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?
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?