capacitor compensation question - Page 1
Forum section: electronics / projects

author: kvresto
author_rank: Regular Contributor
timestamp: October 19, 2016, 11:03:34 am
content: Hi everyone.The pic shows a trimmer cap in parallel with a resistor, and I often see this arrangement in attenuators of some type of input signal, square or sine. Whats this technique called, and how do you design for it. Ive also seen it in CRO probes, and everyone has a name for it.thankskvresto

author: rs20
author_rank: Super Contributor
timestamp: October 19, 2016, 11:15:20 am
content: That pic shows a trimmer cap in parallel with acapacitor? That just allows you to have a fine adjustment on the capacitor value (i.e., if the trimmer is 10pF, then you can adjust just a little bit from 470pF to 480pF.)

author: kvresto
author_rank: Regular Contributor
timestamp: October 19, 2016, 12:25:18 pm
content: actually I was thinking more of this type of circuit, I hunted online to find one, just imagine one or both of the caps are actually trimmers. How do you calculate the values for the capacitors? Also whats the proper name for this type of compensation? just out of curiosity.cheers

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 19, 2016, 12:26:27 pm
content: Trimming or padding.If you need a somewhat smaller value than the trimmer, you can connect some capacitance in series with it. This guy made a handy calculator:http://seventransistorlabs.com/Calc/Range.html#capTim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 19, 2016, 12:29:44 pm
content: Quote from: kvresto on October 19, 2016, 12:25:18 pmOh for one of those, that's a lead-lag compensation network. The R*C time constant of the R || C on top, has to match the R || C below.Simple symmetry.Tim

author: kvresto
author_rank: Regular Contributor
timestamp: October 19, 2016, 12:42:08 pm
content: RC time constant of a RC parallel cct?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 19, 2016, 12:54:56 pm
content: Yeah, so R1 * C1 = R2 * C2. Solve for whichever you need.Tim

author: kvresto
author_rank: Regular Contributor
timestamp: October 19, 2016, 01:04:06 pm
content: ok, no worries, that makes sense.So if the top part R || C = the bottom part R || C, is this cct then "compensated"? from a signal integrity stand point,? ie a square wave will actually look like a square wave, attenuated by the resistive divider.

author: danadak
author_rank: Super Contributor
timestamp: October 19, 2016, 01:06:53 pm
content: This is a tad lengthy but covers basic probe compensation, which is oneapplication of what you are interested in.http://circuitslab.case.edu/manuals/Probe_Fundamentals-_Tektronix.pdfRegards, Dana.

author: rs20
author_rank: Super Contributor
timestamp: October 19, 2016, 09:31:33 pm
content: Another reason, IIUC, for adding capacitors to a resistor divider is to attenuate thermal noise generated in a voltage divider resistors. Adding a capacitor to the output of the divider will "shunt away" a lot of thermal noise, which is a good thing, but the resulting asymmetry leads to a low-pass filtering effect on the voltage divider. Adding a capacitor on the top side resolves the asymmetry, giving a low-thermal-noise voltage divider -- the downside is that the input now presents a capacitative load.

author: kvresto
author_rank: Regular Contributor
timestamp: October 19, 2016, 10:48:29 pm
content: So how would I begin to make a capacitor selection? say its an attenuator for an analog front end, attenuating a large sig(5V-15V) sine/square wave.

author: w2aew
author_rank: Super Contributor
timestamp: October 20, 2016, 12:50:25 am
content: Quote from: kvresto on October 19, 2016, 12:25:18 pmVery common arrangement for 10x oscilloscope probes. At low frequencies, the 10x voltage division is dominated by the 1Mohm input resistance of the scope and the 9Mohm series resistor in the probe. As the frequency goes up, the scope's input capacitance lowers the input impedance, thus the division ratio would move away from 10:1. Thus, a compensation cap is placed around the 9M series resistor, and adjusted so that as the frequency goes up, the 10:1 voltage division is maintained (ultimately dominated by the ratio of the scope's input capacitance and the compensation capacitor).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 20, 2016, 03:18:11 am
content: Capacitor selection depends on the equivalent circuit of the bottom part, which is probably the part you already have. More generally, the top impedance (shown as the R || C) is N times the bottom impedance.This works even if the bottom impedance is much more complicated (which it generally will be, due to stray L and the input impedance of active circuits), of course, constructing a complicated yet precise impedance is a challenge, so you want to make sure the bottom impedance is as simple as possible.Alternately, you can work backwards from what the min/typ/max input impedance should be. For example, a 10x scope probe is 10 megs and tens of pF, which allows the cable and scope input to be approx. 1M || 200pF or so, which means the scope input needs to be designed to satisfy that (which, with JFETs and such, is quite possible).At very high frequencies, by the way, notice that the equivalent isn't a true capacitor. The schematic is a bit of a lie. Transmission lines and resistances take over -- because if they don't, sooner or later the capacitors will themselves become non-ideal, and the impedance bucks up and down all over the place, and you get ringing. (For a probe built from chip R/Cs, this will be well in the 100s of MHz.)Tim

author: kvresto
author_rank: Regular Contributor
timestamp: October 20, 2016, 05:38:21 am
content: Hmmm, just realizing, although there are heaps of references to how a cro probe works, I've discovered how difficult this area is, and we all have put together voltage dividers, I now know its been just luck that my cct have worked above DC.So its a suck it a see approach. Built test and measure. I suspect as no one can provide a step by step design process, its more complicated than it looks.kvresto

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 20, 2016, 06:55:01 am
content: Quote from: kvresto on October 20, 2016, 05:38:21 amI could if you paid me.The impedance of a ladder network (N components) is a continued fraction (Nth level), so you can always subtract out a given (real or imaginary) component, invert, and repeat to achieve arbitrary accuracy.If the ladder network is lumped-equivalent and finite, this gives an exact solution*; for any two-pin (one-port) arbitrary element (that is analytic, as any real two-terminal impedance must be), it will converge.*I don't actually know this, and a proof would either be interesting...or dreadful.A ladder network can be bridged with impedances between non-adjacent nodes, for instance; or, more generally speaking, you can measure the impedance between any two nodes in a general graph network. Neither of which will have the straightforward continued fraction formula for their impedances -- but if an equivalentdoesactually exist, then it implies topological conversion as well, which is really cool.Tim

author: kvresto
author_rank: Regular Contributor
timestamp: October 20, 2016, 09:06:02 am
content: As I said, I suspect as no one can provide a step by step design process, its more complicated than it looks.Its all fun, just another rabbit hole to explore at some point.cheers, and thanks.