# Full and half wave rectifier with Capacitor filter problem

Discussion in 'Homework Help' started by matthew798, Feb 23, 2013.

1. ### matthew798 Thread Starter Member

Jan 16, 2013
38
2
Hey guys.

I think my answer sheet is mistaken...

I have the following problem:

Half wave rectifier, with 47μF filter and 10k load. RMS on primary xformer winding is 120v at 60Hz. Transformer ratio is 8/1. Find Vdc and Vr. (Avergae voltage and ripple voltage)

120*√2 = 169.705v on the primary winding peak

169.705 / 8 = 21.21v on the secondary winding peak

So since it's a half wave rectifier,

21.21 / ∏ = 6.75v is the average voltage (Vdc)

And finally,

Vr = 6.75 / (60Hz * 10KΩ * 47μF) = 239.36mv

Now my answer sheet says that it should be 752.13mv

It would seem that they used Vp / (∫*Ω*f), whereas I used Vdc / (∫*Ω*f). I strongly believe that to be incorrect, because using Vp instead of Vdc does NOT take into account that it is running on a HALF wave rectifier... Which effectively doubles the time it takes for the capacitor to start charging again.

Thanks guys!

2. ### #12 Expert

Nov 30, 2010
18,076
9,684
I believe your mistake starts at Vp/Pi

3. ### matthew798 Thread Starter Member

Jan 16, 2013
38
2
Can you elaborate? Would it be Vp/2 instead? Is that not for calculating Vrms with a half wave rectifier?

4. ### matthew798 Thread Starter Member

Jan 16, 2013
38
2
Never mind I understand it now!

5. ### #12 Expert

Nov 30, 2010
18,076
9,684
That's the rules on the Homework forum. I can't tell you the answer, but I can point you at it.

Aug 29, 2013
1
0
dear matthew798

i agree with what u have written. the formula doesnot account for the difference in the ripple half wave and full wave rectifier. can u plz help?

7. ### donpetru AAC Fanatic!

Nov 14, 2008
185
27
In the case of half-wave rectifier with capacitive filter, the ripple voltage (Vr) is:

Vr = I_avg / f * C

where I_avg - is the average current through the diode, which is equal to current drawn by the load; f = 60Hz, C = 47uF.
Please note, in this case, the RMS current in the secondary winding of the transformer is much higher as I_avg (usually I_rms = 2.5 * I_avg).

You say the transformer ratio is 8/1, so this mean 120/8 = 15Vrms in the secondary winding. We know consumer value as 10k. Then the rectified voltage is U = 0.95 * Us where Us is 15V. Now, let's calculate:

I_avg = I_rms / 2.5 = U / 2.5 * 10k = 0.95 * Us / 2.5 * 10k = 0.57mA.

Vr = I_avg / f * C = 0.57 * 10-3 / 60 * 47 * 10-6 = 0.202V = 202mV.