# half wave rectifiers and ripple factor

Discussion in 'Homework Help' started by micachu, Mar 1, 2015.

1. ### micachu Thread Starter New Member

Oct 13, 2014
15
0
We constructed the circuit below

and when measuring the output we got a ripple factor of abut 25%.

We then added a 10 ohm resistor next to the capacitor as shown,

and the results were a ripple factor of 29%.

I've been stuck and can't figure out why the ripple factor went up just by adding a small resistor in series with the capacitor.

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2. ### crutschow Expert

Mar 14, 2008
13,475
3,361
Why would you think the resistor would have no effect?
The load is drawing current from the capacitor during the time it's not being charged from the diode.
This current has to flow through the resistor in series with the capacitor which causes a voltage drop across the resistor.
This voltage drop subtracts from the capacitor output voltage so the capacitor voltage is lower than without the resistor.
If you view the resistor voltage you will see this.

3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Also, it slows the rate at which the capacitor can be recharged.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Plot the capacitor current with and without the resistor.