# [que]calculation for half wave rectifier

Discussion in 'General Electronics Chat' started by simpsonss, Jul 18, 2010.

1. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
hi,

i'm doing some calculation for the half wave recitifier circuit that built with a 1N4007 diode, a filter cap 10uF/35V or 2200uF/35v , transformer rating is 15-0 1A, load resistance 100 ohm or 10k ohm.

after construct the circuit using 10uf/35v and a 10k ohm. then i calculate the ripple voltage using this formula
Vr = Vp / fCR

i get Vr = 4.72V

then, i calculate the Vo using Vo = Vp - 0.5Vr

i get ~21.24V.

then, i compare my calculation with the waveform on the scope and DMM, as what i get is using DMM i get Vo=22V and from scope Vr = 4.0V

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Now, i would like to go for 10uf/35V as filter cap and 100 ohm as load.

after calculation it seems like i get Vr = 472V. Could it be so high??!!!

can anyone help me on this?

thank you.

2. ### blah2222 Well-Known Member

May 3, 2010
554
33
Well you're 100 ohm resistor is 100 times less resistive than the 10K and after using the ripple voltage equation you will have a result that is 100 times greater than from using the 10K.

4.27 V -> 427 V

3. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
i understand of course the result will be 100 times greater. But i'm confuse about what i would get on my scope? When i'm using 10uF and 100K ohm load the ripple voltage is 4.72V which it is obviously shown in the scope. But what if when i use 10uF and 100 ohm? will it be 472V for the ripple voltage? can a 1/2 resistor stand for this high voltage? this is the point i'm struggle in.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
No voltage cannot reach 472V II Kirchhoff law don't allow such a situation. The output voltage will look almost the same as if we remove capacitor from the circuit.

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5. ### simpsonss Thread Starter Active Member

Jul 8, 2008
173
0
hi, what software u are using for this simulation?

Mar 24, 2008
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