# [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 Distinguished Member

May 3, 2010
581
38
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
4,693
1,321
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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