# selection of capacitor

Discussion in 'Homework Help' started by vead, Dec 15, 2011.

Nov 24, 2011
621
8
i want to convert 240v ac to 12v dc
peak voltage of secondary=16.970v
rms voltage of secondary=12v
output of bridge rectifier=16.970-1.4=15.57v dc

how can calculate capacitor for circuit plz explain with formula and ripple voltage of capacitor thanks.....

Last edited: Dec 15, 2011
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The simple conservative estimate for the filter capacitor is

$C=\frac{I_{load}}{2fV_r}$

f is the input mains frequency [Hz].
Vr is the allowed peak-to-peak ripple voltage.

Vdc≈Vpk-Vr/2

where Vpk is the rectifier output peak value.

This is based on the 'crude' assumption that over each rectified half wave cycle, the capacitor rapidly (instantaneously) charges to the rectified peak voltage then discharges at a constant rate into the load over the rectified cycle interval T/2 where T=1/f.

Nov 24, 2011
621
8
thanks for quick reply but i dont know ripple voltage how can i calculate it

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Take another look at my post.

If Vdc=12V and Vpk=15.57V and Vdc≈Vpk-Vr/2 then what is Vr?

You also need to state the operating frequency.

A word of caution: The ripple voltage will be so high that the simple relationship for calculating the capacitance will over-estimate the required value quite significantly. In this case one might be forced to consider a more accurate estimation method.

Last edited: Dec 15, 2011
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
In such circumstances one would better estimate the capacitance as

$C=\frac{I_{load}\Delta t}{V_r}$

Where

$\Delta t=\frac{\frac{\pi}{2}+arcsin ( 1-\frac{V_r}{V_{pk}})}{\omega}$

ω is the input frequency [radians/sec]

NB:The arcsin() function result would be calculated in radians rather than degrees.

Last edited: Dec 15, 2011
6. ### jimkeith Active Member

Oct 26, 2011
539
99
Simply solve for 1V peak to peak ripple and then select a larger, standard value capacitor--that will be good for most applications

Nov 24, 2011
621
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using this formula
$C=\frac{I_{load}}{2fV_r}$

f=60
v ripple=1v
thus c=1/1000 μf

please check me if i correct

8. ### MrChips Moderator

Oct 2, 2009
12,624
3,451
No, your answer is only a million times too small. The correct answer is C = 1/1000 F = 1000 uF.

Last edited: Dec 15, 2011
9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
In post#1 the OP specifies that the DC output is to be 12V.

If the ripple is 1V p-p the (mean) output would be ~15V DC.

Selecting 1V p-p therefore doesn't meet the OP's stated spec.

To obtain ~12V average output the ripple would have to be more like ~7.1V.

I understand this would be a poor outcome in terms of the overall performance and that a 1V p-p ripple would be more acceptable. In the end it depends on what the OP is actually required to achieve in answering the question - assuming this actually is homework.

Last edited: Dec 15, 2011

Nov 24, 2011
621
8
i want to convert 240v ac to 12v dc
peak voltage of secondary=16.970v
rms voltage of secondary=12v
output of bridge rectifier=16.970-1.4=15.57v dc

need correct equations for my smoothing cap
can you show calculation of capacitor for this circuit