EEVblog #1367 - 5 Types of Oscilloscope Passive Probes COMPARED - Page 1
Forum section: electronics / general

author: EEVblog
author_rank: Unknown
timestamp: January 26, 2021, 11:04:38 pm
content: Dave looks at the pros and cons of 5 different types of oscilloscope passive probes.Switchable x1/x10, Fixed x10, High voltage single ended, DIY Transmission line Z0 probe, and BNC to banana/croc leads.PART 1 of 2

author: Tom45
author_rank: Frequent Contributor
timestamp: January 27, 2021, 01:41:36 am
content: The Siglent 2000X Plus can be set to 21X probe attenuation. It also has 50 ohm termination and probe readout pin support.Thanks for the Zo probe information. I look forward to the rest of the story in part 2.

author: Anthocyanina
author_rank: Frequent Contributor
timestamp: January 27, 2021, 03:42:10 am
content: why is "high capacitance for high frequency work" listed as a con? isn't it a pro of x10 probes that they offer a wider bandwidth and are better for high frequency than x1 probes?

author: HyperSpectral
author_rank: Newbie
timestamp: January 27, 2021, 07:05:45 pm
content: You may have implied that a 350MHz probe on a 350Mhz scope will give 350MHz bandwidth. The actual bandwidth of this system will be slightly less that 250MHz. The calculation of this is done using rise time and the system rise time is the square root of the sum of the squares of the individual rise times. The result of this is that the bandwidth of a scope and probe of identical values is 70.7% of that value.So, in order to approach the scope bandwidth, a much higher bandwidth probe must be used. Note that in theory, the scope bandwidth can not be reached if a probe of finite bandwidth (any actual probe) is used. A 500MHz or higher probe should be used to get realistic results near the 350MHz scope bandwidth.All that said, the quicksand that many people fall into is forgetting that any waveform other than a Sine wave has harmonics starting at 2 or 3 times the fundamental. So, a 350MHz scope will display a 350MHz Sine wave at 3dB loss, but any other waveform will be seriously distorted.

author: madires
author_rank: Super Contributor
timestamp: January 27, 2021, 07:32:59 pm
content: Would a BNC plug with an IR photo-diode be a passive or active probe?

author: Tom45
author_rank: Frequent Contributor
timestamp: January 27, 2021, 07:37:05 pm
content: Quote from: Anthocyanina on January 27, 2021, 03:42:10 amConsider this simplified circuit of a 10:1 scope probe:At DC, the reactance of the tip capacitor is infinite and the probe's input resistance (impedance) is 10 Meg.For AC the capacitor impedance is no longer infinite. As the frequency goes up, the capacitor's impedance drops.Dave calculates the capacitor's impedance and shows the formula.At 500 MHz, a 10 pf tip capacitance has an impedance of only 32 ohms. So the 10 Meg probe impedance is only 32 ohms at 500 MHz.

author: Anthocyanina
author_rank: Frequent Contributor
timestamp: January 27, 2021, 09:07:48 pm
content: oh, i see, and that lower impedance is a con since it may affect how the circuit being measured behaves, and i guess that also limits what voltages you can measure at those high frequencies?thank you!

author: Bud
author_rank: Super Contributor
timestamp: January 28, 2021, 03:44:57 am
content: @19:30 -20:30 talking about the 3ft piece of coax having 100pF capacitance which at 1GHz represents 1.6 Ohm shunt impedance- NO. At RF frequencies coax is a transmission line and when terminated to a matched load (50 Ohm in this case on the scope input) will only have resistive losses in the wire , which were specified right there in the datasheet that you showed , in the next table right below the capacitance/ft data. That datasheet specified only 1 db of loss at 1GHz for 1m length of the coax used..If coax cables had THAT low shunt capacitance at RF, they would be useless at high frequency, which is not the case and is an exact opposite.

author: tautech
author_rank: Super Contributor
timestamp: January 28, 2021, 08:10:56 am
content: Basic errors....a few minutes in, the compensation trimmer on a passive probe is LF not HF although some passive probes have both and then they are in the BNC end compensation box.13.30 that 100x probe is specially designed for isolated channel probes in that all connections on it provide isolation from voltages that might bite whereas your standard 100x probe is little different to any other passive probe.

