Oscilloscope accuracy theory with an attenuated probe - Page 1
Forum section: electronics / beginners

author: lfs2K
author_rank: Newbie
timestamp: November 13, 2020, 01:34:28 pm
content: HelloI've just started studying electronics, and there's this question I've been unable to get a definitive answer for.. I've done electronics as a hobby for a while so I have some understanding of what's what but not sure on the theory & math...We're doing oscilloscope measurements of voltage, and need to estimate the accuracy of the measurement, the formula for which we're reading from the scopes manual.Our scopes manual (DSOX1102G) gives this as±[3% full scale ± 0.1 div ± 2 mV ± 1% of offset setting + 0.25% full scale], (when the scale is >10 mV/div)I've believe I've got it this far. If I read 5V, scale on 5V/div, it would be 5 ± 1.802 VHow should it be calculated with a 10X probe? The voltage at the input of the scope is 1/10 of what it would be with a 1:1 connection.I've understood that if the scope is set in software to scale the probe, it doesn't affect the analog front-end.https://www.keysight.com/upload/cmc_upload/All/1000_series_prog_guide.pdfFrom page 96:"PROBe command specifies the probe attenuation factor for the selected channel. This command does not change the actual input sensitivity of the oscilloscope. It changes the reference constants for scaling the display factors, for making automatic measurements, and for setting trigger levels."http://www.cmicrotek.com/Accurate_current_measurements_with_oscilloscopes.pdfAnd from this, I'm understanding that the constant ±2mV is the ADC:s inherent accuracy, and thus independent from the scale setting.So if I read 5V, scale on 5V/div, should I calculate with 0.5V on the input with 500mV/div scale (like the hardware), and then multiply both by ten? That would yield 10(0.5 ± 0.182) = 5 ± 1.82 Volts.If I just do it from what's on screen, it's the same as the original 5 ± 1.802 V. So which is correct?I mean, the difference is absolutely trivially small, but I'm interested in what's the "correct" way hereThank you

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 13, 2020, 11:00:17 pm
content: Quote from: lfs2K on November 13, 2020, 01:34:28 pmIn general, we add all accuracy of all used devices together, so we have to add the accuracy of your probe. For example, it can be is 3% or 5% (of a value), so it is ± 0,25V.But I'm not sure if there isn't a mistake in a datasheet: "±[3% full scale" may possibly be 3% of a value, because for digital meters we usually describe it's accuracy in relative values to a value but not to a full scale.But if there isn't a mistake then the result is 5 ± 2.052 V.

author: lfs2K
author_rank: Newbie
timestamp: November 14, 2020, 12:46:37 am
content: I agree it would be logical for the accuracy to be specified as a percent of the reading, rather than full scale, but I've already been assured that the first one is correct (5 ± 1.802 for 5V with 5V/div), but the 10X-probe & software compensated situation is a mystery, and that's what I'm trying to find outAnd also how did you get 2.052? Full scale is 8 divisions on the scope, so for 5V/div (0.03×8×5 + 0.1×5 + 0.002 + 0.01×(zero offset) + 0.0025×8×5)

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 14, 2020, 03:53:30 pm
content: I added 5% of a 5V (0.250V) to your 1.802V.

author: lfs2K
author_rank: Newbie
timestamp: November 14, 2020, 08:31:57 pm
content: Oh yeah, I see. Could've worded my question better, I'm interested in the scopes accuracy only, assuming there's an ideal voltage input.This illustrates what I'm trying to figure out:The accuracy is calculated with the previous formula, from the values on screen.The 0.5 at the input could be 5V with a 10X probe, or just 0.5 V, doesn't matter if it's an ideal 0.5 V DC.On the bottom one, the scope is set in software for a 10X probe, so it multiplies the reading.I'm trying to solve which one is wrong, or is the whole idea of thinking it this way fundamentally flawed..

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 15, 2020, 11:50:07 pm
content: If attenuator is in 1:1 then we don't add its 5%.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: November 16, 2020, 01:44:32 am
content: If you add inaccuracy, you are really multiplying accuracy.Lets just use 5% as a stated accuracy to make calculations easy for me.Plus or minus 5% stated accuracy is really 95% accuracy, the result could be stated as being 5% wrong or 95% correct.In your circumstance you are looking at the accuracy of two lumped things.To get the final accuracy, you need to multiply the ACCURACY of the probe by the accuracy of the scope.Lets just say scope accuracy is plus or minus 5% and probe accuracy is plus or minus 5%95% X 95 % = 90.2% AccuracyThe ten times has nothing to do with this calculation.Another way of looking at it is the voltage magnitude of the ERROR of the scope will be one tenth of the 5% error. Assuming an entirely accurate 10X probe, which only lets in 1/10 of the signal (lets ignore any inaccuracy of the probe for now):5% of 10 volts is .5 volts at the source of the signal5% of one volt is .05 volts at the scope input.If you are reading a 10 volt signal and using the ten times (perfect) probe, the signal will be 1/10 at the scope input. The scope input will be 1 volt. Potential error is .05 volts. Both the measured signal strength and the magnitude (not the percent) of the error are divided by tenIn digital scopes multiplying it up to 10X is done digitally (theoretically no error in calculation), in old scopes this was done by analog circuitry, probably the attenuator in the scope, compounding the error.So, discarding any error in the probe, the scope reads one volt and calculates it up to ten. A signal will still be 95% accurate. You can multiply .05 volt potential inaccuracy by ten to get .5 voltsIn CRT scopes the "full scale" meant the deviation of the line on the CRT from what was put into the scope. This was attributed mostly to the inaccuracy in the CRT deflection at the limits of the screen of the CRT.I imagine the "full scale" deviation in digital scopes has to do with the inaccuracy of the A/D converter at it's voltage limits.

