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The below waveform shows peak current occurs at the instant when diode starts conduction. Why is the peak current not happening at the max input voltage point?

enter image description here

Usually the rectifier output caps are large in value, making the output voltage more or less constant.

Current = (Vin-Vc)/(R1) ( neglecting diode drop)

-> As Vin is increasing, shouldn't the current also keep increasing? What am I missing here? enter image description here

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3 Answers 3

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When the diode starts conducting and dumping current into the capacitor, the rate of change of the AC voltage is much bigger than when the AC waveform hits the top of its peak. Given that capacitor current is proportional to \$\frac{dv}{dt}\$ it follows that the biggest peak in the capacitor current occurs when the diodes start conducting. Consider this simpler scenario: -

enter image description here

Image from here showing how when an AC voltage is applied to a capacitor, the maximum current occurs when the voltage waveform is at its maximum \$\frac{dv}{dt}\$ AND zero current flows when \$\frac{dv}{dt} = 0\$.

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You can see in your first diagram that the output voltage is NOT constant because the voltage on the filter caps has drooped between cycles.

It's for precisely this reason that the peak current occurs when the diode starts conducting and has to charge the caps from the minimum droop point to the point where the diode conducts.

After that initial charge the cap is much closer to the peak, and the slope of the sine wave is flatter so the current to bring the caps to the peak voltage level is smaller. $$(I=C*dv/dt)$$

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From the point on where the diode is conducting, the influx current is proportional to the derivative of the voltage across C plus the draining current (which is significantly smaller most of the time and only varying comparatively little as long as the ripple is small compared to the total voltage).

The derivative of the input voltage starts falling at the zero crossing already.

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