author: Bud
author_rank: Super Contributor
timestamp: January 29, 2021, 03:10:50 pm
content: Here is an excerpt on a home made 21:1 probe from "High Speed Digital Design, A Handbook of Black Magic" by Johnson/Graham.The 1kOhm series resistor is not to "isolate high capacitance", but to reduce the rise time degradation L/R due to inductance of the sense loop, see the attached picture. With a 1kOhm resistor the ratio L/R is much smaller.

author: tchicago
author_rank: Regular Contributor
timestamp: January 29, 2021, 03:39:04 pm
content: Quote from: Bud on January 28, 2021, 03:44:57 amBut the scope's input impedance is not 50 Ohm. It is 1 MOhm. Or, you are saying that the RF impedance is 50 Ohm, while DC impedance is 1 MOhm? I don't think this is the case here.

author: Bud
author_rank: Super Contributor
timestamp: January 29, 2021, 06:04:58 pm
content: The 21:1 probe is made using 1kOhm external resistor at the test point and 50 Ohm terminator either switched on on the scope input or using an external termination resistor at the scope input connector if the scope does not have 50 Ohm input feature . Otherwise it will not have 21:1 voltage divider ratio. (1000+50)/50=21. You may be confused because cheap and some mid-range scopes do not have built-in 50 Ohm input impedance feature and only have 1M input. For such scopes you have to use an external 50 Ohm terminator at the scope input.

author: tggzzz
author_rank: Super Contributor
timestamp: January 29, 2021, 08:21:49 pm
content: Quote from: Bud on January 29, 2021, 06:04:58 pmAnd then it isn't 50ohms, but 50ohms//20pF. That capacitance will change the attenuation as a function of requency, depending on the cable length.That's why I like my 1970s Tek 485: it has a proper 50ohm input attenuator.

author: nctnico
author_rank: Super Contributor
timestamp: January 29, 2021, 09:31:44 pm
content: Quote from: Bud on January 28, 2021, 03:44:57 amAdded to that is that both sides of the coax need to see an impedance of 50 Ohms. Just putting 1k Ohm in series is not going to cut it. Terminating the end with 50 Ohm is only helping to get rid of reflections.

author: tggzzz
author_rank: Super Contributor
timestamp: January 29, 2021, 10:23:37 pm
content: Quote from: nctnico on January 29, 2021, 09:31:44 pmEr, no.I suggest you understand section 3 ofhttp://www.davmar.org/TE/TekConcepts/TekProbeCircuits.pdf

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 12:45:01 am
content: Quote from: tggzzz on January 29, 2021, 10:23:37 pmJust look at these screendumps. The first is without 50 Ohm at the beginning of the coax cable, the second is with.And for kicks using a Tektronix P6156 with a 20x attenuator and a probe-to-BNC adapter:The 2nd image looks identical to the actual signal when connected directly. The others don't.Direct connection:In the end you can't beat transmission line theory. It is clear that not putting a 50 Ohm termination at the beginning of the coax cable is a a trade-off to get a reasonable result with minimal attenuation.Edit: fixed inline images.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 01:27:29 am
content: What is the scope, the scope's input impedance, the source, the source waveform, the cable length, the probing technique?At these speeds the details matter. many are 50ohms//20pF, which causes problems.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 01:41:21 am
content: Quote from: tggzzz on January 30, 2021, 01:27:29 amThis is measured using the calibrator output on a Lecroy Wavepro 7300A set to 50 Ohm with full bandwidth. The coax cable (about 50cm) is soldered onto an SMA (with 1k series resistor) which is connected to the calibrator output (BNC). Lead lenghts about 2cm. The test with the Tektronix P6156 probe (with 20x attenuator) is done with a BNC adapter on the probe in order to connect to the calibrator output. No ground leads or ground spring:The P6156 is basically a 1k Ohm resistor in series with a coax cable so it is no surprise the result is the same. Edit: I see something went wrong with the inline images which I fixed now.

author: PartialDischarge
author_rank: Super Contributor
timestamp: January 30, 2021, 02:54:36 am
content: Quote from: Bud on January 29, 2021, 03:10:50 pmI don’t have access to the rest of that book, but it is amazing how misleading the treatment of the risetime in that probe is. The discussion is centered around L/R and it is mentioned that this probe is going to be very fast, nope, the limiting factor is the input C to the scope and the dominating effect is RC with now R=1k. With a 1MOhm scope and external 50Ohm termination, that probe is going to be slow as molasses, with scopes with internal 50 Ohm better but not spectacular.I would not waste time in building that probe when a FET probe is what is needed there

