How to measure voltage drop through resistor with oscilloscope? - Page 1
Forum section: electronics / beginners

author: Chriss
author_rank: Frequent Contributor
timestamp: February 22, 2020, 09:33:09 pm
content: Hi!I put a 1R8 resistor in serie with positive leg of a 2200uF electrolit cap.Than I set my acope probe positive clip to one side of the resistor and the negative clip to the other side of the resistor.I feed the circuit with 12VDC.Put +12v to the resistor and gnd to the minus leg to the cap.The sine wave on the scope was very interesting for me.I didn't expected such of diagram.The voltage started to rise up and suddenly fall back when it reached aroun 5.6v to around 3v and then it started to rise slow up to almost 12v.Can somebody help to figure out what I did wrong?Or is that measuring ok?I also built up in LTS. and MultiSim apps but the result was diferent than my real measuring.Thanks for any help and suggestion.

author: ArthurDent
author_rank: Super Contributor
timestamp: February 22, 2020, 09:40:01 pm
content: You are shorting out the cap from what you described. Watch video.

author: Chriss
author_rank: Frequent Contributor
timestamp: February 22, 2020, 10:22:28 pm
content: So, what would be the correct way to check the voltage drop trough a resistor in the circuit?Let's say in a circuit where with the cap is a res in series?Thanks.

author: rstofer
author_rank: Super Contributor
timestamp: February 22, 2020, 10:41:50 pm
content: Quote from: Chriss on February 22, 2020, 10:22:28 pmIf you insist on doing it with a single channel of a scope, one side of which is grounded to earth through the BNC connector, you need a differential probe. Good ones are VERY expensive.Or, you can use two channels (with no ground clips) and the MATH function to get X-Y. Sometimes you only have X+Y and you have to invert Y but these choices are fairly clear on the menu. Check the user manual for math functions.ETAAll of the above assumes your project has an earth reference somewhere. Without a schematic or a picture of the scope image, that's about all we can say. If you don't have an earth reference then everything above is nonsense. But we need pictures.The X-Y thing will work whether you have an earth reference or not so give it a try.

author: Chriss
author_rank: Frequent Contributor
timestamp: February 22, 2020, 11:37:59 pm
content: Here is my circuit and waveform.My scope and testing circuit are complete isolated from the main power source.I'm trying to achieve to measure the voltage drop through a shunt res to calc the inrush current.It is hard to measure with average dmm but I got an idea somehow to determine the voltage drop withmy scope and the make the math.If that is actually possible to do with a scope.btw. when I try to simulate this circuit and measure it with LTS and MultiSim I got no reading on the scope and dmm also in the app.Only 0v.And here is what LTS simulate when I put the resistor to the positive side.The blue line is the I(R3)...I can do whatever I wish with my real circuit I can't get that value like LTS.Thanks for any help.

author: SilverSolder
author_rank: Super Contributor
timestamp: February 23, 2020, 12:39:03 am
content: Your simulation is showing 0.1ms per division on the graph, whereas the scope is showing 10ms per divison - which is 100x different.Try to get both to show the same scale.I think you might find both are "doing the right thing".

author: Chriss
author_rank: Frequent Contributor
timestamp: February 23, 2020, 12:47:22 am
content: Ok.But what is my signal representing on my real scope?Only a bit over 5v?

author: rstofer
author_rank: Super Contributor
timestamp: February 23, 2020, 01:33:41 am
content: On your real scope: 5V = I * 1.8 Ohms so I = 2.77 Amps inrush current.The maximum theoretical inrush is 12V / 1.8 Ohms or 6.66 Amps ignoring series resistance in the capacitor and source impedance from the voltage source (and they all have some). In effect, you have a total circuit resistance of 12V / 2.77 Amps or 4.3 Ohms. You know where 1.8 of it comes from so we need to account for the other 2.5 Ohms. It's coming from somewhere.Use the second channel on your scope to monitor the source voltage during the inrush period. Perhaps it is sagging a bit.Or all of the numbers are wrong - there's that!

