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I hate to be asking this, but I just don't understand the operation. I've been reading various resources, mostly this: http://services.eng.uts.edu.au/~venkat/pe_html/ch07s1/ch07s1p1.htm though. From what I understand, when the voltage source is connected to the circuit, current flows through the circuit like so:

enter image description here

When this is happening, presumably the capacitor is being charged as well. But even at this point I seem to be misunderstanding: surely the load receives the input voltage as well? So even if you wanted the load to receive say 20V stepped down from 120V, when the source was connected at least, there would be 120V across it. Surely this could damage components rated for 20V? I suppose with PWM you could minimise the time that the load was exposed directly to the input voltage to negligible times, but I'm not sure. I'd just really appreciate it if someone could clear up my confusion about this.

In terms of general operation: one way I can think of the buck converter working is if the capacitor is charged to the 'Output' voltage and then the inductor is used to keep the capacitor near this voltage during the presumably much longer off cycles?

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The basic answer to your misunderstanding is that the load and the input voltage are NOT connected directly when the switch is closed. The inductor is between the two.

Voltage accross a inductor can change quickly, but current only changes proportional to the applied voltage. In other words, the current will ramp up slowly. The load only sees the current coming out of the inductor. If the switch were to remain closed, that current would eventually build up too much voltage on the capacitor. However, the switch is only closed for short periods and with a average duty cycle so that the average current coming thru the inductor is exactly what is needed to maintain the desired voltage on the output capacitor.

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  • \$\begingroup\$ Ah! I don't know how I missed that, thanks! Inductor is in series with the rest of the circuit. I was looking at the parallel capacitor and load and neglecting the inductor more or less completely. Now that that's cleared up, my conceptions on how it 'should' work where the capacitor is charged up 'only so much' until the output voltage seems to be true. \$\endgroup\$ – ColdestShadow Aug 25 '12 at 23:51
  • \$\begingroup\$ I also stuck at the same place as you were. But I have re-readed and gone through some video tutorials. \$\endgroup\$ – Standard Sandun Dec 15 '12 at 17:24

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