How to power two LEDs with a 1 Farad charged Super Capacitor as efficiently as possible?

I have came across a awesome super capacitor Flash LED Driver chip for cameras.

On page 10 of the LED driver datasheet, it says:

How long the LED current is regulated depends on the initial CAP voltage, capacitor value, the charge current, LED forward voltage and the LED torch current setting

My questions are:

How long can two SVL 5630 , 2.8v, 200mA, LEDs remain on when driven with an On Semiconductor, 4 Amp Super Capacitor, flash LED Driver powered by a 1 Farad super capacitor charged to 3V?

Is there a more efficient way to drive these LEDs?

• Isn't the point of a camera flash to pump as much light as you can in a short amount of time? – pipe Mar 12 '17 at 9:18
• define "basic components"; what you need seems to be a switch mode power supply. I'd say ICs that fulfill that role are "basic building blocks of modern circuit design", but that's my opinion, and if I'm going to write this as an answer, chances are you won't accept it because your magical fairy that defines "basic components" doesn't agree with me. So, define things, or your question remains unclear. also, maybe a bit of an attempt of your own would do your question good. – Marcus Müller Mar 12 '17 at 11:17
• Sounds like a good candidate for a Joule Thief. – JvO Mar 12 '17 at 12:57

Less than 8 Seconds

calculations below

THE DRIVER

LED DRIVER FEATURES (Datasheet page 1)

• 2 Channels at 2 A Each in Flash Mode
• 2 Channels at 200 mA Each in Torch Mode

LED DRIVER RECOMMENDED OPERATING CONDITIONS (datasheet page 2)

• LEDA, LEDB current (in flash mode) Up to 2 Amp
• LEDA, LEDB current (in torch mode) 10 to 200mA

The LED driver's datasheet examples use two 0.55 Farad Super Caps.

What makes the flash driver so awesome is how it charges, from a 2-5.5V battery, and balances 2 super caps, and gets the flash ready quickly for the next flash.

The LED driver charges two super caps not one.

The LEDs

You are not powering a flash LED, so a boost regulator would do the job.

Given what the datasheet says and the LED and Super Cap you selected you want to light up two LEDs in torch mode.

The LED you selected is not a Camera Flash LED.

The LED datasheet says the applications are:

• Office lighting
• Retrofit lamps
• Interior automotive lighting
• Backlighting

At 200mA max the LEDs will output about 100 lumens each.

If I assume correctly the number of LEDs does not matter. You just want about 100-200 lumens for as long as a 1F Super Cap will allow.

The Super Cap

You are using one 1 farad super capacitor.

A single digit Farad supercap is used for quick charge discharge, not a storage cell.

A supercap energy cell is in the thousands of Farads.

As an energy cell a supercap is used to bridge power gaps lasting from a few seconds to a few minutes. Like to supply energy between power failure and backup generator starting.

Unless that awesome chip came with some free super caps I doubt you will want to use a supercap in place of a battery. The specific energy (capacity) of a supercap is 10-50 times less than a Li-ion battery. And they cost a lot more than a battery.

Assuming that I have a 1F/5V supercapacitor charged to 3V,

Because the cap you specified is 5V, it is not a super cap. But that does not matter because you are charging it to 3.0v, so it's just a chemistry technicality. Some of the most recent super caps are now 3.0v.

Max super cap cell voltage (of reasonably priced and lower capacity) is 2.75V, typically 2.3 to 2.6V max, which is less than the minimum forward voltage of your 2.8 LED.

Conclusion

Given your selection of driver and non-flash LED it appears you want some light generated by LED(s) powered by a Super Cap.

It appears you do NOT need the features of the flash driver other than to charge the Super Cap. And that may not be necessary.

The driver only lights up one LED at a time, and charges one cap at a time.

The 8 seconds still applies to the selected LED driver.

Calculations

Single LED

Driven by Super Cap and Resistor

Given the assumptions, the number of LEDs does not matter. The selected driver charges and discharges one LED at a time.

Let's say you power a single LED with a 2V forward voltage max. The LED you picked has a current of 200mA max. To power it directly with a resistor, at 200mA, 3V you need a 5.6Ω resistor. 1F x 5.6Ω = 5.6 seconds TC.

But the cap will not discharge like a battery.

With 400mV across the resistor (200mA x 5.6Ω) and 2v across the LED the cap voltage can drop from 3 to 2.6 or 600mV before the LED stops conducting.

600mV/3.0 = 0.2
0.2v x 5.6 seconds =1.12

It will take about 1.12 seconds to drop below the 2V forward voltage.

Why We Use an LED Driver, Not a Resistor

This applies just as well to battery power as it does a Super Cap.

The LED driver uses a switching regulator to driver the LED from the Super Cap. So we just need to calculate the energy provided by the cap, consumed by the LED, and how long it takes to discharge the cap at 200mA.

Energy capacity of SC: 1F* 3v2 / 2 = 4.5 Joules

The LED 2.8v * 0.2A = 0.560V

4.5 J / .560 = 8 seconds

With the selected LED driver, in torch mode, it would be 4 seconds for LEDA and 4 seconds for LEDB (withn two 0.5F caps) which could be lit up consecutively giving 8 seconds.

The above assumes an ideal LED driver with 100% efficiency. So it will be less than 8 seconds.

To power the LEDs with a 3.6v 1000mA hr capacity camera battery would power the LED for ____?

You now have the formulas to calculate that for your self. Hint: Much much longer. You can now also calculate the flash discharge if you desire.

If it is true you just want to light up some LEDs with a super cap now you can see why it makes no sense to charge a super cap with a battery unless you need the short (<400mS) 2A flash discharge.

Why the Driver does not give you the option to power the LEDs with the battery, I do not know. That's what I may have done if it were my project.

• Great, very thorough answer! – metacollin Mar 13 '17 at 6:47

How long can two ...

assuming that the leds (in parallel) are not lit when voltage drops below 2.8v, and assuming a 100% efficient driver. The charges drained from the cap going from 3v to 2.8v is 0.2v * 1f.

the charges going through the leds is 0.4amp * t.

The two must be the same.

so t = 0.5s, shorter with inefficient drivers. you can confirm that easily with a simulation.