Is this a valid way to measure inrush current of an IC?  - Page 1
Forum section: electronics / beginners

author: Jim from Chicago
author_rank: Regular Contributor
timestamp: June 24, 2022, 08:50:48 pm
content: I am trying to measure inrush current to an IC, and fortunately there is a zero-ohm resistor in series with Vin. So I put a 0.1 ohm sense resistor in its place, and probed each side of that resistor with ch1 and ch2 on my scope, according to the attached diagram. So I just power the board and trigger the scope and look at the resulting waveforms on startup. Then I look for the greatest delta between ch1 and ch2 and use V = IR to solve for the current.

author: Manul
author_rank: Super Contributor
timestamp: June 24, 2022, 09:25:42 pm
content: Generally, yes, but... It is important to note, that inrush current may depend heavily on the slope of supply voltage (how fast it rises after power on). So with different supplies it could be quite different. It comes to the question, what exactly do you want to measure. Some theoretical maximum inrush or inrush in specific situation or something else. As always, it is important, that what you measured is what you actually wanted to measure. If these two depart it is easy to get misguided.If you use isolated supply, you can use only one probe. Two probes is also an ok technique. Most scopes can graph the difference (math function -> ch1 - ch2) between channels (aka poor man's differential probe), that could be useful. Beware of channel offset and gain errors, if you want accuracy, check that. Connect both probes to same point and ensure, that they match (difference = zero). Also you need to choose an optimal shunt resistor. Too small or two big will cause you errors. For accurate measurements we use shunts with kelvin connections and current sense amplifiers.

author: LaryPant
author_rank: Contributor
timestamp: June 24, 2022, 10:37:17 pm
content: I agree with Manul's answer, but just remember that scopes are not build to measure precise amplitudes and it might be difficult to accurately measure the inrush current of low power ICs.A common method to measure inrush current is by using an AC current probe connected to a scope, which is, however, still limited by the imprecise amplitude measurements of an oscilloscope.And remember, if the IC is connected to a capacitive load, then make sure that any and all capacitors are discharged before you turn on the power supply to measure the inrush current. Although you may already be aware of thatLary

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 24, 2022, 10:54:38 pm
content: Oscilloscopes (nearly?) always have a "math" option to subtract two channels from each other, and with that function you can see the difference between those two channels as a single trace on the oscilloscope.The shunt resistor is always a compromise between the signal output and the allowable voltage drop.It's quite easy to build a few signal amplifiers with opamps. Experiment with differential amplifiers, and different amplification factors. If you build this neatly and use it with a separate power supply, an isolated power supply or a few batteries, you can make it into a quite usable signal conditioner for a bunch of measurements.

author: TimFox
author_rank: Super Contributor
timestamp: June 24, 2022, 10:58:03 pm
content: To check if your oscilloscope's math function is sufficiently accurate for the high-side current measurement, repeat the measurement (with exactly the same settings on the scope, including subtraction), but with both probes connected to one side of the sense resistor. Of course, that should give zero, but common-mode response will give some result.One problem with this method is that you need to keep both channel A and B on-scale for the digitization to be valid, so the resolution may not be sufficient for a good difference measurement, depending on the relative magnitude of common-mode and differential-mode voltages.

author: David Hess
author_rank: Super Contributor
timestamp: June 25, 2022, 02:43:52 am
content: Measuring the inrush current in that way can work very well.Calibration of the oscilloscope inputs used differentially to increase common mode rejection will give more accurate results but this may not be possible on all DSOs. On analog oscilloscopes this was done with the vertical variable control.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 25, 2022, 02:53:03 am
content: An alternative method is to just put the shunt resistor in the GND lead to your IC, or if you have a floating power supply and circuit you can directly connect the positive output of the power supply to the GND of your scope and measure it with a single probe.

author: David Hess
author_rank: Super Contributor
timestamp: June 25, 2022, 03:09:41 am
content: Quote from: Doctorandus_P on June 25, 2022, 02:53:03 amWhat I have occasionally done is replace the ground return with a virtual ground from an inverting operational amplifier or simple transistor so the ground pin remains at ground while the current can be measured as a voltage at the output of the operational amplifier or collector of the transistor. This can be a little difficult to get working though because any capacitive decoupling tends to make the virtual ground unstable, but the monks taught me the secrets to solve this.

author: Siwastaja
author_rank: Super Contributor
timestamp: June 25, 2022, 06:02:40 am
content: Your principle is correct, it's just that general purpose oscilloscopes have crappy voltage axis. The vertical sensitivity does not go low enough, then the ADC is 8-bit.To solve this, I suggest adding a so-calledcurrent sense amplifier, a small IC which takes the small differential voltage signal and amplifies it by say 50x or 100x. This is now trivial to scope, and you only need one channel. Pay attention to the bandwidth of amplifier ICs if you want to see short peaks. Example of a 2MHz BW part would be LT1999.Dave's uCurrent is basically such current sense amplifier IC made into a more finished product with battery, connectors etc.