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enter image description here

DE is a capacitive load.

In a source the waveforms of the buck converter are given as:

On state

enter image description here

Off-state

enter image description here

I am confused about the waveform of \$v_{dS}\$. On Wikipedia the waveform is shown constant. How can the voltage change if the rate of current is constant? Did somebody see this before and has an explanation for me?

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1 Answer 1

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On the wiki page you linked it shows this circuit: -

enter image description here

Note that they have a capacitor in parallel with a load resistor and you have something called "DE".

The capacitor is important because it smooths out the fluctuations in the output voltage and, as a first approximation, the output voltage can be assumed to be constant. As a second approximation, you will see a saw-tooth ripple superimposed on the constant voltage. A more accurate representation is a little curved at the edges.

There is nothing in your question that explains what "DE" is so hopefully you can take what I've said and interpret it in the knowledge of what "DE" is.

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  • \$\begingroup\$ DE is a capacitive load. \$\endgroup\$
    – hcl734
    Nov 11, 2019 at 14:44
  • \$\begingroup\$ @hcl734 OK so now you have your answer. As a first approximation the output voltage is constant etc... \$I = C\frac{dv}{dt}\$ - with a constant current voltage rises at a rate or falls at a rate with load current (current from the capacitor). \$\endgroup\$
    – Andy aka
    Nov 11, 2019 at 15:38
  • \$\begingroup\$ Yes but why does the voltage over the load $v_{DS}$ reverse in the middle of the on state while the current $i_{DS}$ has the same rate over the whole time? Or is it just the effect of the capacitance delaying the voltage compared to the current. \$\endgroup\$
    – hcl734
    Nov 11, 2019 at 16:06
  • \$\begingroup\$ It looks like the \$v_{DS}\$ waveform in your source is wrong. \$\endgroup\$
    – Andy aka
    Nov 11, 2019 at 16:34

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