Fast rise time effect on oscilloscope probes - Page 1
Forum section: electronics / beginners

author: golden_labels
author_rank: Super Contributor
timestamp: August 28, 2024, 10:16:33 am
content: After readinga message in another topic, I have a question.(1)Quote from: Circlotron on August 25, 2024, 01:17:52 pmI was always under the impression that dielectric strength increases with frequency. So in my mind in signal vs probe confrontation, it would rather be signal that is destroyed.Note I’m asking about the probe part. I’m aware of probe-to-environment isolation at that rise rate being reduced. But I assume — perhaps wrong — Circlotron isn’t talking about the probe being destroyed through discharge to operator’s hand.edit:there isanother similar thread worth reading.(1)Yes, a rare occurance for my person! Celebrate, throw a party this weekend!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2024, 10:50:07 am
content: One insight, and one correction:1. Probes are not simply the huge resistor and capacitor they're often sketched as. Clearly that cannot be: if it were capacitive at high frequency, how can it get flat attenuation into a cable much longer than the wavelength/rise time? Granted, they typically use a lossy cable, which has particular effects, but the equivalent circuit is there's some ESR between probe tip and scope.In short, it approximates a (R1 || (R3+C1)) + (R2 || (R4+C2)) divider network. R1 and C1 are the probe's traditional 9M and some-pF, and R2 and C2 the scope's 1M and 10s pF (and whatever the cable is doing inbetween). But it's not just those, there's ESR R3 and R4. In a traditional probe, R3 might come from the cable, or we can build our own that matches to any length of 50 ohm cable (assuming we compensate for its LF capacitance with the respective value of C1), and thus have an equivalent "lo-Z" probe circuit at high frequencies: C1 and C2 become shorts and R3 = 450 into R4 = 50Ω gets us a 10:1 probe ratio with flat frequency response (above the C1/C2 cutoffs).But now, where is that voltage dropping? As frequency changes, it moves from R1 and C1, onto R3. All of which must be rated for adequate voltage, and R3 in particular must be rated for serious power.2. What is power, anyway?The power content at, say, 10s of MHz, for a square wave of 100s kHz, is quite small. It is perfectly possible to measure such waveforms with a general-purpose probe. I've certainly done it myself, many times.Note that the probe voltage limit is in terms of volts RMS (or whatever scaling) at the frequency. TheCW equivalent voltagefor harmonics of a square wave, for harmonics this high, is tiny.That's not to say Circlotron isfactually wrong*, but it's likely that some other effect precipitated the observation. It seems a case of overgeneralization: assuming that one effect occurred, and applies to all similar cases in general; when in fact the effect was rather specific (cheap probe? anodic migration leading to spark-over?) and quite rare in practice.*Well, to be pedantic: it's seconds (s), not siemens (S). And it can be written µ instead of u. But the latter is a common convenience on western keyboards, and the former is a (sadly all too) common mistake -- both easily cured.It's noteworthy that, at least most cheap probes, are actually a narrow PCB inside the body, with just a couple chip components on it -- the specified R and C values. Being small chips (0805 or so), the voltage rating is woefully inadequate, and failure rapid when exceeded (simply sparks over, or the resistive element burns up, perhaps with ensuing arc shorting, perhaps burning open first). All the same, it's possible such parts handle surge above say 400V for a long time, and perhaps one gets lucky.Since the motivation is personal experience (or, I assume so), I canequally wellclaim that such probe failurenever happens, becauseI've never had one fail due to overvoltage(and believe me, I've tried; I've probed peaks of switching waveforms into the 800V range with seemingly ordinary 10x probes that I almost certainly shouldn't've!).But between those two extremes (at least one instance of catastrophic failure out of N trials, vs. M trials with no failure), clearly you can see there is a wide gulf of possible outcomes and failure rates, of which neither datum really tells us much about the true expectation rate. And that, even if we were comparing apples to apples; but my experience for example covers probably a dozen probes (across a half-dozen types), and theirs, who knows. Let alone environmental and manufacturing variances (e.g. poor cleaning + ambient humidity can encourage anodic migration, etc.).So we must be careful not to misrepresent anecdotes as fact.Tim

author: Circlotron
author_rank: Super Contributor
timestamp: August 28, 2024, 12:54:50 pm
content: Quote from: T3sl4co1l on August 28, 2024, 10:50:07 amThe probe was the standard type that came with my then brand new Rigol DS2072 DSO about 11 years ago. I was probing the primary side of an ignition coil driven by a capacitor discharge ignition. After that I went and got a pair of x100 2.5kV rated probes and never had the slightest problem using those in the same situation. Correlation might not be causation, but I'm not going to tempt fate again.

author: golden_labels
author_rank: Super Contributor
timestamp: August 29, 2024, 10:24:22 am
content: Ananswer in another similar threadshows an example of rated voltage being strongly dependend on frequency.I don’t see, how that would work at low frequencies we’re talking about here,(1)but as Circlotron noted it was 11 years ago and we don’t even know the details of the probe, which died.(1)400 MV/s would be hundred kHz to tens MHz.

author: Circlotron
author_rank: Super Contributor
timestamp: August 29, 2024, 11:35:45 am
content: Quote from: golden_labels on August 29, 2024, 10:24:22 amThis:

author: golden_labels
author_rank: Super Contributor
timestamp: August 29, 2024, 11:45:31 am
content: Ok, so now things are clearer. Mystery solved, my questions answered. Thanks!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 29, 2024, 02:12:59 pm
content: You may want to study Parseval's theorem, which calculates the total power in a resistor (or a signal generally, when applied to any resistor) from the frequency components in a signal. The relative amplitudes of the harmonics of a wave, are given by the wave shape; for a square wave, they go as 1/N, and these are real voltages at the respective frequencies, for anywhere a sine wave decomposition (Fourier transform) is applicable (which is most of the time in electronics). You can quite accurately say that a 100kHz 100V square wave, with 20ns edges and minimal ringing, has an amplitude of ~1V at 10MHz ("~" meaning, give or take the 4/pi and peak vs. RMS (sqrt(2)) coefficients of the exact formula).Of course, we want to sum up all those components and compare them by weight to see what the actual power dissipation of the probe is -- but then we need a model of the probe. We can make some assumptions (like the equivalent network I posted above) to figure this. Or just stuff it into SPICE and measure the RMS, whatever works really.Tim

author: bson
author_rank: Unknown
timestamp: August 30, 2024, 06:29:55 pm
content: A probe is also a loop inductor formed by the probe tip, ground lead and return paths. Much of the overshoot observable with fast LVCMOS and passive probes on a high-bandwidth scope is often due to probe inductance. Just switching from the leaded clamp to a spring clip can make a huge difference.