author: vk6zgo
author_rank: Super Contributor
timestamp: February 23, 2020, 01:36:16 am
content: Quote from: Chriss on February 22, 2020, 11:37:59 pmIf the circuit is actually what is shown in the top picture, the display on the siglent is pretty much what you would expect, except that the time/div seems wrong, but maybe that's just me------I find it hard to decipher the messages on screen shots of DSO screens.With R= 1.8& C= 2000uF, your time constant t= CR is 3.96 msThe common "rule of thumb" says that C may be regarded as fully charged after a time interval of 5CR, or, in this case, 19.8ms.At switch on, C looks like a short circuit, so the applied voltage appears across R.(at the same time, maximum current flows through the circuit)As C charges up, the charge upon it increases, opposing the applied voltage, reducing the current through R, & hence, the voltage across it.After t=CR, (in this case, 3.96ms) the current through, & hence the voltage drop across, R has fallen to approx 37%'of its initial value.After 5CR (19.8ms), the current through, & voltage across, R has fallen to that close to zero that it can be neglected.The charge on C is now "near as dammit" equal to the applied voltage.Howver, as the TV advertiser says, "But wait!---There's more!"Your value for R is very small indeed, so your supply needs to reliably source 6.666 amps.If its reqularion is good enough, or its internal resistance low enough, there is no problem, but if not,these may modify the amplitude, time durations, or both, distorting the display.For your simulation, the "pretend 'scope" must be able to show a time duration of 20ms or more to give you a useable display.

author: Chriss
author_rank: Frequent Contributor
timestamp: February 23, 2020, 10:32:30 am
content: Ok, I think I sorted out everything thanks to you guys.Here is what I did today after I read your posts:I took my esr meter and measured the value of my cap.I took 6 samples of measuring and the average is:C=~2153uF printed C on the cap = 2200uFESR=~ 0.03OhmI did the same measuring with my resistor I use, also 6 samples and took the average:R=1.84OhmI measured the wiring resistance what I use to pow my circuit and it is ~ 3.3Ohm.So, here is my calculation and I come very close to the real situation based on my scope:The max. theoretical I_inrush= 12V/1.84=6.52AThe max. scope based I_inrush=5.60V/1.84=3.04AIf I subtract from the 6.52A-3.04A then I get the current what is consumed but I don't know who eat up my soup.Which is 3.48A.So, now I know does somebody eat my 3.48A but I don't know who.Here is what I did to catch the intruder:12V/3.48A=3.44- this is eating my amperage!I measured the resistance of my wirings what I use to power on my circuit.And it come out does my R_wiring=~3.3.Now when I do this math: R_wiring-3.44= 0.14\$\Omega\$That means my wiring is eating up my 3.48A and there lef 0.14what is probably error in my measuring, errors in my meters, and other parameters of cap, res, psu, scope, scope probes etc.Is this the real story?Am I did it correct?And btw.Here is the new scope pic whee I connected the CH2 to the PSU and catched the inrush current also on the PSU.CH1 is the V through the res.CH2 is the PSU V.I inverted the CH1 for better view.I hope I did my homework correct.

author: rstofer
author_rank: Super Contributor
timestamp: February 23, 2020, 05:32:29 pm
content: Quote from: Chriss on February 23, 2020, 10:32:30 amLab experiments never match the paper version, reality gets in the way. Resistances that weren't considered are the topic of this experiment. You have come a long way with this simple project.Most important take-away: Ohm's Law is a LAW, not a suggestion. You can't treat it like a speed limit. There are situations at the edge of electronics where it might not hold but for everyday work, it's a LAW. If current flow causes a voltage drop, there is resistance hiding around somewhere.I would shorten up the time per division so I could see more of the exponential rise in capacitor voltage. If necessary, I would move the trigger position over to the left. In fact, I would move it very close to the left edge; there is nothing interesting before t=0 (the start of the rising edge).