Driving a 10.2v, 140mA LCD backlight from a 3.7v, 2000mAh battery?

Discussion in 'The Projects Forum' started by JustWood, Jun 16, 2010.

  1. JustWood

    Thread Starter New Member

    Jun 16, 2010
    Hello everyone,
    As I said in my title, I'm trying to drive a 10.2v, 140mA LCD backlight (LED array) from a 3.7v, 2000mAh battery. I have looked into several methods of doing this, and the one that stands out to me seems to be some kind of boost converter. I have never built a boost converter before, and was wondering if this is the best possible method.
    I currently have the LCD operating just fine, and am running the backlight from a 16v power supply through a 50ohm 5W resistor. The resistor is dissipating a lot of heat though, and I am not sure what the limitations of a boost converter are.

    Justin W
  2. SgtWookie


    Jul 17, 2007
    Let's see. 10.2v @140mA = 1.428 Watts output power needed.
    Let's say that you could build a boost converter that was 80% efficient; so 1.428/.8=1.785 Watts input.

    What does your battery voltage measure when it's discharged? I'll guess 3.3v.
    1.785w/3.3v = 541mA input current required. That's a pretty hefty load for a 2AH battery. You might get 2-3 hours of run-time out of it, if it's not powering anything else.

    If you want an intro to switch-mode power supplies, read the comments in this thread:
    and look at the links provided, particularly Ronald Dekker's page and smps.us. If you don't experiment with these things, you will have a really hard time understanding them.