capacitor calculation for ripple for smps i/p and o/p

[deepakchikane]

Dear all..
greetings..!!

i have been suffered with capacitor calculations.. as i m getting confused by reading too mch books..
i want a correct direction to select the capacitor to minimize the ripples... also the tolerances considerations for i/p and o/p...

waiting to hearing you sooon..!!

 

[BradtheRad]

Rough calculations for output smoothing cap:

The capacitor charge drops a certain amount during an idle gap.
The amount of drop depends on the time constant, TC = R * C.
One time constant is the time it takes to change by 63% (by definition).

Suppose your load is 20 ohms. Operating frequency is 6 kHz.
You set a limit of 15% ripple. (A high figure but it shows up well on a simulated oscilloscope.)

Idle time can be as long as 1/6,000 sec. So the capacitor is permitted to fall 15% in 167 uSec.
This translates to 63% in 700 uSec. (To make things easy we use linear calculations.)

Therefore the time constant is .0007 second.
So solving TC = R * C, gives a cap value of 35 uF.

The 10% duty cycle creates an idle time which is close to the maximum, and the ripple is approximately the expected amount for our cap value.

 

[deepakchikane]

Dear..
not satiesfied...

as one of my team mates keeps 940uF/450V CAP after bridge... n i am going to load that capacitor with 750 watts..

how that guy keep above cap after the bridge..??

he waz telling that...
ripple=vpp/loading power *(1/100(freq)*940uF)... SOMETHING LIKE THAT..

But he showed me exactly 17 v ripple by calculations and also practically wen we connect scope..
plz help me out hw..??

 

[BradtheRad]

450V CAP

This suggests a high voltage supply. Supposing it is 450 V... then 750W calculates to 1.67 A.
Effective load 270 ohms.

1/100 is the idle gap time (time between current bursts of rectified 50 Hz mains AC).

A 940 uF cap, and 270 ohms, makes a time constant of .25 sec. Charge level would fall 63% in that time.
Therefore in .01 seconds it falls 2.5%. ( 63 * .01 / .25 ) (Although the discharge curve is not straight, we assume it is, to make things easy.)

2.5% of 450 makes 11.3 V of ripple.

This can only be a rough estimate. It's not a surprise that it's lower than your colleague's figure.

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Consulting another source, here is a simulation:



It agrees better with your colleague's figure.

 

[deepakchikane]

thank u sir..!!

- - - Updated - - -

another one doubt is that,
in dc to dc converter how i can calculate.??
by taking switch frequency of converter..??

consideration criteria for keeping ripple after bridge..??
1% is for the final o/p of converter...

 

[BradtheRad]

another one doubt is that,
in dc to dc converter how i can calculate.??
by taking switch frequency of converter..??

Yes, the frequency is important. At high frequencies you do not need such a large value capacitor.

consideration criteria for keeping ripple after bridge..??
1% is for the final o/p of converter...

If this is mains AC frequency then you need a large capacitor. A round figure comes from the customary advice of 1,000 uF for every ampere you draw.

Or using time constant calculations...
Your ripple spec is 1%, inside a .01 second timespan.
This translates to a 63% drop in .63 sec. Therefore time constant is 0.63.
Suppose load is 270 ohms. Then the smoothing capacitor is 2,333 uF. ( .63 / 270 )

You can get more info from the list of similar threads at the bottom of this page.

 

[deepakchikane]

Dear sir,

i agreed for the above explanation..

i want to calculate the o/p capacitor of my half bridge topology...

switch freq iz 45 khz..
converter worked at 380 v dc...

the o/p is loaded 1.7 amp @400v dc...

the o/p windings of 200v of 2 windings series together to produce 400v output...

please help me out..

i gone similar threads bt my fullfillments not completed..

like c=i*(1/f)/ripple... it coudnt helping me to find optimum solution..

 

[BradtheRad]

switch freq iz 45 khz..
converter worked at 380 v dc...[/B]

the o/p is loaded 1.7 amp @400v dc...

the o/p windings of 200v of 2 windings series together to produce 400v output...

please help me out..

i gone similar threads bt my fullfillments not completed..

it coudnt helping me to find optimum solution..

Method #1.

There is always the method of installing a low value capacitor...
and watching to see if the ripple is too much...
and if it is then try a higher value.
Etc.

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Method #2.

Or, a simulator is also useful for this purpose. 'Let the computer do the math.'

8.4uF. (Duty cycle 10% assumed).



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Method #3.

1% ripple on 400V is 4V. Then the formula:

C = 1.7 * (1/45000) / 4

= 9.4 uF

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Method #4.

The load calculates to 235 ohms. (400 / 17).

We have an idle period up to 1/45,000 sec. This is 22.2 uSec.

Ripple allowed is 1% during that time. Translates to 63% in .0014 sec. The time constant is .0014.

Divide by 235 ohms, gives us 6 uF for the output capacitor.