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enter image description here

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?

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  • \$\begingroup\$ Have you tried simulating it? \$\endgroup\$
    – winny
    Jun 19 '21 at 11:39
  • \$\begingroup\$ "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. \$\endgroup\$ Jun 19 '21 at 11:50
  • \$\begingroup\$ Yes, sorry. I didn't word it properly. I will correct it right now. \$\endgroup\$
    – Newbie
    Jun 19 '21 at 11:51
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Yes you are correct.

When the switch opens, the inductor attempts to keep the current flowing through itself, it generates a back emf which is created in an attempt to keep the current flowing.

When a back emf is generated by an inductor, the voltage at the end of the inductor that the current is flowing towards must rise or, if this voltage is held constant (as in your circuit example) then the voltage at the other end of the inductor must fall. So when the switch opens, the voltage at the left hand end of the inductor is driven down, forward biasing the diode and holding the diode's cathode at about 0.7V below ground. The current, which is being forced to flow in the inductor (current in inductors must change magnitude slowly) flows in a square, down through the load to ground, through the forward biased diode and back up through the inductor to complete a circuit. The current also adds charge to the capacitor.

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    \$\begingroup\$ Thank you for your answer and clarification \$\endgroup\$
    – Newbie
    Jun 19 '21 at 11:59

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