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:10:12 am
content: Quote from: nctnico on January 30, 2021, 01:41:21 amIt sounds like the 50ohm at the source is connected calout-50ohmToGnd-1kohm-cable-scope. In that case the effect you see without the source 50ohm might be due to the cal out behaving properly when correctly loaded.Alternatively, I suspect the scope isn't 50ohm input but is really 50ohm//15pF. In that case you will see the effect stated in the P6156 manual, viz:QuoteThat one fooled me once in the past!As you say, you can't beat transmission line theory

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 11:28:30 am
content: Quote from: tggzzz on January 30, 2021, 10:10:12 amNo because with the second test with the coax, the cal output still sees a 1k Ohm resistor at it's output and yet the resulting signal looks fine.QuoteNo; I have used a 3GHz scope to do the test. I'm damn sure it has proper 50 Ohm inputs! Please check the specs before making these kind of statements.I suggest you to do a simulation with transmission lines in Spice.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 11:46:19 am
content: Quote from: nctnico on January 30, 2021, 11:28:30 amIn that case I don't understand your circuit - please supply a photo or schematic.QuoteI've seen "50ohm" scopes that simply put 50ohm in parallel with the 1Mohm//15pF.Looking at the pictures of the scope, there are many connections that can be used to communicate calibration info from theLeCroyprobe to the scope. The scope might then be using DSP to adjust the display accordingly. If theTekprobe doesn't supply that info, then obviously the correction cannot be done.I'll do a simulation once I have your schematic.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 02:17:57 pm
content: No. Again, you are making assumptions which aren't valid. Just take the schematic (1k in series with the source at the beginning of the coax cable and a 50 Ohm terminator at the end) from the Tektronix paper you linked to and simulate it using a lossy transmission line model. You'll see that you get steps indicating an impedance mismatch (just as transmission line theory predicts).

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 02:24:38 pm
content: Quote from: nctnico on January 30, 2021, 11:28:30 amHere is evidence that youdon'thave a 50ohm input.The pictures below are with my Tek 485 and HP10020a 1.5GHz Z0 probe. The key point about the 485 is that it hastwoinput attenuators: one is the traditional 1Mohm scope attenuator, the other is a traditional RF 50ohm attenuator. An RF relay connects the input toeitherone or the other attenuators.The first picture shows the source: the Tek485 cal out (<1ns risetime) correctly loaded with a 50ohm through terminator that ought to be good to a GHz or so. The probe (500ohm input) is directly connected to the source, without leads. Timebase is 5ns/div.The second picture shows the trace when the scope's (real) 50ohm attenuator is used. I think you will agree that is a nice enough waveform.The third picture shows the trace when using the scope's 1Mohm input with an inline 50ohm through terminator. Hence the input "seen" by the cable is really 50ohm//20pF. You will note the dip you see in your traces. Where do you think that has come from?The fourth picture is used to highlight the probe cable length, by switching to the 50ohm attenuator so the input impedance is 25ohms. You will note the position of the step coincides with the dip in the third picture, thus demonstrating that the dip is due to the 20pF input capacitance.Incidentally, those familiar with using TDRs will realise that a dip in a trace is due to capactance, and a peak is due to inductance.BTW, a Z0 probe doesnotuse a lossy transmission line; it uses standard coax.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 02:55:44 pm
content: Just do the simulation and you'll see I'm right. You will get a step in the signal if the beginning of the coax cable isn't terminated.I'm 100% sure the Lecroy 7300A has proper 50 Ohm inputs. Proof of that is that the 'dip' doesn't show with a direct cable connection. A 3GHz oscilloscope would be utterly useless without proper 50 Ohm inputs. People would have come to Lecroy's headquarters with pitchforks, tar and feathers if they screwed it up.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 02:59:29 pm
content: Quote from: nctnico on January 30, 2021, 02:55:44 pmDo the measurements, and you'll see I'm right!Or, how do you explain those measurements?

