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I am designing a LED driver.

Power source - 18V-42V (I can draw max of 6mA-7mA from source) LED string - 4 series LEDs of 6 strings in parallel ( total 24 LEDs)

I need min of 200mA of peak/instantaneous current through LEDs with 20ms or 30ms ON every 1sec duration.

I am thinking to design like current sink method which will charge a big capacitor and I can draw current from it to drive LEDs.

Please let me know if anyone is having any different idea.

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    \$\begingroup\$ What forward voltage are the LEDs? \$\endgroup\$
    – Andy aka
    Commented Jan 9, 2020 at 12:08
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    \$\begingroup\$ Sounds reasonable. Work out the charge you need per pulse (dQ = amp-seconds = Coulombs) and decide the voltage sag you can tolerate on the capacitor to get its minimum value (C = dQ/dV). \$\endgroup\$
    – user16324
    Commented Jan 9, 2020 at 12:37
  • \$\begingroup\$ @Andyaka Vf= 2.65V to 3.5V \$\endgroup\$
    – Micro
    Commented Jan 9, 2020 at 14:07
  • \$\begingroup\$ @BrianDrummond Yes I am doing the same thing, but problem is I can not offer big size cap (except super cap) on my expected board size. \$\endgroup\$
    – Micro
    Commented Jan 9, 2020 at 14:09
  • \$\begingroup\$ Can you change power source to ensure it exceeds power drain? WHy ? 18V * 6mA = 48mW is inadequate \$\endgroup\$
    – D.A.S.
    Commented Jan 9, 2020 at 16:31

1 Answer 1

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Given:

Power source - 18V-42V (I can draw max of 6mA-7mA from source) LED string - 4 series LEDs of 6 strings in parallel ( total 24 LEDs) I need min of 200mA of peak/instantaneous current through LEDs with 20ms or 30ms ON every 1sec duration. Vf= 2.65V to 3.5V

Translation:

Input:

Vi = 30V nom. ±12V (±40%)
Ii = 6mA to 7mA (nom.)
Pi = 180 to 210 mW (±40%) (nom.) , 108mW (min.)
Ri = 30V/6mA= 5kΩ (nom.), 18V/6mA= 3kΩ min. (est. for MPT)

Output:

24 W. LEDs in Array , Pulsed at 1Hz rate 2~3% duty cycle

Vf= 3.1 V ± 0.4V (depends on quality of parts and ratings)
If= 200 mA, 20ms min. to 250 mA (est.), 30ms max.

LED Array = 4S6P white (CCT, CRI, LPW unknown)
Vo = Vf * 4 + Io * Rs
Io = 6 * 200 mA (min.) (Rs=0 if buying binned parts ±0.1V)
Po= Vo * Io = 12.4V (nom) * 1.2 A (min) ~ 15 W nom.
Po avg.= Po * duty cycle , d.c. = 2~3%

  • Po avg. = 300 mW (nom) = 15 W (nom) * 20ms/sec
    • d.c.= 2% min)

Output Power, Po avg exceeds input Pi: Problem.

Conclusion

  • change power source or requirements or efficacy of LEDs
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