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How to decide value of Capacitor in Power Supply design?

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mananshah93

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Hi to all!!!
I am designing a 7.5V 2A power supply using LM338. Now it's a basic circuit like---AC 230---->Transformer 12VAC 10Amp.--->Bridge Rectifier--->Filter(using Capacitor)--->LM338 (setting R2/R1)--->Vout with Capacitor....
I am posting the circuit diagram ,can any one tell me how to decide the values of all capacitors...

My specification---mains 230V 50Hz AC, T1-12 VAC-10 Amps..,I want Vout=7.5V and 2Amp.
See figure ..
Can you tell me how to determine values of C1 to C5....also diodes D1 and D2 are there.... If I am not wrong D1 is for protection,and what about D2???
 

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MikeMl

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Most Helpful Member
Here is a proceedure for calculating C1.
C2, C3, C5 are specified by National on the Data Sheet.
D1 and D2 are specified by National on the Data Sheet or in their Appication Notes.

C4 shouldn't be there at all. It actually makes the dynamic step response of the supply worse than if it wasn't there. The need for C4 is an "old wives tale" OWT.
 
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Willbe

New Member
For large values of C1, the diode bridge needs a sufficient I squared T rating. A 1N400x can stand about 7 Amps-squared-Seconds.
 

mananshah93

New Member
Here is a proceedure for calculating C1.
C2, C3, C5 are specified by National on the Data Sheet.
D1 and D2 are specified by National on the Data Sheet or in their Appication Notes.

C4 shouldn't be there at all. It actually makes the dynamic step response of the supply worse than if it wasn't there. The need for C4 is an "old wives tale" OWT.
I am sorry, but I do not get this "C2, C3, C5 are specified by National on the Data Sheet.
D1 and D2 are specified by National on the Data Sheet or in their Appication Notes."
 

MikeMl

Well-Known Member
Most Helpful Member

mananshah93

New Member
Here is a proceedure for calculating C1.
C2, C3, C5 are specified by National on the Data Sheet.
D1 and D2 are specified by National on the Data Sheet or in their Appication Notes.

C4 shouldn't be there at all. It actually makes the dynamic step response of the supply worse than if it wasn't there. The need for C4 is an "old wives tale" OWT.
Hi! Mike..thank you for your assistance...I read the article regarding value of C1...but I could not understand some thing...

Capacitor Value
The required capacitance for a given load current and ripple voltage is determined (approximately) by the formula [1]...

C = ( I L / ΔV) * k * 1,000 uF ... where
I L = Load current
ΔV = peak-peak ripple voltage
k = 6 for 120Hz or 7 for 100Hz ripple frequency
Since all my calculations above were done at 100Hz ripple current, this can be checked easily, so ...

I L = 1.44, ripple = 2V p-p, therefore C = 5,040uF"


My power supply requires 2Amp' load at 7.5 volt..Transformer--230V-->9V or 12V AC---with 27VA...
how to calculate value of IL,ΔV and k for 50 Hz.
 

MikeMl

Well-Known Member
Most Helpful Member
...
My power supply requires 2Amp' load at 7.5 volt..Transformer--230V-->9V or 12V AC---with 27VA...
how to calculate value of IL,ΔV and k for 50 Hz.
If you are using either a full-wave bridge rectifier, or a Center-Tapped transformer with two diodes, the ripple frequency is twice the line freq, hence 100Hz. That's why the formula is for either 100Hz or 120Hz. No one uses a half-wave rectifier...

The key point is that C1 must be big enough so that its voltage doesn't sag below the sum of the supply's output voltage plus the regulator's dropout voltage. C1 gets a pulse of current from the rectifier twice per 50Hz cycle, i.e. every 10msec. It charges to a peak determined by the transformer secondary voltage minus some IR drop and the forward drop of either one or two rectifier diodes. For the next 9msec or so, it is discharged by the load current on the supply. Charge is Q=C×ΔV= i×Δt. Rearranging gives C=i×Δt/ΔV.

You know i (supply load current) and Δt (about 9msec at 50Hz), so if you know ΔV (how much C1 can sag) you can solve for C. :D

Guessing that the dropout voltage of your reg is about 2V, and you want 7.5V out, that means that the minimum input voltage to the reg is 9.5V. If you start with a 12Vrms tranny, the capacitor will charge to 1.4*12-2V (peak is 1.414×rms, 2V drop in two diodes)= 15V. ΔV=15-9.5=5.5

Finally, C=2×0.009/5.5 = 0.0033 FARADs!!! = 3300uF

Left as an exercise for the student, figure C1 for a 9V secondary on your tranny.
 
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mananshah93

New Member
Thanks ....Mike....oh! it was a good stuff..........

Now moving to 9V secondary voltage of a transformer...and taking drop out voltage into consideration...9V would not be enough...
still calculating...
1.414*(9-2)=9.898V....then ΔV=9.898-9.5=.398

and C=45226uF :eek:
 

stevez

Active Member
I may have missed something here but I'll share my thoughts so I learn something if I am mistaken.

You do need to be sure that the transformer secondary characteristics and C1 are such that the lowest voltage to the input of the regulator at any moment is at least the desired output voltage plus the differential required across the regulator (possibly the dropout?, I am not sure).

Another area of concern might be the ripple voltage on the output of the regulator. The regulator does have the ability to smooth out some ripple - the datasheet suggests a ripple rejection of 60 dB. The ripple rejection won't fix the problem described in the paragraph above but might help to keep the value of C1 down. For many applications the resulting ripple is low enough but it seems worth considering during the design phase.

Another item - the transformer secondary voltage will change slightly from no load to full load. It is also affected by primary voltage. Make sure you factor these things in if it matters. An example might be that the rating of your secondary is based on a 120 volt primary but the actual voltage where it will be used is 110 volts - you might want to plan for that.

For some applications this stuff isn't worth the worry and certainly other applications demand even further insight and analysis.
 

grim

New Member
Your transformer is a bit over rated too, so it won't give the voltage you expect, as it isn't being fully loaded. you want 15W out and have a tranny rated at 120VA - thats a lot of extra cost and weight, you don't need. :)
 
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