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 03:12:43 pm
content: Quote from: nctnico on January 30, 2021, 02:55:44 pmHere's the simulation, and you are wrong!No 50ohm load at the source.With a pure 50ohm you get a perfect trace.With 50ohm//20pF you get the dips you see in your trace.Your scope isn't pure 50ohms

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 03:17:58 pm
content: That isn't a lossy transmission line model! Model it with a lossy transmission line using real coax cable parameters and you'll see I'm right and your simulation is wrong. The 'regular' transmission line model is an over simplification. In the pspice I use I don't even get the 'dip' if I put 22pf in parallel with the 50 Ohm resistor.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 03:37:33 pm
content: Quote from: nctnico on January 30, 2021, 03:17:58 pmOf course it isn't a lossy transmission line model; it is only 2m of RG174 or similar!Resistive divider Z0 probes use standard 50ohm coax, not the lossy lines used in 10Mohm probes.If you try to make a Z0 probe with lossy cable, you are misunderstanding the theory and will get suboptimal results!

author: Tom45
author_rank: Frequent Contributor
timestamp: January 30, 2021, 04:50:53 pm
content: Quote from: tggzzz on January 30, 2021, 03:12:43 pmCould you elaborate on your two simulations? The circuits shown seem to be the same for both. So what is different between them?

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 05:12:45 pm
content: Quote from: Tom45 on January 30, 2021, 04:50:53 pmThe value of Cscope; factor of 106difference.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 07:06:32 pm
content: Quote from: tggzzz on January 30, 2021, 03:37:33 pmYou are misunderstanding. A lossy transmission line model has nothing to do with lossy coax! A lossy transmission line is a Spice simulation model with a better approximation of real coax so you will also get more realistic results. And you'll also need to model the probe head itself more accurately (including parasitic capacitances). You'll see that the latter will introduce a dip.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 07:31:05 pm
content: Quote from: nctnico on January 30, 2021, 07:06:32 pmYour second sentence directly contradicts your third sentence.Explain the measurements I have shown.Prove your contention by showing us your simulation thatmatches your description and measurementsdoes not match my description and measurementsI'm not holding my breath.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 08:04:34 pm
content: Quote from: tggzzz on January 30, 2021, 07:31:05 pmNo it doesn't; your statement only shows a lack of understanding Spice models for transmission lines. You seem to get stuck at the word 'lossy' but it is used in an entirely different context which has nothing to do with coax. The lossy transmission line model can be used to model any kind of transmission line which does not have to be coax at all.QuoteAll your measurements show is that your probe shows seemingly correct behaviour. However, your probe is a 1:10 model with a 450 Ohm series resistor instead of a 1:20 model with a 950 Ohm series resistor. Crafting the attenuator cartridges for low-Z probes is an art in itself. Maybe HP is better at that than Tektronix but a simple 1k resistor in series with a piece of coax is not going to cut it. Try that first!

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 09:51:23 pm
content: Quote from: nctnico on January 30, 2021, 08:04:34 pmIf you really think that the issue is related to 1:10 vs 1:20 then there is little point in continuing this conversation.Please explain why Z0 probes use standard coax, not the very resistive coax used in 10Mohm probes.Please provide your simulations that justify your conjectures. If you choose not to, please starts why not!Come on; it took me 15minutes.I sure it won't take you much longer than that!

author: radar_macgyver
author_rank: Frequent Contributor
timestamp: January 30, 2021, 10:07:04 pm
content: Could someone with a VNA look at the S11 of their scope input in 50 ohm mode? I don't have one at home.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 10:11:04 pm
content: Quote from: tggzzz on January 30, 2021, 09:51:23 pmNot the issue but a much larger impedance mismatch. Please try using a 1k resistor attached to a piece of coax.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:16:26 pm
content: Quote from: radar_macgyver on January 30, 2021, 10:07:04 pmA sensible suggestion. I doubt nctnico will make such a measurement on his LeCroy.The Tek 485's input attenuators (plural) can be seen in the schematic.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:17:34 pm
content: Quote from: nctnico on January 30, 2021, 10:11:04 pmNonsense.Stop avoiding the issues and provide measurements or simulations!Please explain why Z0 probes use standard coax, not the very resistive coax used in 10Mohm probes.Please provide your simulations that justify your conjectures. If you choose not to, please starts why not!Come on; it took me 15minutes.I sure it won't take you much longer than that!

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 10:24:06 pm
content: See the earlier posting. I added a simple simulation schematic and results. But keep in mind that this simulation does not model the probe head and parasitic capacitances / inductances. However the effects of the improper termination of the coax are clearly visible!

author: Tom45
author_rank: Frequent Contributor
timestamp: January 30, 2021, 10:39:11 pm
content: Quote from: nctnico on January 30, 2021, 10:24:06 pmMaybe I missed something, but which termination are you talking about when comparing the two traces? At the scope end of the coax, or at the probe end?

