Adding LED Headlights to a Riding Lawnmower

Setting Sun

ledAs the days get shorter it is harder to get all the yard work done before the sun sets, leaving me with a half finished lawn. Once we roll back an hour with daylight saving time, it is almost impossible to get anything done after work. The ‘lights’ on my lawnmower are nothing more than a couple 1156 bulbs behind a plastic cover that blocks most of the light; these lights are not effective in dim or dark conditions, hence my need for a better lighting solution.

If you are looking to add some lights to your lawnmower please read on, perhaps my project can be of some help to you.

The Project

The first thing I needed to do is determine the type of electrical system my mower uses. I have a Briggs and Stratton engine so the resources I used were:

https://www.briggsandstratton.com/~/media/Frequently%20Asked%20Questions/Engine/PDFs/alternator_specifications.pdf

and

http://www4.briggsandstratton.com/miscpdfs/RNT/alternator_id.pdf

Using these resources I was able to determine I had a dual circuit charging system, with 12 VDC output to charge the battery and unregulated AC power for the headlights. This complicated matters slightly because now I could not simply connect my new light to the existing circuit; any automotive LED light would require 12VDC not 14VAC, but more on that later.

The information on the current output of the AC lighting coil was not published by Briggs and Stratton so I needed to use some reverse engineering to get a safe maximum wattage. Based on the two 1156 bulbs (which each consume around 27 watts @ 12VDC) I expected the circuit to be able to supply around 50W but because they were driving the bulbs with unregulated AC power, I needed to take some measurements.

Basic Circuits

To determine the power consumed by a single bulb I needed the voltage across the bulb and the current through it. This required the following two measurements:

Measuring voltage and current consumed by 1156 bulb.
Measuring voltage and current consumed by 1156 bulb. iR is your bulb.

I recorded 14VAC across the bulb and 1.54 A current through it using a standard hand-held multimeter. This means the power consumed is :

P = VI = 14V(1.54A) = 26W\

We have two of these bulbs so that works out to 43.12W of power available.

Shopping for Lights

I went looking for an LED light that requires 43W or less, I found this light on eBay that requires 18W and was reasonably priced.

LED_lights

Back to the AC voltage issue……

The LED light is designed to run on DC not AC so we need to convert it. There are a number of ways to do this but the most cost effective and practical would be to use a full bridge rectifier which looks like this.

full-bridge-rectifier

I am not going to explain how a full bridge rectifier works in detail because there are plenty of resources on the web for this. Here is a link to a Wikipedia article if you really want to know:

http://en.wikipedia.org/wiki/Rectifier

What you will be mainly concerned about is the current capacity of the diodes in the bridge. I selected my diodes based on what I had laying around in the lab; I happened to have a whole box of FR304 Diodes that can handle 3A which works out to be P = VI = 12V(3A) = 36W\ . If you need to buy some diodes I would recommend Digikey , Mouser , or Ebay which all have  a great selection. You are looking for a rectifier diode, the forward current capacity is the primary specification you are shopping by and the second point to consider is forward voltage drop which you should keep as low as possible. If you want to order something go with the MUR420G it is cheap, readily available and handles 4A.

**NOTE – You could also order a bridge rectifier as single device (GBU4A) it is cheaper than buying 4 diodes and saves you soldering.

The output from the rectifier will look something like this:

voltage rectification

This will work for your lights but the inclusion of a capacitor across the output will make the voltage much closer to constant DC. You can go with almost any electrolytic capacitor that is 4700 uF or above, they are easy to find at the same locations as mentioned above.

So this is the final circuit you will be building :

full bridge with cap

Test and Install

This circuit will effectively convert the AC voltage from your lawnmowers lighting coil into a DC voltage that your LED lights require. Once you have the circuit all built and installed you should test the output to ensure it is delivering DC. I used an oscilloscope and signal generator to test my circuit but any multimeter will work as well. You want to connect the circuit to your lighting coil, then measure for AC voltage. There should be less than 1V AC, if not you have wired something wrong. Next measure DC voltage, here you should see between 11VDC and 18VDC (depending on engine speed). If everything checks out, attach your LED light(s) and mow away.

James Ciesla

James Ciesla is an IT professional who specializes in data management, data analytics, and IT strategy.

6 thoughts on “Adding LED Headlights to a Riding Lawnmower

  1. Did this conversion using a GBU 8A 50v bridge rectifier,a 4700 25v 85 degree electrolytic capacitor and 2 x 1156 68 SMD led bulbs and am de”lighted” by the results. Is it advisable to mount the rectifier on a heay sink or is the heat generated not an issue? Thank you for a very informative and immensely helpful article.

    1. I am glad to hear it worked out well for you! The LED bulbs you used sound like they are quite powerful!

      As for the heat sink, the preferable method would be to mount the rectifier directly to the lawnmower frame. Due to the size of the steel used on the frame it could be considered an infinite sink. If that is not an option due to the rectifier case design or space considerations you could go without a heat sink, bridge rectifiers are pretty durable components thermally.

  2. Great idea; I have the same system lawnmower and wanted to improve the lighting also. I have only ONE blue wire going to the lights; so how exactly would you wire in the new full bridge rectifier? Thank you.

    1. On the input side to the rectifier you would take the blue wire and connect it to the rectifier input side (it does not matter which terminal in the diagrams above). The second input will come from the lawnmower frame.

      On the output side the polarity matters because it is now DC. So you would find the +12V output from your rectifier (use a multi-meter if you are having trouble) and connect it to your new lights. The black ‘negative’ or ‘ground’ wire for your lights will be connected to the lawnmower frame. I hope this helps, good luck!

  3. Could you dump the output from the rectifier into the battery (along with the normal charging circuit) then wire the lights off the battery?

    1. I would not recommend connecting the rectified output from the lighting coil directly to the battery. The output of the lighting coil varies with engine RPM so if you were to run the engine at full speed (which you should while mowing) you would be dumping 18V or more into the battery which could shorten it’s life. The best setup would be to connect the lights to the battery and use a voltage regulator (boost-buck style) that not only converts the lighting coil’s AC into DC but also regulates the output/charging voltage to be between 12-14V at all RPM. That is a little beyond what I am proposing in this article but it would deliver superior results.

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