0
\$\begingroup\$

I'm very much a beginner at creating my own circuits and using an Arduino so please excuse any stupid mistakes and oversights.

First of all here are the components I want to use in the circuit: Power Bank, LED, Capacitor, Solid State Relay.

What I want to do is light a very high power LED for 1/2 second or so using a capacitor, a boost converter (from 5v to 20v), and the innards of a power bank. I want to use two solid state relays in the circuit. The first one will close and allow the capacitor to charge and then when the capacitor has full charge the Arduino will close the second one and cause the light to flash brightly for a split second. After reading a few articles about powering LEDs from a capacitor it seems I need a resistor in the circuit but I'm not sure what size.

Also, I'm using these two online calculators to get an approximate time that the LED would stay lit for:

https://www.electronics2000.co.uk/calc/capacitor-charge-calculator.php https://www.rapidtables.com/calc/electric/Joule_to_Watt_Calculator.html

  • I'm giving the first one the values of 20 volts and 33,000 uF
  • On the second if I enter 6.6J and 0.5 seconds I end up with 13.2 watts. My LED draws 750ma at 17v so 0.750 x 17 = 12.75 watts

Am I right in thinking then that the 33,000 uF cap should power the LED for 1/2 second when fully charged?

Here is my wiring diagram: enter image description here

\$\endgroup\$
  • \$\begingroup\$ You would be better wiring up a current limiter than using a resistor. \$\endgroup\$ – Trevor_G Feb 8 '18 at 19:26
  • \$\begingroup\$ ALso it depends on your definition "power the led"... it wont stay above 17V long. \$\endgroup\$ – Trevor_G Feb 8 '18 at 19:36
  • \$\begingroup\$ It is better to start with what you want to see "specs" (no pun intended). single or strobe Flash? Duration (variable?), current, Lumens? then choose a logic level NFET to switch low side rated for >5x the current or use LED boost voltage CC driver with enable. This is the better way. Then no need for huge caps. \$\endgroup\$ – Sunnyskyguy EE75 Feb 8 '18 at 20:26
  • \$\begingroup\$ Yes, the on-time and on-off operations should be monitored from the Arduino (that;s what it's for). Output connected to the gate of the N-MOSFET (no need for optocoupler but an optocoupler won't hurt) and you need only one MOSFET or relay. The boost convertor is charging the cap while the led is off. The cap could be IMO, 1000uF but not as huge as 33000uF. You don't need resistor X. \$\endgroup\$ – Fredled Feb 8 '18 at 21:00
  • \$\begingroup\$ @Tony Stewart The effect I want is what the capacitor would give me. I want a huge flash of light that slowly gets dimmer until it goes out (not a strobe, just a flash every few minutes). I want a majority of the brightness to last for about 1/2 second and the amount of time it takes to diminish doesn't matter. I don't know what a NFET is or where to buy one. Could you link to one for more clarification? Also is this the type of LED driver your referring to? linear.com/product/LT3952 It looks a bit complicated to wire to me, like I said, I am very much a novice :) \$\endgroup\$ – Jake L. Feb 8 '18 at 21:38
0
\$\begingroup\$

In general for calculating the resistance needed for to protect your LED you can use this formula:

$$ R = \frac{(V_{source} - V_{LED})}{I_{MAX}} $$

In your case, your source for the LED is the capacitor, which fully charged will be at 20V. The voltage drop across your LED is 17V, and its max current is 750 mA, so

$$ R = \frac{20 V - 17 V}{750 mA} = 4\Omega $$

Practically, your resistor should be slightly larger because your boost converter may charge the capacitor to slightly over 20 V or your LED may not quite handle 750 mA. Also, your resistor needs to dissipate

$$ P = R*I^2 = 4\Omega * (0.75A)^2 = 2.25 W $$

which is significant.

\$\endgroup\$
  • \$\begingroup\$ Jake. Make sure you understand the impact of Page 14,24 & in Cree spec and tolerances for Vf and affects of above assumptions where Vled is clearly not constants is tolerance and temp. ~3V swing for temp and tolerance easy \$\endgroup\$ – Sunnyskyguy EE75 Feb 9 '18 at 0:06
  • \$\begingroup\$ So basically as the voltage to the LED gets higher the temperature will increase and so will the amp draw right? Will the temperature be that great of a factor since I'm only lighting the LED for a fraction of a second? \$\endgroup\$ – Jake L. Feb 9 '18 at 0:25
0
\$\begingroup\$

Doing it your way requires some careful calculations.

re-calc and assumptions

  • RC=T decay is defined by 63% drop (1-1/e)
  • wheras drop 20V to 16V (dim) = only 25% drop in cap voltage,
  • so 63%/25% * T is needed for T=0.5s burst to near half power at 16V
  • meaning 2.5T = RC
  • while R=ESR + Rs, if Rs= 4 ohms and LED ESR ~2 ohms @0.75A, R=2+4=6 ohms

So don't do it this way, because.

  • C= 2.5T/6 = 210 mF
\$\endgroup\$
  • \$\begingroup\$ There is obviously a much better way to do this than a big $100 cap, but not this way. \$\endgroup\$ – Sunnyskyguy EE75 Feb 8 '18 at 23:35
  • \$\begingroup\$ Also unless you have some sort of thermal design, any open loop method can lead to thermal runaway with rising currents with rising temps due to Vf drop \$\endgroup\$ – Sunnyskyguy EE75 Feb 9 '18 at 0:04
  • \$\begingroup\$ I have a 100W 12V LED that works great off an ATX supply with a CPU heatsink. But I would suggest a >25A FET switch with pulse control to the gate with a 1nF decay cap. using a surplus 19V laptop charger. (65W) \$\endgroup\$ – Sunnyskyguy EE75 Feb 9 '18 at 0:15
  • \$\begingroup\$ Hmm...Using a laptop battery is a good idea. However I need it to fit in a small space. A circular hole 1 1/4" in diameter. And I would also need the battery to power the Arduino and the 5v power bank is so convenient for that :) \$\endgroup\$ – Jake L. Feb 9 '18 at 0:30
  • \$\begingroup\$ So what your saying in your above answer is that if I used a capacitor it would have to be really really big to get that 0.5 second bright burst? It would still work though right? It just wouldn't last very long. \$\endgroup\$ – Jake L. Feb 9 '18 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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