# Buck Converter Voltage waveform at the free wheeling diode cathode and anode

I am trying to understand the voltages at the free wheeling diode's cathode and anode in a buck converter.

During the ON-state (as in the above circuit diagram), I assume the voltage at the diode's cathode would be like a constant voltage (We can measure the switching frequency waveform at this node). And at the diode anode, we would observe 0V in the case that node is taken as the ground reference.

Whereas in the OFF-state, since the diode would be forward biased, what would be the voltage at the diode anode and cathode?

At the diode cathode, we obviously would see the switching frequency (the LOW period in the switching frequency waveform). Am I correct? But since the diode is forward biased, would the voltage at the cathode go 0.7V (whatever the forward drop of the diode is) below the anode? So, if the anode was at 0V previously in the ON-state, would the voltage at the cathode be -0.7V during the OFF state? Because, since the diode conducts, we need the anode voltage to be 0.7V greater than the cathode. And since the anode is tied to a 0V reference, the cathode would be at -0.7V during the OFF period of the switching frequency, right?

Can someone explain whether my understanding is correct?

• Have you tried simulating it? Jun 19, 2021 at 11:39
• "at the on state, I assume the voltage would be a square wave" no, sorry. It's a constant value for the duration of the ON state, that's what "ON" means in this context. Jun 19, 2021 at 11:50
• Yes, sorry. I didn't word it properly. I will correct it right now.
– user220456
Jun 19, 2021 at 11:51