# Motorcycle LEDs

#### Defynormal

Joined May 8, 2022
1
Hi all.

I've converted all lights on my motorcycle to LEDs, but I don't see the point in power being wasted with the resistors currently wired in. So my question is:

Can I use a voltage reducer in the circuit? Or will that just increase the amps leaving the watts output to the LEDs the same?

#### ElectricSpidey

Joined Dec 2, 2017
2,146
The resistor is the voltage reducer.

Assuming all of the LEDs are different sets, you would need to use an efficient DC to DC constant current module to drive each set of LEDs instead of the resistor.

Not really a great option unless the wasted wattage is a problem...other than in your mind.

And if the LEDs in any set are in parallel...then things get even more complicated.

#### DickCappels

Joined Aug 21, 2008
8,714
In Taipei many years ago I noticed that when cars stopped at red lights at night, almost every one turned off their headlights and only left parking lights on until the light turned green. My friend explained that this was to save gas. Gas was cheap at the time, but these days, maybe it is worth some measures to reduce wasted power. I doubt it would amount to more than pennies per month, but it is an option.

You can make a switching type current source to reduce the power lost in the resistor. One thing to consider is that resistors, if chosen correctly are much more reliable than switching power supply circuits. Have you ever driven ia motorbike n city traffic at night, in the raid, without lights? I can't recommend it.

The one above is probably good for two or three watts. I run it at 800 mW because I want to conserve battery life.

Start the analysis by imagining that Q3's gate is held at battery voltage and that Q3 is conducting current through L1, the LED, and R1. Whenever the current through R1 is sufficient to cause a high enough voltage drop across R1 to cause significant base current into the base of Q2, Q2 turns on, thereby turning off Q1, which in turn results in Q3 turning off. The hysteresis provided by the positive feedback through R3 to the base of Q2 assures that Q2 remains on until the current through R1 decreases to below the trip point. Current, decreasing with time, continues to flow through L1, via D1 and current, diminishing with time, continues to be supplied by C1 to R1 as well results in the current through R1 gradually decreasing. When it has decreased sufficiently for Q2 to turn off (taking hysteresis into account), Q2 does turn off, thereby turning on Q1, which in turn turns on Q3. When Q3 turns on, current again begins to increase through L1, the LED, and R1. The threshold at which Q2 turns on is now higher because of the hysteresis from R3, and when that threshold is attained, Q2 switches off and the cycle begins anew.

The fact that the loop is closed around LED current is significant. The light output of an LED is nearly linearly proportional to current and is specified in data sheets, while light output as a function of voltage is fairly nonlinear and is not a controlled parameter. It should be clear that the LED current is equal to the base-emitter voltage of Q2 divided by R1. In the case of a 2.2 ohm resistor, this came out to 0.6V/2.2 ohms = 272 ma, which is close to what was observed. See Figure 2, below. The base-emitter voltage of Q2 reduces approximately 1.8 millivolts for each degree of temperature rise. This "thermal drift" is wholly acceptable as a trade off to obtain the simplicity and economy of the circuit. Yes, the circuit could be made to have very low drift at the cost of increased complexity, but a drift of 16 millliamps over temperature will not be noticed visually.

Another switching constant current source was made to charge 100F capacitors but it can be used to drive LEDs instead.

The circuit above can source 500 milliamps. Current is controlled by the current sense resistor between pins 6 and 7 of the IC (0.2 ohms for 500 ma).

Of course you can also use a switching type voltage source and a small resistor in series with the output to power the LED.

Joined Jan 15, 2015
6,593
My bike is a somewhat retro 1992 Harley Davidson Electra Glide. I replaced the headlight, side spots and tail/brake lights with LEDs. Using LED lights for turn signals requires using parallel resistors with the LED lamps. The reason being the turn signal control module is designed around a specific current load. The reality here is using LED turn signals does nothing to save power. Since the power used for turn signals amounts to really nothing I still run incandescent bulbs for my turn signals. Again, the resistors merely add load so the turn signal module sees the correct load. The resistors are typically about 6 Ohm 10 Watt. One for the Right and one for the left sides. Other than turn signals none of the other LED lights require changing anything. They are just 12 Volt LED lighting designed for motorcycle use. Most of your question depends on motorcycle and how it is configured?

Ron

#### Rich2

Joined Mar 3, 2014
241
Without the load resistor (or bulb) they will usually stay lit, so all you have to do is fit an astable 555 circuit on each side of the bike with maybe a power transistor switching the LEDs.

Or use the 12v feed from the indicator switch to power the astable (probably better)