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 10:48:20 pm
content: Quote from: Tom45 on January 30, 2021, 10:39:11 pmAt the beginning of the coax there is resistor R4 which is either 52 Ohms or (as shown in the schematic) has such a high value that it is basically left out. The resulting signal is measured at the end of the coax across the 50 Ohm resistor which resembles the oscilloscope's input signal.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:49:51 pm
content: Quote from: Tom45 on January 30, 2021, 10:39:11 pmNo, you haven't missed something; nctnico's statements are ambiguous and unclear.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 10:51:39 pm
content: Quote from: tggzzz on January 30, 2021, 10:49:51 pmCome on. All this time we have been talking about where the termination should be and now it is unclear?Can't you read a simple diagram? I'd almost pull the troll card.For your own sake please don't post until you have tried a 1k resistor at the end of a piece of coax.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:53:15 pm
content: Quote from: nctnico on January 30, 2021, 10:48:20 pmI can't see that, and have no inclination to search for how you have "revised history".Post the simulations here, where we can see them in the current context.Address the issue of why Z0 probes use standard coax, not highly resistive coax.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 10:54:20 pm
content: Nobody is talking about highly resistive coax. Why do you keep bringing that up?Scroll a few posts up and you see the diagram and simulation results.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:55:11 pm
content: Quote from: nctnico on January 30, 2021, 10:51:39 pmWhere, exactly, is your simulation?Two people (at least) are confused by your postings, so you are being unclear.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 10:56:18 pm
content: Quote from: nctnico on January 30, 2021, 10:54:20 pmYou introduced them.

author: Tom45
author_rank: Frequent Contributor
timestamp: January 30, 2021, 10:59:15 pm
content: Quote from: nctnico on January 30, 2021, 10:48:20 pmThanks. You had two simulations but didn't say which resistor was being changed. I didn't know whether you meant R2, R4, or R3.

author: tggzzz
author_rank: Super Contributor
timestamp: January 30, 2021, 11:14:32 pm
content: Nctnico's simulation and measurements are very different: note the difference between steps and notches.My simulation and measurements of a scope with a 50ohms//20pF input show the same principal features.Conclusion: my model matches reality better than nctnico's.

author: nctnico
author_rank: Super Contributor
timestamp: January 30, 2021, 11:36:35 pm
content: No, you are just cherry picking based on a single measurement with no control experiment to back it up (which is exactly why I initially also tested with a Z0 probe with the same attenuation so I have two different tests showing the same result).If you look closely at the signal I recorded with the Wavepro 7300A you'll see there is a step in there as well. This is missing from your simulation because the model for an ideal transmission line you are using isn't good enough. If you copy my circuit and start adding the parasitics you'll see you get closer to what the 7300A is showing.Again: try the 1k resistor at the end of a piece of coax!

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 12:07:50 am
content: @nctnico, why did you do the simulation with a75 Ohmcable? [ sqrt(0.378e-6/67.3e-12) = 74.944 ]Of course there will be reflections when terminated with 50 Ohm.

author: nctnico
author_rank: Super Contributor
timestamp: January 31, 2021, 12:11:41 am
content: Quote from: gf on January 31, 2021, 12:07:50 amI thought I'd got the parameters for a 50 Ohm cable but I see that is not the case and with this mismatch there will be reflections back into the 1k resistor.

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 12:16:25 am
content: Quote from: nctnico on January 30, 2021, 11:36:35 pmI have amuchbetter control experiment than yours: Imeasurea pure 50ohm input and also a 50ohm//20pF input.QuoteLet's keep our eye on the ball, not irrelevancies. The key bone of contention is whether the WavePro's input is 50ohm or 50ohm//15pF.My model (the 50ohm//15pF) clearly demonstrates the notch seen in your waveform (and my 50ohm//20pF waveform).Your model clearly doesnotdemonstrate the notch.My measurements clearly demonstrate the notch seen in your waveform - and the absence of a notch with a pure 50ohm input.So, should we trust models or measurements? I prefer models backed up with measurements.Over to you to produce a model that demonstrates that notch you see in your measurements....

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 12:17:44 am
content: Quote from: gf on January 31, 2021, 12:07:50 amSnort!It is noteworthy that he claims his model matches measurements when there are such gross errors!

