Simple Connecting Series LEDs to Parallel

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That's 8 LEDs I need 7 also I can't modify the 6 already connected in series they're part of a display :)

run a parallel set of 6, with 5 painted black.

Need to know the current through the 6 LEDs.

Do you have a manufacturer's name and part number for the LEDs? The data sheet would allow you to calculate the resistance and current, given a 2 V drop across the LED. Then you could calculate the resistance need to give you an 10 V drop across the resistor with the same current as is going through the 6 LEDs.

Another option is to use a series of zener diodes instead of a resistor.

Thank u :) unfortunately I don't have any details of the LEDs they are in a display so I don't have much access to them I can only measure the voltage going into them it varies from 8 to 14v I think. I added a 10k resistor this works however if u lower the brightness of the display then all the LEDs go off and you can't see the display at all only the 7th led gets power, I wanted them all to be equal if possible. The 6 LEDs originally connected in serial are connected as they are no resistors, etc

So try my 2nd suggestion. Use a series of zener diodes with the 7th LED.

-D-D-D-D-D-D-

-Z-Z-D-Z-Z-Z-

Or something similar.

Sorry I don't know details about the circuit and the application to give really a good advice. But as you mentioned that there are no resistors visible the power supply is might be working as constant current source - which is common for the operation of LEDs.

So if you connect one LED with series resistor parallel to the string of the 6 LEDs you divert current from the string 6 LEDs into the parallel path - this results in less brightness of the 6 LEDs and shifting in the light temperature or "light color".

So if you want to attach another LED to the string I would at first try to open the supply to the LED string and insert the 7th LED - if the current source is able to provide the higher required output voltage - excellent the problem is solved.Furthermore it is recommended to measure the current through the diodes in case yo need to design the circuit for the added LED.

If the direct insertion into the circuit does not work as the current source is not able to provide the required output voltage and the LEDs are only dim - the only way is to seek a supply voltage in the set which you can load with the required current (you have already measured) through the LED. To operate the LED you will need a current source which you can simply design according the current source application circuit in http://www.national.com/ds/LM/LM117.pdf - refer to the current source circuit diagram on page 18 (bottom). Then you can avoid the color and brightness problems.

By adjusting the value of the series resistor you can set the brightness of the additional LED to be equal to the other 6. For the selection of the LM 317 package you will need to figure out the current requirements as there differnt packages available. For low currents you could use a SMT package, if the current is higher you might need to divert to the TO220 package.

Good luck

Just buy a flashlight.

Without more info, try this:

use a 10- 50 kOhms potentiometer in series and start with the highest resistance.

(use the slider connection and one of the other pins) slide or turn slowly until you find the light intensity acceptable. (if you slide too far the led will become very bright, hot and dead)

Disconnect the potentiometer and measure the resistance with a Ohm meter.

Put a resistor of about the same value in the place of the potentiometer and enjoy the light. Cherry

There are lots of LED resistor calculator sites on the web.

Example: http://ledcalc.com/

use 4 and 4 and paint one of the 4 black

The Answer to your Question is:

The total Resistance (Rt) of the 6 LEDs in Series with Resistance (R) and one LED in Parallel with Resistance R is:

Rt= 7/(6R). It comes from following Calculations:

Series Parallel

RT = 1/6R + 1/R.

To simplify the Parallel Equation lets multiply both sides of Equation by 6R, then

6R x RT= 6R/6R + 6R/R = 1 + 6 =7 Thus

RT= 7/6R

To equalize the current and voltage, you could add a parallel resistance after each series LED of the same value as the LED resistance. This will give you 6 L shape sectors of the same resistance therefore they should all lit with at the same level, if that is what you are looking for.Hope this solves your problem.

Rafael L. Lama

Profesional and License Mechanical Engineer

http://casadominicanadepuertorrico.ning.com

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That should be 1/Rt = 1/6R + 1/R.

All the above answers assume the addition of extra LEDs will not overload the supply. If the string of 6 LEDs you currently have was obtained as such, consider getting a second string - with power supply if needed - then hide 5 of them.

This is for a blackberry led mod, the additional led is wired from the LCD display to the touchpad so the touchpad illuminates when the screen does, when the screen contrast is adjusted all the LEDs adjust however the only problem was when it's adjusted about 25% only the touchpad led lights and the screen goes blank lol Thank u I will try the Zener diode and the pot suggestions, I got the same mod working perfectly on an older handset using a 2k resistor however this LCD uses more power, I can't take apart the LCD itself to connect anything to the LEDs already in there...

Thank u uli_newbuilder for ur thorough post. I can cut the ground track of the LCD LEDs and add an additional led there all 7 LEDs work and contrast settings work with all 7 LEDs being just slightly dimmer which isn't actually noticeable however I wanted to try and do this without modifying the phone too much and with it being reversible :)

Just some observations really.

