Resisters for LEDs_Parralle vs Series

[DLMatthys]

In a lot of circuits that I build for my Starships I've always used LEDs in parralle with a resister soldured on each LEDs positive side. This has always worked for me without failure since I have forgone the use of all incadesant filliment lamps since the turn of the century.

I'm reading up all the time about LEDs and circuits to learn more than I did in High School electronics the more knowleged writers and experts say not to use LEDs parralle, but rather in series. I'll not argue that since I am not an engineer versed in theory but yet quite experainced with the hot iron. One thing I do know about series circuits is that if one fails the whole series string fails. So I stay clear of series circuit hookups.

So I've tried the calculators to get my series resistance factor...and I even did the math on paper...It aint working out if I'm intending to use more than one LED of the same kind and rating. Since I always have used one resister on each one LED in paralle. I want to try it in series instead. One resister to drop the power for multiple LEDs in series.

Any ideas what I'm doing wrong? Help me sort this out.

Here is the JAVA script calculators that I use the most:

http://ourworld.compuserve.com/homepage ... en/led.htm
http://www.ledsupply.com/reca.php

What value should I use in the space titled LED Voltage when using more than one LED in series? (I've aways entered the forward voltage value for the one LED; ie 3.6 volts)

What value should I use in the space Total Voltage?
I always input that value for the Wall Wart transformer I use. (output set at 6 volts 700ma)

When I enter more than one LED (1) in the space Total LEDs the calculations do not work. The JAVA script error says "Total voltage must be > (greater) sum of LED voltages" If I used a 9 volt Wall Wart instead...I get a value for a series resister. Higher volts...I do not want to use.

Final question. What happens when I use one (1) Resister in series soldured to the positive of two (2) LEDs wired in parralle? I know already it works without a problem, done it for years and years but what happens to the circuit current in that configuration (in theory)?

[Sparky]

Final question. What happens when I use one (1) Resister in series soldured to the positive of two (2) LEDs wired in parralle? I know already it works without a problem, done it for years and years but what happens to the circuit current in that configuration (in theory)?

In a parallel circuit with one resistore the sum of the current for each member of the parallel branches is passing through the resistor. If you have 3 LEDs that each need 20mA the resistor has to pass 60mA.

This means the resistors wattage limit needs to be considered when using one resistor. Also if the resistor fails all the LEDs will go out. You have a series setup there even though all the LEDs are in parallel, their circuits all have to go through the resistor. The resistor is the bottel neck, and the point of failure which will cause all LEDs to go out.

If you have some LEDs going on and off (switchable say) a single resistor feeding the switchable LED and the non-switched LEDs will not be dropping the right amount of power for the circuit. That's a good reason to have one resistor per LED, or better yet use a regulator to just provide the right current/voltage for the LEDs.

[Sparky]

Here is the JAVA script calculators that I use the most:

http://ourworld.compuserve.com/homepage ... en/led.htm
http://www.ledsupply.com/reca.php

What value should I use in the space titled LED Voltage when using more than one LED in series? (I've aways entered the forward voltage value for the one LED; ie 3.6 volts)

What value should I use in the space Total Voltage?
I always input that value for the Wall Wart transformer I use. (output set at 6 volts 700ma)

When I enter more than one LED (1) in the space Total LEDs the calculations do not work. The JAVA script error says "Total voltage must be > (greater) sum of LED voltages" If I used a 9 volt Wall Wart instead...I get a value for a series resister. Higher volts...I do not want to use.

This gets into real basic Ohms law, Kirchhoff's current law, series circuits, and parallel circuits.

Before people start jumping in with the, diodes are not linear acting components. Please realize that the JAVA script programs and the hand method all assume steady state conditions. They won't work if the LED starts changing its operating voltages and currents, and worse if the power supply value you input starts fluctuating.

So continue with the simple assumption that the LED values and power supply values you have will be until the circuit is thrown out.

The first rule to think about is; in a series circuit, voltages add and current remains constant.
So the current flowing through the series loop, has to be the same for each component. If you have LEDs they'd better all want the same current.

Current can't be different for each item in the chain. Kirchhoff's Current Law.

The voltage given by the power supply will be divided up amongst the individual components.

So if you have 2 LEDs that want 3.6 volts at xx mA each, the LEDs in series without a resistor will want 7.2 volts @ xx mAs. The voltages have added, the current has not. If you add one more LED the whole chain will need 10.2 volts.

This blows past your 9-volt power supply's, voltage limit.

Now many people point out that wall warts measure a higher output voltage than their listed output. You need to check this under a load. The dropping resistor will not, absolutely not, protect your LEDs if the power supply suddenly starts putting out more voltage than you used in the calculations.

Many power supplies will show a higher voltage when there rated, with no load present, as soon as you hook up something, the voltage drops to the rated level. I suggest hooking up some LEDs with a resistor as would be needed if the power supply was working at its rated level, then measure the voltage coming out of the wall-wart. This will let you see how the wall-wart is really running under the loaded condition.

The end result from the initial analysis though is that with the high voltage requirements of blue and white LEDs, 3.6 volts. you will need a higher voltage power supply if you want to use a series circuit, but the current rating will not need to be greater.

The main difference to the end user in using parallel vs. series:
Parallel circuits are more robust when considering component failures.
Parallel circuits will want more current less voltage,
Series circuits will want more voltage less current, a current driver regulator will work great on this circuit improving efficiency (works to keep the LEDs constant brightness even as a battery's votlage drops).

Power (voltage*current) will remain the same for either configuration, no free power or perpetual energy systems here.

[DLMatthys]

Now I totally get it. You're the right man with the right answers Sparky!

I never had problems with one resiter for two LEDs parralle. Never any heat nor malfuntion with 5mm red 20ma LEDs used in Trek bussard collectors. The one resister saves a lot of space.

Thanks. I'll be staying on the parralle course!


Update: Tried the calculator with the sum of current draw of current of 2 LEDs. (20ma + 20ma = 40ma) Results all good.