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Discussion Starter · #1 ·
I have recently built a lap timer with the following components;

RadioShack 276-0145 - Phototransistor x2
RadioShack 276-0143 - Infrared LED x4
Maplin UG01B - Multi Voltage Regulated AC/DC Adaptor

The Phototransistors work great and are powered directly from the lap timer but the LED's keep dying on me.

Currently the LED's are wired in series and powered by the adaptr set to 9V (I get nothing from 6V).

What ohm resister do I need and how should it be wired??
 

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Your wiring system is probably the main ingredient in your problem.

NEVER wire LEDs in series, I don't care what anyone says, the load will be spread un-evenly amongst them.
Then, as soon as one blows; if it goes closed circuit, the same resistive load is spread across one less LED . . snowball effect.

Wire them in parallel with a resistor for each.
The shift between 6 and 9 volts should not make much difference except brightness, certainly not be working at 9, and yet nothing at 6 volts. They should just be dimmer. Do you watch/test while looking through a digital camera?

So, check what current draw the LEDs are meant to have; and wire them individually, in parallel, with a resistor to each, put the resistor on the positive side of the circuit.

Here's a quick spreadsheet for calculating resistor value.
Spreadsheet for LED wiring

As example of circuit layout, here is one end of the circuit board layout of my bridge.


If you need for example 20ma at 9 volts, type in 9 as supply voltage, LEDs are almost all 1.2 volt, type in 20 for the LED current rating, and a value of 390 ohms pops up.
The beam is typically fairly narrow, so be sure the LEDs are shining directly at your sensors.

I use a 5mm chinese infra-red LED rated at 20ma on my bars, but drive them at about 23ma, using a 12 volt supply and 470 ohm resistors, and they are bog solid.
 

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With due respect you can wire LEDs in series, just like resistors and the same current must pass through each. This does not mean they will produce the same output but that is another story. The rest I agree with.

You MUST use a resistor in series with the LEDs. 1.2V (I suspect the voltage drop may be a little higher but not to worry, 20mA might be a bit high in any case) drop across each LED = 2.4. If you start with 9V then 9 - 2.4 = 6.6V

You want about 20mA so R = V/I = 6.6/0.02 = 330 ohms.

cheers
rick1776
 

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Rick, you are just plain wrong, You don't "MUST" wire them in series at all.

Wiring LEDs in series is a ****e practice, and every electronics tech. I talked to before I did my light board said the same thing.

I have built a 3 figure number of this circuit and had one LED fail in about 4 years. Not bad for over 500 LEDs in use.

That LED part number has a tech spec. that says it's forward voltage drop is 1.28, but users have commented that it is actually 1.2
It also has a current capacity of 100ma, but that figure is generally based on pulse/short use. For hours of use like we do in slot racing, I recommend sticking to the 20-25ma range, and that is plenty to drive the matching transistor at a distance of around 100 - 125mm.

Phil, I said IF, and the converse is also true. If any LED fails and goes open circuit when you wire in series, the whole bar goes out, and then you have to take them out of circuit and test every one on a resistance one at a time, a very tricky task if you have them soldered into a circuit board.

Freewheeler, stick to what I said and you will be just fine.
 

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slots,

get the fur down, where did I say you you MUST wire LEDs in series??? I said you must wire a resistor in series with the LED, please reread the post. In no way did I say you were wrong. In fact I prefer to wire in parallel with each LED having its own limiting resistor. You can however wire them in series as long as you have enough drive voltage to overcome the pn junction losses for each LED, which as you have stated off the data specs is 1.28V. So a volatge of 9 volts will easily accommodate two LEDS in series.

So 9 - 1.28 -1.28 = 6.44

R = 6.44/0.02 = 322. Or like I said before 330 will be OK.

By all means wire them in parallel or series, which ever is easier, or fits in the best if room is tight. Both will work. Honest *****.

cheers
rick1776
 

· Rich Dumas
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Before you do anything else make sure that you know what the voltage from the power supply really is. The operating voltage will probably be at least a little less than the open circuit voltage. I once bought what was supposed to be a 12 volt regulated power supply to run some LEDs and found that it was actually 13.8 volts. If I had done the calculation based on 12 volts the LEDs would probably have burned out at once.
 

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Discussion Starter · #8 ·
Many thanks gents for all your input
we have done a bit more maths and came up with a answer 78 ohms , so we used 82 ohms which did not work ,
we then went to 75 ohms which seems to work OK and 68 ohms was brighter still , so in the end we have
settled on the 75 ohms for the time being and see how we get on.
.....just say thanks again for the reponse ever little bit of info helps
cheers gents..
 

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Just as a point of interest I use the 5 volt supply from my laptop USB ports to power my LEDs with a series resistor. It's easy to butcher a USB cable and obvious which is the plus and minus. They tend to use red and black.

 

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How many LEDs are you running in series? It would seem like 5 or 6 with a dropping resistor of only 75 ohms. Just occurred to me, are you running AC or DC9V?? I have a nagging suspicion you may be using an AC supply. Which weird as it sounds will work....just not that will. Check the power wall wart. Does it say AC9V or DC9V.

cheers
rick1776
 

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I would tend to agree with Slots, wire LEDs in parallel with individual ballast resistors. that is how I was taught (30+ years ago) as an electronics technician and still work to.
It is a case of what is best practice v. what you 'can' do.
 
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