1
\$\begingroup\$

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.

One LED driver on 30V

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.
enter image description here

  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).
\$\endgroup\$
1

2 Answers 2

Reset to default
2
\$\begingroup\$

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: -

enter image description here

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.

\$\endgroup\$
4
  • \$\begingroup\$ 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 \$\endgroup\$
    – AaronD
    Feb 23, 2015 at 17:50
  • \$\begingroup\$ @AaronD: if I understand correctly, each second the voltage drop is equal to the current divided by... \$\endgroup\$
    – machineaddict
    Feb 24, 2015 at 13:02
  • \$\begingroup\$ @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; \$\endgroup\$
    – machineaddict
    Feb 24, 2015 at 13:04
  • \$\begingroup\$ That looks about right. \$\endgroup\$
    – Andy aka
    Feb 24, 2015 at 13:31
1
\$\begingroup\$

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.

Now then, about ripple:

Because you have a regulator - specifically a current regulator, as opposed to the more common voltage regulator, but it's still a regulator - you really don't need to care that much about ripple. You have a minimum voltage to do what you want (dropout voltage), you have a maximum voltage from the rectifier, and in this case you have a well-known operating current.

Choose a cap or build a cap pack that can handle ripple current equal to your running current, and will still be above the dropout voltage just before the next rectifier pulse while supplying that current.

Once you figure that out, double the capacitance or more to account for component tolerance and product life expectancy (electros lose capacitance over time), order it, and test it.

\$\endgroup\$
3
  • \$\begingroup\$ One module contains 10 LEDs. And there are 10 modules (100 LEDs). I measured the voltage on each module and it varies from 30V to 31V. Basically, they are regulating themselves (the current). And about the ripple: I had some LED based bulbs and they were flickering. I want to have zero flickering, I can test them with an oscilloscope. But I wanted to have an idea about the capacitor, instead of testing different capacitors for that value. \$\endgroup\$
    – machineaddict
    Feb 24, 2015 at 8:46
  • \$\begingroup\$ Consumer-viable (cheap) LED bulbs are probably regulated about as well as cheap LED rope lights: they're not. Just like the rope lights that I looked at, I would not be surprised at all to see enough LED's in series to add up to just less than the peak AC voltage, and a low-power resistor to take up the slack. In other words, they ditch the cap entirely, add a small resistive load, and make the rectifier out of LED's. This will only work at all at the AC peaks, hence the flicker. On the other hand, your circuit actually has a regulator. \$\endgroup\$
    – AaronD
    Feb 24, 2015 at 15:40
  • 1
    \$\begingroup\$ You will NEVER eliminate the ripple voltage. To do that would require infinite capacitance, which you don't have. :) So you will always have ripple voltage. The question is, "How much is acceptable?" Because of the regulator, the answer is actually, "quite a lot," compared to what you might have been thinking. My answer attempts to have you derive that actual limit, and then create some wiggle room between it and where your circuit actually operates. \$\endgroup\$
    – AaronD
    Feb 24, 2015 at 15:47

Your Answer

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

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