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 12:24:03 am
content: Quote from: nctnico on January 31, 2021, 12:11:41 amIf the cable is perfectly terminated with its Z0 at the end, then there won't be any reflections at the end, and no Z0 termination at the beginning were necessary. If the end is mis-terminated, though, then an (additional) Z0 termination at the beginning helps of course to prevent the reflected signal being reflected back from the beginning to the end again (or at least to reduce the amplitude of the 2nd reflection if the termination at the beginning is not perfect either).

author: nctnico
author_rank: Super Contributor
timestamp: January 31, 2021, 12:25:01 am
content: Quote from: gf on January 31, 2021, 12:24:03 amYou are right about this. Still it doesn't explain the results I'm seeing.

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 12:40:21 am
content: Quote from: nctnico on January 31, 2021, 12:25:01 amDo you have a VNA? Would you consider measuring S11 (or the impedance) of the 50 Ohm scope input? Is it purely resistive and independent of frequency?A direct coax connection between signal generator and scope is also source-terminated by the generator, thus a mis-termination at the scope input may not matter too much. For a Z0 probe w/o source termination of the cable it certainly does matter.Is your cable 50 Ohm (or accidentally 75 Ohm too, like in the simulation)?

author: nctnico
author_rank: Super Contributor
timestamp: January 31, 2021, 12:51:14 am
content: Quote from: gf on January 31, 2021, 12:40:21 amI could use a network analyser but that only goes to 300MHz. Cable should be 50 Ohms but since I got the same results with a Tektronix probe it is reasonable to assume the cable I used is 50 Ohms. Some further investigation is needed as I would be highly surprised (and dissappointed -not for being wrong but due to equipment not being as good as I expect it to be-) if it turns out the oscilloscope is the cullprit after all.

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 01:04:43 am
content: Quote from: nctnico on January 31, 2021, 12:51:14 am300 MHz is better than nothing. A parallel capacitance of a few pF should already show up at this frequency.Another question: Does an external 20dB attenuator (a good one, with proper frequency rating) in front of the scope input make a difference?

author: StillTrying
author_rank: Super Contributor
timestamp: January 31, 2021, 01:26:24 am
content: I usually find that LT's lossless transmission line has bugs, even after a new OS and latest version of LT it still has them.Changing the value of a completely unrelated V or I source changes the TL's output.

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 10:25:49 am
content: Quote from: nctnico on January 31, 2021, 12:51:14 amHere are somevery quck and dirtyresults from my NanoVna and Tek 485. They are sufficient to show that you can measure the difference at 300MHz.Sweep is from 100MHz to 900MHz, marker at 340MHz.The first is connected to the scope's proper 50ohm attenuator.The second is connected to an inline 50ohm attenuator.The third is connected to the inlineattenuatorload connected to the scope's 1Mohm//20pF attenuator. [EDIT: inline 50-ohm load]It will be interesting to see your results with the LeCroy.

author: 2N3055
author_rank: Super Contributor
timestamp: January 31, 2021, 11:21:21 am
content: I guess that MSOX3104T has true 50 Ohm path....Note: This was with loaded master calibration. If I were to calibrate it would have been more accurate, but for this purpose it is enough..

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 11:29:24 am
content: Quote from: 2N3055 on January 31, 2021, 11:21:21 amYup

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 01:58:04 pm
content: Quote from: tggzzz on January 31, 2021, 10:25:49 amInline attenuator (if yes, how many dB?), or just inline terminator?