I would be astonished if you can find a white LED with a forward voltage of around 2V: they're nearly all at least 3V. So I suspect that the original set of LEDs is two parallel strings of 3 series LEDs, or 3 parallel pairs of series LEDs.

Think of an LED as a current driven device which has a (relatively) fixed forward voltage; so the fact that you see voltages of between 8 and 14 Volts across the backlight module suggests that the module is not being driven by a current controlled source, but that it is being driven by a voltage source, and, that there are series resistors incorporated into the backlight.

Let's assume that the LEDs have a forward voltage of 3.5 Volts and there are three parallel sets of two series LEDs with a series resistor

Then when V is 14 volts, each resistor has 7V across it: so to get an LED current of 20 mA you would need resistors of 7/0.02 = 350 Ohms

Lets assume that the resistors are standard values of 390 Ohms: then when V = 14V
LED current = 7/390 = 18 ma
and when V = 8V
LED current = 1/390 = 2.6 mA

Your best bet is to add a fourth parallel string of 390 Ohms and two LEDs

If you add a single LED you want the current to be just less than 20 mA at 14V: so
R=(14 -3.5)/0.02 = 525 Ohms

I'd choose 560 Ohms then at max the LED current is 10.5/560 = 18¾ mA
But at min the LED current is (8-3.5)/560 = 8 mA

So your keypad will not have dimmed nearly as much as your screen.

You said something about adding a 10K resistor somewhere. With a 12V supply the max current you can get through a 10K resistor (without any other series components) is 1.2 mA: you'd see an LED light up at that but it wouldn't be very bright.

Regards

1. << calculate the resistance of LEDs >>

Resistance is not measured for design of LEDs or Lamps, but Voltage/Wattage for lamps and Fwd Voltage drop and wattage in case of LEDs. To calculate the series resistance in case all the LEDs are less than the required number to be connected to a power supply in excess:

All the data is given by manufacturer. See the current at your optimum requirement and calculate the resistance by Vps - Vdrop by total LEDs in series = the excessive voltage to be dropped by series resitance.

2. << ... if I do this obviously power would go through the easiest least resistance route I.e the single parallel led, ...>>

If your power supply is capable of delivering the power of your circuit the thing which will happen is not the other circuits will be deprived of current reqired but your SINGLE LED [without a proper series resistor] will flash-out.

You can connect any number of LEDs [or lamps] of similar current ratings in series across a supply with two conditions:

a. Supply voltage should not exceed the voltage required by lamps. In this case a series dropper is required

or less, LEDs may not coduct at all or dimm and lamps will dimm.

3. << ... however there is a brightness option ... >>

Correct ! you can always connect a series resistor or an additional LED or a diode to dimm.

Try this:

It used to work with red / green LEDs. It's a relaxation oscillator, you drain much less current than with resistors or zeners for the same brightness. Hope you have the place for 3 more components. The transistor was a BC548, which gives you a switching voltage of 7 to 8 V. Maybe an SMD version will fit. Suggest you breadboard the ckt first starting with around 1 kohm and 1 microF. Your present power supply is switched, too, hope it wil not start blinking.

Unless I'm missing something this seems quite simple. Most LEDs have their current set by some resistance in series with them. Each LED has a, relatively, constant voltage drop across it. Anything you put in parallel with the string you already have will not affect that strings current, unless the additional current required by the parallel single LED + current limiting resistor exceeds the limits of the power supply. Since most LEDs glow with relatively low forward current, and normally have voltage drops of approximately 1.2 to 2 volts, you can calculate 2 resistance values using 10ma and say 25 ma (look at typical Max. current ratings for the LED you want to use). Now pick a potentiometer that will encompass this range from its center point and adjust it to get a similar brightness as the other 6 LEDs. Now you can substitute a standard value fixed resistor for the pot. or leave it as is, if the pot. is small enough.

If I've assumed something wrong about your application, please correct me.

Please point me to the spec. of a white LED with a forward voltage drop of less than 3V.

I did say "normally." You are correct. Different colored LEDs can have higher voltage drops. The point was, this isn't a hard solution to calculate if you know the characteristics of the LED you want to use.

So you are performing a mod to a Blackberry device. Extra lamp for some reason.... I can't add anything to what the other contributors have suggested. Seems like a lot of effort just for the hell of it. Whatever lights your candle I guess. Just mess about blindly. You will either succeed or fail. If you fail and the Balckberry is killed then all you are doing is bringing forward the inevitable. RIP Blackberry.

Thanks folks the pot method worked, about 14k seems to work best with all LEDs working even on the lowest contrast setting. Looks great about half way :)