• @KMC No, if the probe capacitor is over-compensating then it will be a little bigger. I'm thinking of a series compensation capacitor here - maybe you are thinking of a different compensation capacitor setup. Anyway, that all misses the point of what my answer is trying to deliver. I've added a picture for clarity. I think you might be referring to this type of setup – Andy aka Sep 3 '20 at 15:54
• Yes. I was originally thinking $Cp$ in parallel instead of $Ct$ in series as my adjusting capacitor, but thanks to your clarification I think I get why I had the logic reversed and mixed up. If the adjustment (that cap screw) is done at the tip or Ct (in series) then the higher capacitance leads to over-compensation (over-shooting), and if adjustment is done after the cable (in parallel to scope's internal capacitance) then its the other way around. I was confused as different online sources use Ct or Cp under the same use of term "over-compensation". – KMC Sep 3 '20 at 16:39

On top of the good explanation given by Andy aka, let me add some insight using transfer functions. I'll use the fast analytical circuits techniques or FACTs described in the book I published a while ago. By calculating the various time constants of the circuit when set in different conditions (zeroed excitation and nulled output), you determine the transfer function in the twinkle of an eye:

What you want with this circuit is a flat ac response across the oscilloscope bandwidth. To meet this result, you have to make sure the zero and the pole cancel each other meaning they are located at the same frequency. If not, the attenuation and the phase are not constant across frequency as shown below where $$\C_1=C_2\$$:

Since the scope input capacitance is fixed (let's assume 12 pF), what value should you adjust $$\C_1\$$ to obtain the flat response you want? The below Mathcad sheet shows that a 1.33-pF capacitor will do the job:

When adjusting the scope probe, you can either have $$\C_1\$$ higher or lower than 1.33 pF. When $$\C_1\$$ is lower than 1.33 pF, the zero is at a higher frequency than the pole. For instance, for a 0.5-pF value, the zero is 35.4 kHz while the pole is located at 14.2 kHz. In other words, the pole dominates the response and you see rounded edges on the scope. If the cap. is now set higher than 1.33 pF, e.g. 3 pF, then the zero is set at 5.9 kHz and the pole is higher at 11.8 kHz. The zero now dominates the response and the result is a differentiated signal.

It is possible to run a quick SPICE simulations with the three different values and see the results:

As shown below, when the cap. is 1.33 pF, the signal nicely goes through with a divide-by-ten attenuation and when the cap. is 3 pF (zero lower than pole) then you see the differentiation. Finally, when the cap. is too small (0.5 pF), the pole is at a lower frequency and dominates the response: the edges are rounded.

I have found this video which nicely shows how the time-domain response changes when the pole and zero distance changes. A good illustration to understand what is going on here.

• Minor correction, in the first figure the third equation says the zero is when Z1(s) = 0, but I believe it should be when Z1(s) = inf. This lines up with illustration in that same figure where the bottom left says Z1 --> inf – Michael Sep 3 '20 at 17:44
• Oh, good catch! Oui, this is when the impedance approaches infinity that the output is nulled. I will correct it right away, merci! – Verbal Kint Sep 3 '20 at 19:54
• No problem, BTW, I am working my way though your book and am very pleased with the low entropy equations. In undergrad I vividly remember huge, unwieldy transfer functions where no physical meaning is obvious. The forms discussed in the book are very intuitive (and fast of course!). – Michael Sep 3 '20 at 20:35