author: radioing
author_rank: Newbie
timestamp: January 31, 2021, 07:27:47 pm
content: Hi all, brief theory about the Z0 probe:Z0, gamma (R, L, C, G)1k tip Zin -> ================================ Zl (termination, oscilloscope side, 50 ohms)|<-------------l ------------->|The input impedance Zin of the lossy transmission line (nonzero R & G line parameters) is given byZin = Z0 * [Zl + Z0*tanh(gamma*l)]/[Z0 + Zl*tanh(gamma*l)],wherecharacteristic impedance of cable Z0 = sqrt[(R + j*omega*L)/(G + j*omega*C)],propagation constant of cable gamma = sqrt[(R + j*omega*L)*(G + j*omega*C)],load (terminating) impedance is Zl,l is cable length,j is imaginary unit,tanh is tangent hyperbolic,andL, C, R, G are cable parameters (per one meter),omega = 2*pi*frequency.And now:1. When the cable is lossless (R = G = 0), the Zin formula can be rewritten toZin = Z0 * [Zl + j*Z0*tan(beta*l)]/[Z0 + j*Zl*tan(beta*l)],wherecharacteristic impedance of cable Z0 = sqrt(L/C) and it is pure real number,beta = omega*sqrt(LC) and it is also pure real number.From this Zin formula, it results that when we terminate the lossless line with load Zl = Z0,the impedance seen at the cable input (Zin) is exactly Z0.As there are no reflections at the Zl (reflection coefficient is 0 if Zl = Z0), there can't be alsoreflections at unmatched cable input returned toward Zl. Due this fact, any voltage measured atthe cable input will be also measured at the cable load Zl, of course delayed due to nonzero cable length.The situation changes when the Zl is not eq. Z0 (but hopefully close). In this case, there are reflections at Zl (osci side) and there are alsoreflections at the cable input (1k tip side), because the reflected signal going back through cable sees 1k + some parasitic C and L.Note that at Zin side, there are nearly total reflection due to such impedance discontinuity.How to improve this situation:A. Of course, try to match Zl to Z0 (better osci etc.;-).B. Terminate cable at Zin side so that reflections spreading from Zl see impedance close to Z0 and are not reflected back.The solution is to plug 50 ohms resistor at the cable input. In this case, the Zin changes to 25 ohms approx., so the division ration doubles.To increase sensitivity back, the 1k can be replaced by 470 ohms resistor (but the load of measured circuit increases).2. When the cable is lossy (nonzero R and/or G) and loaded (terminated) by pure real impedance Zl (osci 50 ohms)reflecting only L and C parameters of cable.There are the reflections at the the Zl and there are also reflections at the cable input if it is also not matched (like in prev. example).Thus, the signal measured at the Zl is jammed by multiple reflections from Zl mismatch and cable input impedance mismatch.The situation is even worse as the parameters R and G are significantly frequency dependent and the transmission line becomes dispersive,i.e., the signal will spread out in time.There are too limit cases:a. The cable is not too lossy (up to some desired frequency, e.g. cable with good dielectric, short cable,...).Then the matching of complex Z0 and pure real Zl is good enough so that the reflections coming to Zl measurement point are smallin amplitude and does not disturb the original signal too much.b. The cable is too much lossy.In this case, the reflections are attenuated by cable itself (imagine cable with 10 dB att -> first reflection is attenuated by 20 dB (supposed total refl. at Zin point) -> 0.1 amplitude)and they are also hopefully marginal at the measurement point Zl.What can be done to dump reflections coming to Zl measurement point?A. Match Zl to complex Z0 -> hard to do as Z0 varies with frequency.B. Use better cable.C. Use long cable to increase reflections attenuation -> not good idea, we lost the sensitivity.C. Match cable at the cable input to reduce reflections of reflections from Zl.

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 07:32:08 pm
content: Quote from: gf on January 31, 2021, 01:58:04 pm0dB Yes, I'm an idiot: it is an inline load.Original edited to avoid more readers wasting their time. But thanks forreadingit!

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 07:58:03 pm
content: Quote from: tggzzz on January 31, 2021, 07:32:08 pmI just asked, because if it were an attenuator (say 20dB), then I would have expected better decoupling (i.e. that the capacitive load attached to it would not affect the input impedance so much).EDIT: Btw, could you please also also sweep S21 of the Z0 probe (from the tip to the end of the cable - thus including all the parasitics), in order to get its frequency response, and also let the Nano calculate the corresponding time domain impulse/step response [I think the Nano can do that, can't it]? Just curious - particularly what effect the parasitics have.

author: nctnico
author_rank: Super Contributor
timestamp: January 31, 2021, 09:44:24 pm
content: I did some testing with the Lecroy 7300A but didn't got much wiser. First of all it turns out my Mini-circuits directional coupler is broken and the other one I have is a cheap one from Ebay with only goes to 200MHz. With the latter the SWR line (using Anritsu MS4630B network analyser) isn't flat (from 1.00 @30MHz to 1.05 @200MHz) when connected to the 7300A's input. But I get the same result with the R&S RTB3004.I get exactly the same trace on both the Lecroy 7300A and the R&S RTB3004 in 50 Ohm mode using two different generators. A 12dB inline attenuator doesn't change the signal shape at all on the Lecroy 7300A.All in all nothing conclusive yet.BTW: the NanoVNA seems like a very useful tool instead of having to pull out a spectrum analyser and RF generator to make more in-depth measurements. I'm going to order one.

