help? voltage drop 16-12v to 4v

Question

Help appreciated! Is the attached voltage drop diagram correct? I need to convert a 16-12v battery for use with a 4v LED lamp. I've bought an led desk-lamp which was sold with a DC supply rated as follows:

4v 0.75A 3W.

I want to use it with a 16v battery (fully charged) whose voltage drops eg down to 12v over a few hours. IE, do I need to buy 1x16 ohm and 1 x 5.33 ohm resistor?

Thanks so much!

jw0752 · Accepted Answer

Hi John, If you get one of these you can convert the 12 volts to the 4 volts without wasting as much energy as the use of resistors would entail. This is basically what Don recommended but marketed as a finished product. You should also put a current limiting resistor in series with your LEDs. Using your numbers a 5 ohm resistor would limit your current to less than .75 A at 4 volts. Once you power the converter you will need to adjust the potentiometer so that the output is the 4 volts you require before hooking up the LEDs. http://www.ebay.com/itm/7-32V-to-0-8-28V-DC-DC-Boost-Buck-Converter-12A-Constant-Current-Volt-Regulator-/301484905256?pt=LH_DefaultDomain_2&hash=item4631e69328 John

D_Hersey · Answer

There are two approaches to take.

If you use a resistive ballast the calculation goes like this:

P=IV

3 = 4*I

correct, 750 mA.

The drop across the ballast would be 16V - 4V (supply - LED drop) 12V

V=I*R

12=0.75 * R

R = 12 * 4/3 = 16

P=I*I*R

=9W

get about a 16 ohm resistor rated for about 20 W. You put the led and ballast in series, than lay the string across the supply. Your schematic is wrong.

Also, this soln has a drawback. Light consumes 4W, ballast 9W. Efficiency is poor.

michaelwylie · Answer

Typically you would drive the LED with a circuit like so: where you set the voltage across the diode to 4 Volts, and set the current in the circuit to 750 mA. If you perform the calculations, the resistor will be 16 ohms. The issue here, as Don stated, is that a large portion of your power will be consumed by the 16 ohm resistor (P = I^2 * R = 9 W). If you are fine with that, then you're done. If you want the battery to last longer you'll require a different solution.