How to calculate a electrolytic capacitor based on current and voltage to have the smallest ripple

I have built a circuit involving LEDs powered on 230V AC.

The LED current driver that is powered on 30V and contains 10 LEDs of 1W each. 10 of these modules are linked in series and are powered from 313V and it draws 80mA. The 313V is obtain with a bridge rectifier and a capacitor connected to the main line. 1. How do I calculate minimum capacity of the electrolytic capacitor based on current and voltage to have the smallest ripple?
2. How do I know the wattage on AC line? I know that the DC wattage is around 25W (313V*0.08A).
• Feb 23 '15 at 17:30

How do I calculate minimum capacity of the electrolytic capacitor based on current and voltage to have the smallest ripple?

Smallest ripple is when capacitance is theoretically infinite. You have to accept that there will be ripple and you have to decide how much this ripple can be: - Taken from here

How do I know the wattage on AC line? I know that the DC wattage is around 25W (313V*0.08A).

The wattage taken from the AC power line is probably a few percent more than what is consumed in the DC circuits you have attached. Power in = power out + losses in rectifier and smoothing capacitor.

• Actually, it's even easier than your diagram suggests. He's using a constant-current load, so the discharge is linear. V/s = I/C Feb 23 '15 at 17:50
• @AaronD: if I understand correctly, each second the voltage drop is equal to the current divided by... Feb 24 '15 at 13:02
• @Andy: if I have a 220uF capacitor that results in 7V ripple. Is that correct? Vp = 310V; RL = 4Kohm; C = 0.00022F; Dt = 0.02s; Feb 24 '15 at 13:04
• That looks about right. Feb 24 '15 at 13:31

So...you have 10 constant-current circuits in series, so that each of them can contain 1 diode. You might have some issues with that because they must all have exactly the same current simply because they're all in series, but they'll all be set slightly differently because no two components are ever equal.

Therefore, one circuit (which one is completely random) will do all the work of regulating the entire chain while the rest become much closer to short-circuit because they're not quite getting the current that they're set for. Now that one circuit is dropping nearly ALL of the voltage left over between the LED string and the power supply, it's going to get stressed possibly beyond its design.

It would be a much better idea to take your single circuit and just give it all 10 diodes in series. In total: 2 transistors, 2 resistors, and 10 diodes. R2 (current-sense) and the lower transistor can stay exactly as they are, R1 (transistor bias) should be increased by a yet-unknown amount, and the upper transistor should be able to handle the entire residual voltage at the intended current.