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 10:25:32 pm
content: Quote from: gf on January 31, 2021, 07:58:03 pmYes, exactly, and your question was very sensible.QuoteYes, I could - but not today.The hp10200a spec is 1.5GHz with <0.7pF tip capacitance.

author: tggzzz
author_rank: Super Contributor
timestamp: January 31, 2021, 10:28:36 pm
content: Quote from: nctnico on January 31, 2021, 09:44:24 pmIt is worth getting one, but you have to choose which variant. The differences can be found by googling.Mine has the annoying feature that the calibration dara is not saved across power downs. It has to be recalibrated every time it is turned on.

author: gf
author_rank: Super Contributor
timestamp: January 31, 2021, 11:17:29 pm
content: For some time I'm after theV2 Plus4model but is seems to be out of stock at the moment in all stores.

author: nctnico
author_rank: Super Contributor
timestamp: February 01, 2021, 12:43:17 pm
content: Quote from: nctnico on January 31, 2021, 09:44:24 pmI did some further investigation. It seems that the problem is not the oscilloscope input but the interaction between the signal source and the probe itself. If I put an attenuator or 50 Ohm feed-through between the Lecroy 7300A calibrator output and the probe the signal improves. Something else to keep in mind is that the Lecroy 7300A has more than 8 times the bandwidth compared to tggzzz's Tek 485 so it will show narrow spikes much larger.

author: tggzzz
author_rank: Super Contributor
timestamp: February 01, 2021, 01:12:00 pm
content: Quote from: nctnico on February 01, 2021, 12:43:17 pmMore likely "In that case the effect you see without the source 50ohm might be due to the cal out behaving properly when correctly loaded."https://www.eevblog.com/forum/blog/eevblog-1367-5-types-of-oscilloscope-passive-probes-compared/msg3438928/#msg3438928The Tek 485's cal out signal is most entertaining in that respect: it looks like an exponential risetime quantised into discrete levelsQuoteAny energy reflected by an improper termination will be independent of the scope's bandwidth. Thus there is an argument that the area under the spike will be the same.

author: nctnico
author_rank: Super Contributor
timestamp: February 01, 2021, 01:57:56 pm
content: Quote from: tggzzz on February 01, 2021, 01:12:00 pmBut isn't the whole point of probing to see the signal as it is as much as possible? Requiring a signal to be 'correctly loaded' is not always possible or even desireable (think about the good old PCI bus for example as an unterminated high speed bus).

author: tggzzz
author_rank: Super Contributor
timestamp: February 01, 2021, 02:14:14 pm
content: Quote from: nctnico on February 01, 2021, 01:57:56 pmYes, and in your experiment youwereseeing the signalcorrectlywhen it didn't have the 50ohm termination. It is just that the source's output was different when not loaded with 50ohms.Here's the Tek 485's cal out when driving 1Mohm//20pF via a 2m lead. The timebase is 100ns/div. Not exactly a 1ns risetime - or rather each stepis1ns risetimeThe point is to measure the signalin normal operation without disturbing it. If the signal is meant to be unterminated (e.g. a PCI bus), then of course you wouldn't terminate it because that wouldn't reflect (ho ho) normal operation.

author: nctnico
author_rank: Super Contributor
timestamp: February 01, 2021, 02:58:58 pm
content: Quote from: tggzzz on February 01, 2021, 02:14:14 pmNo. In my experiment I terminated both ends of the coax of the Z0 probe which gave me the correct signal shape. I didn't load the calibrator output. I'll try to do some more modelling when I have the time.

author: tggzzz
author_rank: Super Contributor
timestamp: February 01, 2021, 06:17:59 pm
content: Quote from: nctnico on February 01, 2021, 02:58:58 pmI really do wish you would provide a schematic or photo. That would enable us (and probably you) to understand what is happening.If you have had a 50ohm termination between the 1kohm resistor and the 50ohm coax (in addition to at the scope end), then that is clearly an incorrectly implemented Z0 probe.What does the manual say the cal out load should be when measuring fast risetimes (it won't matter for the 10kHz low frequency probe calibration)?