-1
\$\begingroup\$

I am hoping to power a single high power LED with a forward voltage of 1.4V and a forward current of 1.4A using a 5V lithium ion battery.

From what I understand, there are several approaches:

  • using a simple resistor, in series
  • use a step down switching regulator
  • build a constant current source using transistors

As I require the design to be as efficient as possible, I have ruled out the simple resistor in series.

My research suggests that building a constant current source using transistors would still result in quite a poor efficiency, so I have ruled this out.

I have come to the conclusion that a step down switching regulator would be the best solution as it can cope with the variable input voltage that would result from the lithium battery discharging, and is capable of high efficiencies whilst supporting PWM from a micro-controller.

I have found that the LT3477 can support input voltages of between 2.5-25V, and can output up to 3A. The datasheet for this device contains the following diagram:

Schematic

The above image shows that the LEDs experience a 300mA current; I was wondering how to modify the circuit to increase the current to 1.4A.

Any advice is greatly appreciated, as are any better ways of approaching this problem. I would preferably like to stick to using a 5V power brick, so I can easily power a Raspberry Pi at the same time. PWM is not essential. The LED has a wavelength of 940nm (infrared).

\$\endgroup\$
  • \$\begingroup\$ Li-ion batteries don't come in 5V quanta. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 5 '14 at 21:40
1
\$\begingroup\$

That part will work well for you. The equation given by Bruce is correct for setting the average LED current to 1.4A. This is probably the best way to do it. However, as noted in a comment to your original post, your battery voltage is probably not 5V. If you are generating the 5V bus with a switching regulator from a single-cell Li-Ion, then you will suffer the losses of both the LED driver and the upstream regulator. Efficiency would be improved by running the LED driver directly from the battery.

\$\endgroup\$
  • \$\begingroup\$ I plan on using the following battery pack: "EC TECHNOLOGY® New 22400mAh Portable". It is apparently a lithium ion battery that provides 5V for charging of mobile phones and tablets, so would this have a built in regulator? \$\endgroup\$ – user50774 Aug 6 '14 at 12:48
  • \$\begingroup\$ Yeah, it does. If you were dead set on improving your efficiency, you could try to open up the pack and tap directly into the battery. Chances are the internal regulator is at least 80-90% efficient, so you'll gain something but not that much. \$\endgroup\$ – user49628 Aug 6 '14 at 16:29
0
\$\begingroup\$

Rsense sets the current limit. Its voltage is regulated at 100mV, so to get 1.4A you need a resistance of 0.1V/1.4A = 0.07 Ohms.

\$\endgroup\$

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.