# Ripple current in synchronous buck converter

I am designing a synchronous buck converter and I am at the bread board stage in my design. I wanted to check the current waveform through the inductor, but I do not have a current probe available. I have a differential probe so I decided to add a 1 Ohm resistor in series with my inductor and measure the differential voltage across it, which should be the same as the current. Below is my waveform, I am wondering why the current waveform is not as triangular as I usually see for a switch mode power supply. Is it because I added the 1ohm resistor in series for an inductor (6.8uH) that only has only around 0.020 Ohms of DCR?

EDIT:

I switched to a ferrite core that should have substantially less core loss and I tried a few different FETs and it didn't seem to make much difference. Below is another waveform, this time CH2 is at the switching node. One thing I forgot to mention is that I am using a half bridge driver that adds a small amount of dead time to keep the two FETs from having any shoot through. Although I can't imagine that being the reason, it's probably worth mentioning. Also, I now have the scope triggering properly and the time scale is now correct.

I also put the diff probe across the inductor, and below is the waveform. I've never noticed the slope, is that due to the series resistance? Or potentially the Vds of the FETs changing because the current ripple is large?

• 1 ohm is of course too much, but there is something else. Why does it seem that the current leads the voltage? Really weird... – Gregory Kornblum Sep 19 '15 at 14:10
• That is the input PWM, not the switching node voltage. When it is low, the high side FET is on and the di/dt is positive. – Brett Prudhom Sep 21 '15 at 14:16

No, adding a 1 $\Omega$ sense resistor in series with your inductor will not cause steps in the current waveform. Adding the resistor is like adding winding loss, and that will only cause an exponential curvature, with $\tau$ of L/R, in the current ramp. If you look closely, you can see the curvature in the current ramp in your picture.

A step in the current waveform can be caused by core loss, but that step would go the other way. Here's what core loss would look like:

See the step at the switch point? That's an extreme example, and tends to be hard to see in low perm cores. Anyway it's the reverse of what your picture shows. So, unless you have managed to reverse time, it's not core loss. (Note: it is possible to reverse apparent time by scope aliasing. So, with aliasing, the inductor current could be of inductor with core loss, or as mentioned below, could have step caused by inductance in the sense resistor.)

It looks like there is about 3A in the inductor, so about 10W in the sense resistor. Power resistors like that tend to be inductive either by construction or geometry. A parasitic inductance in series with the sense resistor could cause an apparent step in the voltage across the sense resistor, since it would make an inductive divider. But, that step would look like the core loss step.

Differential probes usually have at least 40dB of common mode rejection, and sometimes as much as 60dB. Really unlikely that it's because of the probes, unless they are damaged.

Is it possible that Ch2 of the scope has been scaled and added to Ch1? That's really what it looks like. Digital scopes and math functions. It looks suspicious, especially since the waveforms don't line up.

Instrumentation:

It would be a big improvement to reduce the value of the sense resistor (as others have said). One way to do that would be to make a current probe using a current sense amp. With a current sense amp it would be easy to use a 0.1 $\Omega$ sense resistor, and maybe with some trouble get down to 10m$\Omega$. Something like a LT1999 could work if you need bidirectional sensing. If the current is always positive you could get more bandwidth using something like a MAX9643. For bidirectional sensing and wideband use a wideband instrumentation amplifier could work, something like a AD8421. Using a much lower value sense resistor would also mean a much lower parasitic inductance.

• I switched to an inductor which should have very little core loss compared to the first. Please note the edit I added to my original post. I believe you are correct that it is not an issue with the diff probe. – Brett Prudhom Sep 21 '15 at 18:49
• @BrettPrudhom - That's a much better photo than the first. It looks like the aliasing has been taken care of. The current waveform looks like core loss because of the step, but it could also be caused by parasitic inductance in the sense resistor. And yes the slope on the inductor voltage waveform is caused by drop across the sense resistor. – gsills Sep 21 '15 at 19:01
• Do you think maybe it's also because the ripple is so large that it's varying the voltage drop across the FETs? As best I can tell from the FET data sheet it shouldn't change that much but that would make sense to me as well. – Brett Prudhom Sep 21 '15 at 19:14
• @Brett - Hm, that's surprising. What does Vin look like at the power modulator input? – gsills Sep 21 '15 at 19:31
• @Brett - I mean the power rails, from the drain of the control FET to the source of the sync FET. Also, what FETs are you using. Do you really have ~5V gate drive? – gsills Sep 21 '15 at 19:39

Ideally the current through the inductor would not be able to change instantaneously as it does in your trace so something is not quite right.

One thought is that the common mode rejection of the differential probe is not very good so what you are seeing is some of the switching node signal mixed with the current sense voltage.

One test is to connect both ends of the probe together but leave them connected to the resistor - this way there will be the same common mode but no differential signal, you should get no signal, if you do it is a limitation of the probe.

Which end of the inductor do you have sense resistor? At the DC output end there shouldn't be much AC voltage.

What does the 100ms at the bottom indicate? Is that the sweep rate? I would have expected one cycle to be in the microsecond region, not 100ms! What frequency are you running at?

One ohm is a bit large for a current sense resistor - you are losing a couple of volts p-p across that resistor. Ideally it would be lower but your signal would then be lower as well.

Here are some instructions for simple AC current probe that would be much better than a resistor.

• Hi Kevin, I checked the CMR of the diff probes and it seems alright. I got a nice 0V with little noise when I connected the two sides together on either side of the sense resistor. I started with the sense resistor on the output side, but have also tried it on the switching node side and get the same results. I never even noticed the '100ms' and I believe it might have something to do with my problem. I am switching at 400kHz according to my signal generator, but according to this scope the switching frequency is around 5 to 6hz. I have to think that I actually am switching at 300khz – Brett Prudhom Sep 21 '15 at 14:04
• and the scope is incorrect or else I would most likely be frying this inductor. I've never used this particular scope before, and I have had triggering issues so I am running it in "auto mode" where it seems to sweep. Any idea what's wrong with the scope? I think I will try a different scope. – Brett Prudhom Sep 21 '15 at 14:06
• Please notice the updated waveform in my original post. Thank you. – Brett Prudhom Sep 21 '15 at 18:47

Looking at your second scope capture, it appears that the triangular current ripple features approximately di/dt = 1 A/us rising and -1A/us falling. In the hypothesis that the resulting approx. square wave you see superimposed on the triangle is entirely due to the current sense resistor inductance:

pk-pk square wave = 2 L di/dt = 1V ==> 500 nH

I would guess you're getting some common-mode interference there somewhere. That obviously cannot be the pure current signal because the di/dt is extremely large.