How does smoothing an AC voltage with a capacitor work?

[phantomvommand]

Specifically for part (i) and (ii), I get the rough idea that when the voltage of the AC source is decreasing, the capacitor can discharge to "cushion" the drop in voltage. However, I have some questions about when this occurs.

1. There is an assumption that the capacitor will be charged to maximum when the AC voltage reaches its peak. Why is this assumption (largely) true?

2. How should I solve part (iii) and part (iv)?
For part (iii), I know that the period of discharge is approximately 1/f = 0.02, and RC = 0.5.
The capacitor should have Voltage = 20(e^-0.02/0.5), and the maximum voltage is 20V. The difference of the 2 values is 0.78, which is twice the correct answer of 0.39V. (BTW I am assuming 20V is the max voltage, not the RMS value, do correct me if I'm wrong.)

For part (iv), I think I am not very clear on what "conducting" means. I took "conducting" to mean "allows current to flow through", isn't that just 50% of the time?

All help is appreciated, thank you!

 

[BvU]

Hi,

1. There is an assumption that the capacitor will be charged to maximum when the AC voltage reaches its peak. Why is this assumption (largely) true?

If the voltage source can deliver the necessary current, then the capacitor is charged up to the maximum voltage minus the (small) voltage drop over the diode.
Before we delve into iii and iv, please post your sketches and describe them in some detail. Then your questions about them will be easier to tackle.

 

[phantomvommand]

Hi,
If the voltage source can deliver the necessary current, then the capacitor is charged up to the maximum voltage minus the (small) voltage drop over the diode.
Before we delve into iii and iv, please post your sketches and describe them in some detail. Then your questions about them will be easier to tackle.

Thanks for the reply. Above are my sketches. Due to smaller RC value for the 100 ohm case, the capacitor discharges much much faster. I think that part (iii) is asking for the difference in potential between the point where the potential of discharged capacitor = potential of AC source, and the maximum point.

 

[BvU]

For part (iii), I know that the period of discharge is approximately 1/f = 0.02, and RC = 0.5.
The capacitor should have Voltage = 20(e^-0.02/0.5), and the maximum voltage is 20V. The difference of the 2 values is 0.78, which is twice the correct answer of 0.39V. (BTW I am assuming 20V is the max voltage, not the RMS value, do correct me if I'm wrong.)

The 20 V is given is said to be peak-to-peak, so the amplitude is 10 V and there is your factor 2.

For part (iv), I think I am not very clear on what "conducting" means. I took "conducting" to mean "allows current to flow through", isn't that just 50% of the time?

The diode conducts if the voltage at the anode (input) is greater than the voltage at the cathode (output )

So you are supposed to set up some equation and estimate the time this is the case.

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