author: lfs2K
author_rank: Newbie
timestamp: November 16, 2020, 06:36:41 pm
content: Thanks! That explains alot, but still in my case, the scope's accuracy isn't specified as a linear relation, it has a constant.My question is really if, when using a software-multiplying digital scope, with a 10X probe, the error (just for the scope) should be calculated with the hardware values OR the ones on-screen.I've just been told to ignore what's happening inside the scope, make sure the software compensation is the same as the probe's attenuation, and just read the values off screen. But the values aren't the same if you think about it more closely..There's two different outcomes, depending on how it's done:

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: November 16, 2020, 09:49:31 pm
content: Are they referring to the trace on the screen or their voltage measurements printed on the screen?I think they are referring to the actual trace which in your scope is analog, as stated in specs.Full scale would be 40 volts at 5 volts per division on your scope (tell me if I am wrong) your scope has 8 divisions. I have scopes with 8 and 10 divisions.The specs I have read are 3% of full scale +.25% full scale, I am ignoring the DC offset.Copied from the spec sheet:DC Vertical Accuracy is ± [DC vertical gain accuracy + DC vertical offset accuracy + 0.25% full scale]DC vertical gain accuracy is [3% full scale] (> 10 mV/div)Vertical offset accuracy is [± 0.1 div ± 2 mV ± 1% of offset setting] (we will ignore vertical offset accuracy since we are not using it).At 5 volts per division:40 volts (full scale from -20V to +20V) times 3.25% is 1.3 voltshttps://www.keysight.com/us/en/assets/7018-05520/data-sheets/5992-1965.pdfAnyway I think the error should refer to the actual input value (which is 5V in you original post), not the downsized value.If you bring down the full scale the accuracy would much better.Maybe this is the point.Besides that it is hard to SEE the 1.3 volt error on the screen on a 40 volt full scale.You are putting a 5 volt standard on the scope input and try to read it off the scope scale and see if it inside the calculated potential error .Which it should be.Scope automatically puts in the 10X probe value thru the pin.

author: lfs2K
author_rank: Newbie
timestamp: November 16, 2020, 10:11:00 pm
content: Yes I know the scale is on way too high to get a good reading, it's just an example.Can the ENTIRE "Vertical offset accuracy" row be ignored if there's no offset?Since even if offset is zero, there's still [± 0.1 div ± 2 mV] on that row, which adds to the total..Does it "auto-zero" itself somehow so the accuracy would only be ±[DC vertical gain accuracy+ 0.25% full scale] ?

author: lfs2K
author_rank: Newbie
timestamp: November 16, 2020, 10:51:19 pm
content: Actually I just stumbled across the answer I was looking forguess I didn't do enough googling at firsthttps://edadocs.software.keysight.com/kkbopen/understanding-voltage-measurement-accuracy-for-infiniivision-oscilloscopes-584425356.html"The second term, ±2.0 mV, is another zero error that never goes away. It dominates the accuracy at smaller vertical scales. If you are using a probe, multiply this by the probe attenuation factor.Example: using a 10:1 voltage probe, it is now 20 mV"So the (multiply both, hardware values) way is correct

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: November 16, 2020, 11:34:36 pm
content: OK I read the keysight stuff.Now you know far more about oscilloscope error values than I ever did.Surely more than I wanted to, I guess that is why education is sometimes painfulOUCH!!! No wonder I dropped out of engineering school when they told me that my degree would not guarantee that I would drive a train.

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 17, 2020, 06:10:21 pm
content: Quote from: Wallace Gasiewicz on November 16, 2020, 01:44:32 amBut in metrology - we add accuracy. If there are two devices involved in some measurement - to get the summary accuracy we add the accuracy of both metering devices. When the divider is used as 1/10 divider (but not 1:1) - it is one device, and oscilloscope - is another device.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: November 18, 2020, 01:48:41 pm
content: Accuracy is usually, (sometimes?) stated in plus or minus percent. This is really stated possible ERROR.Lets say you have one device that is plus minus 5% . Take the worst scenario and say that it is off 5% (not likely, but possible)Then it is 95% of real value (or could be 105%). The next device is off possibly another plus minus 5%.The first device gives a negative 5% worst case reading (or resulting output) .... of 100 units. which would be 95 unitsThen the second device gives a negative 5% worst case reading.95% times 95 (the -5% worst case read value on the instrument) is not 90, it is slightly more at .95 times 95 or 90.25Or another way:5% of the worst case of a measurement of a 100 unit value would make the measured value 95.5% of that 95 value is slightly less than the unit 5. It is 5% of the worst case 95 value, which is 4.75, making the possible worst case result 90.25If the worst case error is the other way and the measured value is 105 (the +5% error), +5% error of this incurred in the second device would be 5.25, making a measured value of 110.25 possible. 105% times 105% is 110.25%This .25 has little practical meaning if few devices are in series, so merely adding or subtracting the serial 5% errors is pretty close.But if 10 devices are in series 95% to factor 10 is a much different number than serially adding or subtracting the 5% error. Something like 62 vs 50 on the minus worst case scenario or in the plus worst case scenario 162 vs 150.The LIKELIHOOD of any of these worst case scenarios is very small however.

author: Vovk_Z
author_rank: Super Contributor
timestamp: November 18, 2020, 09:55:45 pm
content: I agree with you in general, there is some difference between strict mathematics and fast metrology calculations.