Calculating Peak Capacitor Voltage Across a Half-Wave Rectified AC

[p75213]

Hi,
I am trying to figure out the peak voltage across a capacitor when charged by 1/2 wave rectified AC. The formula for instantaneous voltage is : V_c=V_s(1-e^-t/RC) where:
V_c=Capacitor voltage,
V_s=Source Voltage,
t=time
RC = the capacitor time constant

So I thought I could integrate with respect to t that formula from t=0 to t=1/4 of the period of the AC cycle. Something like this:
∫V_pSin(ωt)(1-e^-t/RC)dt where:
V_p=Peak Voltage

However that can't be correct as I got some ridiculous answer when I used an online integral calculator to do the work.

 

[mfb]

The formula for instantaneous voltage is : V_c=V_s(1-e^-t/RC) where:
V_c=Capacitor voltage,
V_s=Source Voltage,
t=time
RC = the capacitor time constant

This assumes a constant source voltage.

What is the time-derivative of V_c?

So I thought I could integrate with respect to t that formula from t=0 to t=1/4 of the period of the AC cycle.

Be careful with the time of maximal V_c.

 

[donpacino]

take a look at this picture. The answer for peak voltage is quite simple!

http://www.circuitstoday.com/wp-con...half_wave_rectifier_with_capacitor_filter.jpg

 

[p75213]

This assumes a constant source voltage.

What is the time-derivative of V_c?
Be careful with the time of maximal V_c.

dV_c/dt = V_pSin(ωt)*1/RC*e^-t/RC+(1-e^-t/RC)*V_p*ω*Cos(ωt)

Don't ask me to plug values into that or graph it. I've already tried and can't get anything to either calculate a result or graph it.

 

[NascentOxygen]

RC = the capacitor time constant

Are you involving the time constant of the capacitor's charging path, or the capacitor's discharging path? The two are independent.

The fact that you seem to believe you need maths tells me you aren't yet understanding the circuit's principle of operation.

 

[p75213]

Are you involving the time constant of the capacitor's charging path, or the capacitor's discharging path? The two are independent.

The fact that you seem to believe you need maths tells me you aren't yet understanding the circuit's operation.

The charging path is what I am interested in. Your right - I don't understand the circuits operation.

 

[NascentOxygen]

The charging path is what I am interested in. Your right - I don't understand the circuits operation.

In electronics, no amount of maths will save you if you don't first acquire a thorough understanding of how the circuit works. You need to examine a set of waveforms typically found around this circuit, either by doing the lab experiment or by finding a textbook or web resource. Then, the maths follows the practical.

Try googling. It's all there waiting for you.