I have an application where I have to drive 40 high-power LED's in short pulses using a LED light controller. I got the part after the light controller dimensioned, so now I just need to choose a suitable PSU to supply the light controller. The PSU output voltage has to be in the range of 35-48Vdc and be able to supply around 550W of power within 20us.

I found a couple of 48Vdc 600W PSU's that will meet the first two requirements, but what parameter should I look after to be sure that the PSU can deliver the needed power within just 20us?

Clarification: A single pulse starts with a 20us delay (low), then there is a 300us period where the pulse is high followed by a 19700us delay (low). My main concern is to make sure that the PSU can deliver the power needed immediately after the initial 20us delay.

  • \$\begingroup\$ How about posting some datasheets? \$\endgroup\$ – Matt Young Aug 13 '12 at 15:36
  • \$\begingroup\$ farnell.com/datasheets/319845.pdf or farnell.com/datasheets/1301934.pdf or farnell.com/datasheets/93443.pdf are examples of PSU's I have looked at. \$\endgroup\$ – mola Aug 13 '12 at 15:55
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    \$\begingroup\$ What is the average power you need? In other words, what is the time delay between pulses? \$\endgroup\$ – Szymon Bęczkowski Aug 13 '12 at 15:59
  • \$\begingroup\$ @SzymonBęczkowski The pulse width is 300us and the delay is 19700us, so the average power is only around 8W. \$\endgroup\$ – mola Aug 13 '12 at 17:29
  • \$\begingroup\$ Depending on your wire length, you may need a local reservoir anyway, or the inductance will kill your pulse. \$\endgroup\$ – apalopohapa Aug 14 '12 at 8:01

\$ 550W \times 300 \mu s = 165mJ \$ (energy per pulse)

\$ 165mJ \times 50/s = 8.25W \$ (average power, 20ms pulse period so 50 pulses per second)

35–48Vdc range needed so use a 48V 10W PSU and a bank of capacitors for storing energy. Caps will slowly charge up to \$ V_{max} = 48V \$ and during the pulse, they will be discharged down to \$ V_{min} = 35V \$. Energy needed in caps \$ \Delta Ε = 165mJ\$. The caps will give this energy while discharged from \$V_{max}\$ to \$V_{min}\$:

\$ \DeltaΕ = C\dfrac{(V_{max})^2}{2} − C\dfrac{(V_{min})^2}{2} \$

solve for C

\$C = 2 \Delta Ε((V_{max})^2−(V_{min})^2) = 300 \mu F \$

(330μF closest cap value)

If you use a bigger cap, the voltage will be more stable, plus a cap this size costs almost nothing. Remember to get one with at least 63V rating.

  • \$\begingroup\$ See my clarification. \$\endgroup\$ – mola Aug 14 '12 at 7:26
  • \$\begingroup\$ Where do you get the 50s from in your average power computation? \$\endgroup\$ – mola Aug 14 '12 at 7:27
  • \$\begingroup\$ From your answer it seems like my problem is not solved by choosing a specific PSU but rather by adding a charge reservoir to the circuit - this I would like to avoid. A PSU must have some sort of internal reservoir; otherwise a cap would always be needed, correct? If yes, is there a PSU parameter that describes the size of this charge reservoir? \$\endgroup\$ – mola Aug 14 '12 at 7:32
  • \$\begingroup\$ I doubt a 22 \$\mu\$F/63 V electrolytic will have an ESR and ESL low enough to get 1 A/\$\mu\$s rise times from. An X7R will do better on ESR, but I don't know about ESL. And they're expensive. \$\endgroup\$ – stevenvh Aug 14 '12 at 7:46
  • \$\begingroup\$ @stevenvh You are right, but I think it is enough to add a small film cap in parallel with electrolytic just to give the initial current. \$\endgroup\$ – Szymon Bęczkowski Aug 14 '12 at 8:13

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