# Designing LED photography strobe power supply

Discussion in 'Power Electronics' started by CommanderLake, Oct 10, 2020.

1. ### CommanderLake

199
6
Oct 2, 2012
I've played with a couple of Cree CMA3090 LED's for things like a light that lights up the clouds and a high speed strobe and even using one as a flash connected to a DSLR hot shoe, but one is slightly too weak for that task so I want to try using 4 of them to make a strobe I can use connected to a DSLR with a cable for the trigger and a low voltage power supply.

The only thing I need help with is the power supply, I don't want to mess with mains voltage so it will have to use a 12-24v supply but I need to boost that to 50v for the LED's and the total current could be up to 16 amps which is 800w!
The flash will be up to 1 maybe 2 ms in duration.
Obviously I don't want it to draw 800w from the low voltage supply so it will need a huge amount of capacitance to back it up, supercaps maybe.

What kind of power supply would I need to provide such a pulse without the output voltage dropping?
If I make a boost converter that can provide the current for the duration of the pulse without needing a big heatsink the input capacitor would have to be huge and the low voltage supply would need to be current limited so its not overloaded by charging the capacitor.

Then there's the task of choosing the controller IC for the supply, just look how many there are on Digikey!

2. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
50 V is already above the level considered to be safe by many standards (e.g. UL: 42.2 V (DC)).
If you need 50 V for the LEDs buy a 50 V power supply. Alternatively use the LEDs in parallel (each with its own current limiting resistor) at a lower voltage.

You can use capacitors to stabilize a power supply against short current surges as in your application. However, capacitor can provide only very, very short current pulsese because they quickly lose energy and the voltage will drop considerably.
I = 16 A
V = 50 V
t = 2ms
Assuming an acceptable voltage drop of 10 % this means dV = 5 V (dV = Delta Volts)
The relevant equation for the voltage/current/time/capacitance relationship here is generally C = (I × t) / V or taking into consideration the change in voltage, dV, C = (I × t) / dV
With the above values you end up with C = 6.4 mF = 6400 µF. A single 8600 µF electrolytic capacitor (better 2 × 4700 µF or 4 × 2200 µFin parallel) could be used. Higher capacitance values are required if the allowable dV is less. You do the math.
Why capacitors in parallel instead of a single big capacitor? An electrolytic capacitor has an internal series resistance (ESR = equivalent series resistance). So called low-ESR types have around 20 mΩ (give or take). At 16 A this results in a voltage drop of 0.3 V which will not be available for the LEDs and heat up the capacitor. Using 4 smaller capacitors in parallel reduces the current through each capacitor to 1/4, i.e. 4 A instead of 16 A. The resulting voltage drop will be 1/4, too, i.e. 0.08 V. You have less power dissipation in the capacitors and more power for the LEDs.
Note that you will need some means of current limiting (either a series resistor or built-in current limiting of the power supply) to keep the inrush current in check. Inrush current is the current that flows initially when the capacitors are discharged and the 50 V power supply is initially turned on. The capacitors look like a short circuit to the power supply until the voltage across the capacitors rises due to the charge current. This inrush current can be very high and possibly overload the power supply or a fuse if not limited.
With capacitors you will always experience a voltage drop. You can keep it arbitrarily small by increasing capacitance, see above.

Don't build your own power supply. Frankly, you seem to lack the experience and dealing with 800 W, even if only peak power, is not what I recommend for beginners. Buy a suitable off the shelf power supply and add capacitors (considering my above rantings).

How are you going to trigger the discharge and how are you going to turn the LEDs off after 2 ms? If you fail to turn off the LEDs in time, the energy stored in the capacitor(s) may destroy the LEDs.

3. ### CommanderLake

199
6
Oct 2, 2012
Maybe this isn't the bast place to ask this, we don't seem to be on the same level.

4. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
What is confusing you? Maybe I can detail my explanation where it is required.
You're free to ask anywhere, but without some basic theory - and what I wrote is not much in terms of theory - you will not get along unless you buy an off the shelf unit that does what you envision.

5. ### CommanderLake

199
6
Oct 2, 2012
I'm not a novice I was looking for suggestions for a DC-DC converter but I found better help from TI.