 ### Network # Questions about simple power supply

P

#### Paul Blitz

Jan 1, 1970
0
Harry Muscle said:
What kind of cap and what value should I use to smooth out the DC
before I feed it to the regulator (LM 340T12)? Is there a formula for this
cap value? Is anything else required?

The "finger-in-air" figure I have used for many years is "2000uF per amp".
So, if you need only 100mA, use 200uF, and then put it into your regulator.
Remember to rate the voltage high-enough (eg 24v ac transformer often gives
about 30v under low load, multiply by 1.4 (if using full bridge rectifier)
so I'd select 50V working).

Paul

I

#### Ian Bell

Jan 1, 1970
0
Paul said:
The "finger-in-air" figure I have used for many years is "2000uF per amp".
So, if you need only 100mA, use 200uF, and then put it into your
regulator. Remember to rate the voltage high-enough (eg 24v ac transformer
often gives about 30v under low load, multiply by 1.4 (if using full
bridge rectifier) so I'd select 50V working).

I use the good old formula C * dV = I * dt

Where C is the smoothing capacitance, dV is the desired ripple voltage (peak
to peak), I is the desired load current and dt is 5mS for full wave
recification.

IAn

H

#### Harry Muscle

Jan 1, 1970
0
Ian Bell said:
I use the good old formula C * dV = I * dt

Where C is the smoothing capacitance, dV is the desired ripple ( ak
to peak), I is the desired load current and dt is 5mS for full wave
recification.

IAn

Thanks, that means that if I want a ripple of no more than 10V, current of
0.5A I would have:

C*10 = 0.5 * 0.005
C*10 = 0.0025
C = 0.00025
or C = 250uF

Is that correct?

Thanks,
Harry

I

#### Ian Bell

Jan 1, 1970
0
Harry said:
Thanks, that means that if I want a ripple of no more than 10V, current of
0.5A I would have:

C*10 = 0.5 * 0.005
C*10 = 0.0025
C = 0.00025
or C = 250uF

Is that correct?

Thanks,
Harry

Looks OK to me. Remember the bottom of the ripple needs to be comfortably
above the minimum input to the regulator. Personaly I would have thought
10V ripple a bit high.

J

#### John Popelish

Jan 1, 1970
0
Harry said:
Thanks, that means that if I want a ripple of no more than 10V, current of
0.5A I would have:

C*10 = 0.5 * 0.005
C*10 = 0.0025
C = 0.00025
or C = 250uF

Is that correct?

You also have to have a capacitor with a ripple current rating that is
bigger than the actual ripple current, or it will overheat. Larger
caps generally have a higher ripple current rating, so you may have to
go up a bit just to get within this rating.

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