# HELP HELP HELP

Joined Feb 20, 2006
2
a half wave rectifier operatin from a 50hz 24volt AC supply is connected to a 100 Ω load
caluclate the value of the smoothing capacitor required to maintain a peak to peak voltage ripple at 2volts

Joined Jan 10, 2006
614
Bloody time constants, I'd have to go way way back to my tech stuff to refresh my tired memory, and I can't really be bothered.

OK, try this site.....

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by helpmeplease@Feb 21 2006, 10:35 AM
a half wave rectifier operatin from a 50hz 24volt AC supply is connected to a 100 Ω load
caluclate the value of the smoothing capacitor required to maintain a peak to peak voltage ripple at 2volts
[post=14227]Quoted post[/post]​
Calculate the elapsed time from one peak of the sinusoidal to the point on the increasing slope of the next sinusoidal where the voltage drop is 2V. If you want to be accurate you have to do this, otherwise an approximate value can be obtained from the time space of two rectified sinusoidal peak.

Read this page on capacitor discharging and use the formulaes to calculate the capacitance from the discharging time calculated above, load resistance and voltage drop of 2V.

#### Nirvana

Joined Jan 18, 2005
58
Well if you don't consider the junction voltage of the diode (assumed to be Silicon), then C = Q/V
which (as Q = IT) is: C = IT/V, now I in this case is V/R which is 24/100 = 0.24A. T is 1/F = 1/50 = 0.02, and the required voltage is 2. Therefore with this information we get C = IT/V = (0.24 x 0.02)/2 = 0.0048/2 = 0.0024F which is 2.4mF.
If you considered the junction voltage simply use the above calculations but take the junction voltage away from the input voltage and use that to calculate I. For silicon the junction voltage is about 0.7 volts, for Germanium its 0.2 volts. Hope that helps you.
I got 0.00233 F or 2.33mF when using a silicon diode.
Nirvana.