3
\$\begingroup\$

I am building my first power supply unit (my first project too) and need help in figuring the size of filter capacitor (and type, though I presume Electrolytic) I need to use after the rectifier diodes.

Specs:- 2 channels of 1.5VDC - 9VDC adjustable outputs with max 2.5A over both channels

Components sourced:- 229D12 Transformer - output is 12.6VCT (3.8A) at 60 Hz (65% = 2.5A) .... 4 x 1n5404 rectifier diodes (rated at 3A with Vf=1V) .... LM350 adjustable voltage regulator (rated at 3A with Vf=1.25V)

What I know so far:- 60Hz rectified is 120 cycles/sec so T=0.00833sec .... 12.6V - (2*1V diode drop) = 10.6V rectified - 1.25V (Vf of LM350) = 9.35VDC max .... Max current is going to be 2.5A (limited through LM350 and 3.74 Ohms of resistance)

I have tried sourcing the formula to calculate the size of capacitor needed, and I have seen many formulas regarding capacitors. Each one of them confused the heck outta me with funky symbols and abbreviations that I have yet to learn and each of the formulas was for something other than what I was wanting.

I speak basic English, understand average Mathematics, but still have no idea on most of the symbols and abbreviations used in EE. Please keep your answers simplified for this noob. :D

I appreciate any help ya'll can give.

\$\endgroup\$
4
  • \$\begingroup\$ 12.6V RMS is more like 17.8V DC when rectified (not counting diode drop, so more like 15V after the diodes)... \$\endgroup\$
    – Majenko
    Commented Dec 21, 2014 at 23:39
  • \$\begingroup\$ Don't know why I was thinking P-P Specs updated ... 1.5VDC - 12VDC :D I may just put in a 12VDC linear regulator and feed the LM350 adj. off that. \$\endgroup\$ Commented Dec 21, 2014 at 23:43
  • 1
    \$\begingroup\$ Also, I note that you don't list your heatsink for the LM350. If you use a 12 volt preregulator, and try to supply 1.5 volts at 2.5 amps, the LM350 will dissipate about 25 watts, and that will require serious heatsinking. \$\endgroup\$ Commented Dec 22, 2014 at 2:35
  • \$\begingroup\$ I see your point. I figured I could think of thermal protection after I knew what components I would need, but at 25W heat dissipation there is a definite fire risk with my PSU. Back to the drawing board ... \$\endgroup\$ Commented Dec 24, 2014 at 18:37

1 Answer 1

3
\$\begingroup\$
  1. Define your maximum ripple voltage: ΔV (also called dV, V ripple). Look at image below and I guess you should figure it out.

  2. Use this formula to determine minimum capacitor capacity.

    \$C = \frac{I}{Vpp * 2f}\$

    where: C is capacitor capacity in Farads (1F = 1000000uF), I - current, Vpp (or delta V) - voltage change on capacitor when it is being discharged, f - frequency before bridge

    This method can be found in wikipedia article about ripple, just in diffrent form (formula on wikipedia calculates ripple at given capacitor).

enter image description here

Image source: wikipedia article mentioned above + me

  1. Use capacitor with voltage rating at least a bit higher than your transformer amplitude, not RMS voltage. Amplitude for sinuoid after full bridge is:

    \$Vpp = Vrms * \sqrt{2}\$

In your case it will be about 17,8V, so you cant use 16V capacitor.

Also notice, that transformer open circuit (no load) voltage may be much higher than nominal.

\$\endgroup\$
9
  • \$\begingroup\$ The voltage will be even higher with a light load, and higher again if the mains voltage is on the high side, possibly around than 22 or 23V. 25V is the minimum safe rating, and 35V would be better. \$\endgroup\$ Commented Dec 22, 2014 at 1:37
  • \$\begingroup\$ You are right. Transformer open circuit voltage is higher than nominal voltage. Thanks for pointing that. I added that in my answer. \$\endgroup\$
    – Kamil
    Commented Dec 22, 2014 at 1:45
  • \$\begingroup\$ @Kamil: \$Vpp = 2(Vrms * \sqrt{2})\$ \$\endgroup\$
    – EM Fields
    Commented Dec 22, 2014 at 9:36
  • \$\begingroup\$ @EMFields Oh dear, I meant Vpp after bridge. I corrected that. \$\endgroup\$
    – Kamil
    Commented Dec 22, 2014 at 13:52
  • \$\begingroup\$ @Kamil: Part 1. Vpp has a very specific meaning in the context at hand, that meaning being the sum of the absolute values of the negative and positive peaks of the waveform exiting the transformer. From the OP's: I speak basic English, understand average Mathematics, but still have no idea on most of the symbols and abbreviations used in EE. it's imperative that he not walk away from the discussion with misinformation which will later cause him grief. \$\endgroup\$
    – EM Fields
    Commented Dec 22, 2014 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.