Rectifier Diode & Capacitor Ripple Current ratings.

Thread Starter


Joined Jan 8, 2013
Hi all, I want to design a linear power supply for nothing more than my own gratification. I have used some guidance from the 1980 National Semiconductor Voltage Regulator handbook, but want to double check everything, so i know that it is reasonable and to get that warm fuzzy feeling before I buy the components

The desired output specs are 15VDC, 0.6A constant current load.
Assuming my Regulator cct has a 4V Drop out, the minumum DC input shall be no less than 19VDC

My planned Bill of Materials are:

18V 25VA Transformer
4x 1N5401 Diodes (3A) as a Full wave bridge rectifier (FWBR)
1x 63VDC 2200uF Capacitor

In Section 8 of the Nat. Sem. Handbook, it says as a guide, Secondary Transformer current is 1.8 times the Load current for a FWBR + Capacitor filter.
Therefore my Secondary current is 0.6*1.8 = 1.08Arms.

Circuit to be driven from a 18Vrms Transformer. Calculated an approx 1.151Ohm ESR (calculated from the Primary DC Resistance, the turns ratio and Secondary DC Resistance)

1.08Arms * 1.151 ESR = 1.24Vrms Drop (Transformer spec doesnt quote the Inductance so cannot calc Xl or the Z, so ESR is closest i can do)

Diode currents should be assumed at 3 times the DC load current, so 1.8Aavg. 1N5401 is rated for 3A continuous, 100V Reverse rating. Should be suitable.

The Handbook also states that the RMS Capacitor current should be assumed 2-3 times the DC Load. in this case 1.8Arms Ripple Current.
I have chosen a DC rating greater than 2x the Peak No load voltage of the transformer. I.e. 63VDC
With a bit of hit/miss i think a 2200uF Capacitor is suitable, based on the ripple voltage, calculated from Art of Electronics, Horowitz & Hill, Full wave bridge rectifier ripple voltage is;

Vripple (pk to pk) = Iload / 2*F*C

= 0.6 / 2 * 50 * 2200uF
= 2.727 V pk 2 pk.

Then to add up voltages and to calculate the minimum Secondary RMS voltage required if the 230VAC Transformer was supplied down to 220VAC.

Vf = 0.8V (worst case scenario)
FWBR Effeciency = 92%
Nominal Line Voltage 230, min 220
Vt = Transformer ESR drop (1.24Vrms)

Vt + [(Vout + Vdrop + 2*Vf + Vrip) 0.92 ] * (230/220) * 1/SQRT(2)= (from National Semi Handbook)
1.24 + [(15 + 4 + (2*0.8) + 2.727) *0.92] * (230/220) * (1/SQRT(2)) = 17.102Vrms minimum (so 18Vrms Transformer should be fine)

Does this all look sensible/logical? Any thoughts/comments would be greatly appreciated. :)


Joined Mar 14, 2008
You do realize that an 18Vrms transformer has a peak output of 1.4 * 18Vrms = 25.2Vpk, so the no load DC output with a bridge rectifier will be about 25.2Vdc - 1.2Vdc = 24vDC.
And since a transformer with no-load typically has a higher output voltage than rated (to allow for its winding resistance drop), the actual no-load output will likely be a couple volts higher or about 26Vdc.

It might be preferable to use a 15Vac transformer with a low-dropout regulator (one that has a volt or so maximum drop) to minimize power loss.

Thread Starter


Joined Jan 8, 2013
Yes i appreciate the no load voltage will be greater, but really im only concerned about the trough voltage of the ripple.

Greater peaks will mean greater Pdisp. but this is more a of an exercise to have a go at designing my own circuit and going through the pitfalls etc. of the design.