Hi.
I'm not as circuit savvy as I thought I was.
I'm trying to replace the indicators on my motorcycle with LED jobs, but rather than buying off the shelf I want to build my own as I want to incorporate sidelights and brakelights in there as well.
What I am trying to do is wire three rows of five amber LEDs in each indicator on some vero board. The bikes voltage is around 14v (to a maximum of 15v). I've bought the LEDs and some resistors and I've experimented with some different ways of wiring them up but I can't seem to get the required brightness. I can get one LED at a time nice and bright but the others are dimmer. If someone can just show me the way to connect them for maximum efficiency I can figure out how to put in the other lights.
Thanks in advance.

 

do you mean something like this?
http://www.dutchforce.com/~eforum/index.ph...8&hl=motorcycle

-fire

 

hi

furnish us with the following info:

1. the LED you are using, is that a standard type or the ultra bright type

2. what is the lowest voltage status of your bike battery.

brightness of the LED will depend on the current w/c you will feed via a resistor. you can either wire it in parallel or series.

moz

 

Firestorm - yes and no. What I am trying to do is turn my front indicators into LED indicators and running lights, and turn the back indicators into LED indicators, tail lights and brake lights.

Moz - yes, the LEDs are ultra bright ones. The amber is 1000 - 2000mcd, the white and the red are both 2000 - 3000mcd.
They are the square ones to give a wider spread of light.
I want to extract as much light out of them as I can. I have enough space inside the housings to fit a row of 5, a row of 6, a row of 5, a row of 6 and a final row of 5 reading across from the top. I was thinking of having the three rows of five in the front as amber and the two rows of six as white. In the back I was thinking of having two rows of six amber, two rows of dimmer reds and one row of brighter reds.
I don't know what the lowest voltage of the battery would be. 0v I guess. It's a 12v system.

 

All I need to know really is the best way to wire 5 or 6 LEDs together, will they need individual resistors or just one, are they best in series or parallel and can someone draw a diagram to illustrate it for me?
Cheers.

 

The forward voltage drop across ultra-bright leds can be anything up to 3V at maximum current.
The battery voltage will range from roughly 12V off charge to 15V while the engine is running.

I'd aim for two LEDs plus a resistor in each series chain. Any more will start to show a drastic brightness change with battery voltage changes.

You don't say what the LED continuous forward current is, so I'll guess 30mA.
At that, and assuming 2.5V typical forward drop on each LED, the voltage across the resistor will be anything from 7V to 10V. 30mA @ 10V needs about 330 Ohms.
This will still give 21mA at 12V.

If the forward voltage drop or current is less, you need a higher value resistor.

The reason for only 2 leds per chain is the drastic change in current as the battery voltage changes. e.g. if you used 3 leds giving a 7.5V drop, the voltage left across the series resistor is 4.5V (at 12V) to 7.5V (at 15V).
Allowing 30mA at 7.5V needs about 250 Ohms, which only gives 18mA at 12V supply.
This many be just about acceptable, but the brightness change when the engine is started will be very noticable.

My advice is to use two leds plus a resistor in series for each section of the overall light, then use as many of these in parallel as required to make up the total array.
This should give reasonable consistent light & if any LED fails it will only affect one other in the array.

To work out the resistor value, look up the LED forward voltage & continuous current rating, which will probably be different for each colour LED.
Double the voltage & subtract from 15.
Divide this by the LED current rating in amps; e.g. if the LEDS had a forward voltage of 2.3V & current 24mA:
15 - (2.3 + 2.3) = 12.6 (Voltage across resistor)
12.6 / 0.024 = 525 (Resistance to give required current)

So, 525 Ohms would give 24mA on a 15V supply. Use the next higher standard value, 560 Ohms.
The resistor power would be 0.024 * 12.6 = 0.302 W so a minimum of a 0.5W resistor would be OK.

 

Originally posted by rjenkins@Jan 1 2006, 04:09 AM
The forward voltage drop across ultra-bright leds can be anything up to 3V at maximum current.
The battery voltage will range from roughly 12V off charge to 15V while the engine is running.

I'd aim for two LEDs plus a resistor in each series chain. Any more will start to show a drastic brightness change with battery voltage changes.

You don't say what the LED continuous forward current is, so I'll guess 30mA.
At that, and assuming 2.5V typical forward drop on each LED, the voltage across the resistor will be anything from 7V to 10V. 30mA @ 10V needs about 330 Ohms.
This will still give 21mA at 12V.

If the forward voltage drop or current is less, you need a higher value resistor.

The reason for only 2 leds per chain is the drastic change in current as the battery voltage changes. e.g. if you used 3 leds giving a 7.5V drop, the voltage left across the series resistor is 4.5V (at 12V) to 7.5V (at 15V).
Allowing 30mA at 7.5V needs about 250 Ohms, which only gives 18mA at 12V supply.
This many be just about acceptable, but the brightness change when the engine is started will be very noticable.

My advice is to use two leds plus a resistor in series for each section of the overall light, then use as many of these in parallel as required to make up the total array.
This should give reasonable consistent light & if any LED fails it will only affect one other in the array.

To work out the resistor value, look up the LED forward voltage & continuous current rating, which will probably be different for each colour LED.
Double the voltage & subtract from 15.
Divide this by the LED current rating in amps; e.g. if the LEDS had a forward voltage of 2.3V & current 24mA:
15 - (2.3 + 2.3) = 12.6 (Voltage across resistor)
12.6 / 0.024 = 525 (Resistance to give required current)

So, 525 Ohms would give 24mA on a 15V supply. Use the next higher standard value, 560 Ohms.
The resistor power would be 0.024 * 12.6 = 0.302 W so a minimum of a 0.5W resistor would be OK.

[post=12825]Quoted post[/post]​

Thanks Rjenkins, thats very useful.
Sorry It's taken me so long to reply, I've been offline for a while.
Cheers!