# Using 12v 50w motorcycle light with two 7.4v RC batteries

I recently purchased two motorcycle lights (12v DC 50w) and only needed one. Now I have some 7.4v 4500mah 45c batteries leftover from my RC hobby a while back. I would like to create a 3d printed enclosure for these items, making a torch (with an old laptop heatsink).

I have read lots of google posts, however I still do not understand whether or not my LED will draw too much current if I purchase a step down transformer like this: https://www.banggood.com/DC-9-35V-to-DC-1V-35V-80W-Automatic-Step-Down-Module-Boost-Buck-CC-CV-Power-Converter-Module-Adjustable-Voltage-Regulator-p-1510115.html

I would like to know if the above mentioned method (using two batteries and stepping down the voltage) would work or if using a single battery and stepping up the voltage would also be possible (and what driver would be suitable for this).

Please note that these batteries are identical.

• 12V 50W is quite high for a LED. Are you sure it's not a standard incandescent lamp? – Natsu Kage Feb 2 at 11:13
• @NatsuKage Apparently they do exist globalsources.com/si/AS/Kimh-Electronic/6008850291553/pdtl/… though specs say 3.1A @ 12V DC 30W CREE LEDs, so maybe only 30W actual. Note this light is "Multi-volt operation: wide operating voltage range from 10 to 30V DC" so perhaps can be powered directly by 4S lipo! – Bruce Abbott Feb 2 at 12:27
• @BruceAbbott Thanks, although I knew they existed. I just doubt that they're really LEDs in this case. Most bulbs you can buy off shelf are incandescent. – Natsu Kage Feb 2 at 14:01
• Can you provide a bulb link? Is it really LED? Is it really 50W or ? – Russell McMahon Feb 2 at 21:43

This will allow a much improved answer.

IMPORTANT: If buck only, there will be a minimum Vin-Vout spec and at Vout=12V you may need Vin of > 12V. Worst case, if (Vout-Vin)min was 4V (unlikely) the driver would be unusable for your purpose. So, the IC matters.

LiIon or LiPo cells are rated as 3.6V nominal per cell, 3V min (higher is safer) and 4.2V max.
So 2 2S 7.2V batteries in series will; give 12 - 16.8V.
Without knowing the IC used it is not possible to be sure whether it is buck only or buck boost. The single inductor , single IC suggests buck only (based on other common Chinese modules) The two pots presumably set CCCV limits.
Buck-boost with a single switch and single inductor is possible with Vin to Vout polarity inversion.

IF correctly rate the converter has the power rating to do what you want.

The ability to set CV & CC (Vout max and I out max) makes the driver well suited to LED use.

Maximum dissipation in the driver can be expected to be ABOUT Power_in x 10-20% or about 5 - 10 Watts (maybe less)

A less good but instant solution:

You appear to be in New Zealand.
For "playing", the use of a series/parallel resistor array will allow you to set up a driver immediately using series resistor to the LED. This is less efficient than using the driver, dissipates much more battery energy as heat, but allows you to be experimenting within time-to-buy-from-Jaycar-NZ hours.

A resistor voltage dropper will reduce the supplied voltage as Vbattery drops, but will compensate somewhat by increasing current somewhat as VLED drops. Overall a series resistor is an OK solution for experimenting.

Jaycar NZ sell eg these resistors.
4.7 Ohm 10 Watt
1 Ohm 10 watt
1 Ohm 5 Watt $0.60 /$0./42 The 5 Watt resistors cost less per Watt than the 10 Watts - 60 cents NZ each or 42 cents with a "trade card". Buying 10 x 1 Ohm, 5W resistors allows playing (and lots of heat).

For Vin = 16.8V, Vout = 12V, Imax = 4A (48 Watts)
Use 16V. Rdropper = V/I = Vdrop/I = (16-12)/4 = 1 Ohms.
4 x 1 Ohm resistors in 2S2P arrangement = 1 Ohms.
More resistors in other arrangements allows you to experiment. All resistors dissipation = V x I = 4V x 4A = 16W = 4W/resistor.
As Vbattery drops current will fall

Worst case resistor efficiency = Vout/Vin_max = 12/16.8~+ 75%.
A converter may achieve 80-90% and maybe better. Maybe.

• I have a feeling the torchlight will become too hot to handle with that much loss through resistors. They want it made from a heatsink after all. xD – Natsu Kage Feb 2 at 14:11
• @NatsuKage Extra heat is undesirable. But a 50W LED will dissipate upwards of 50% of the input energy as heat (the balance goes out the front as light) so at 50W in you get 25W + dissipation and probably in the 35W range. The resistor dissipation is something you'd like to avoid, but it's not ever going to run cold. – Russell McMahon Feb 4 at 15:19
• Thank you for your response. I just came back from my school camp and will purchase the driver. I'll keep you posted on the results. – Sean McC. Feb 7 at 10:14
• @SeanMcC. Are you in NZ? – Russell McMahon Feb 7 at 11:55