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SparkyCal

Mar 11, 2020
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I am working on my guitar pedal and I need to figure out what size of a resistor I need to step down the 9 volt power supply in order to light up three LEDs that are run as follows:

All three anode leads are run in series to one another
All three cathode leads are run in series to one another as well.

Is that configuration correct? If so, what value of resistor do I need? The specs ons the LEDs are as follows:

First LED: Purple 20 m/A, 3-3.2V

Second LED: Pink, 20 m/A, 3-3.2V

Third LED: White, 20 m/A, 3-3.2V

Thank you!!!
 

DBingaman

Jun 27, 2021
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Not sure what you mean by 'all three anodes in series' and all three 'cathodes' in series. Can you draw a sketch? If the three LED's are wired in parallel, due to differences in voltage drops not all of them will come on at the same brightness level. If they are in series, 9V is not enough to overcome the 3-3.2V at 20mA level.
 

DBingaman

Jun 27, 2021
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I am working on my guitar pedal and I need to figure out what size of a resistor I need to step down the 9 volt power supply in order to light up three LEDs that are run as follows:

All three anode leads are run in series to one another
All three cathode leads are run in series to one another as well.

Is that configuration correct? If so, what value of resistor do I need? The specs ons the LEDs are as follows:

First LED: Purple 20 m/A, 3-3.2V

Second LED: Pink, 20 m/A, 3-3.2V

Third LED: White, 20 m/A, 3-3.2V

Thank you!!!
If all three are in parallel you can try (9V-3.0) = 6V over total current 60mA or a 100 ohm resistor. But again because each one has slightly different turn on voltage curves some may be brighter than others. The better way would be to have each LED has it's own resistor with a value of 300 ohms. This will provide around 20mA to each one with equal brightness to all three.

This is the best solution. It also allows you to adjust each LED brightness by simply changing the resistor value. Just be sure not to exceed LED max current.

LEDfrom9V.PNG
 

DBingaman

Jun 27, 2021
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What exactly do you want to accomplish? All three LED's turn on when pedal is depressed?
 

SparkyCal

Mar 11, 2020
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Hi- your suggestion to have each have it's own resistor is, I agree, the better solution. Yes- I just want all three to turn on when the pedal is clicked.
 

Harald Kapp

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Not sure what you mean by 'all three anodes in series' and all three 'cathodes' in series.
I agree, it is either all anodes together and all cathodes togehter, which would make it a parallel connection. Or it is a serial connection, then it is anode to cathode.
bethuse each one has slightly different turn on voltage curves some may be brighter than others. The better way would be to have each LED has it's own resistor
I disagree: LEDs are current driven. In a parallel circuit as proposed by you the difference in turn-on voltage may lead to different currents through each LED and thus different brightness. Also power dissipation is 3 times that of a series connection.
If, on the contrary, the LEDs are put in series, current is the same for each LED and thus brightness matches much better. Plus power dissipation is reduced.
But: in this case the voltage drops across the LEDs sum up to more than 9 V and a series connection will not light them up at all (or only very dimly) on a 9 V power source. This is no issue with the proposed parallel circuit.
A power saving intermediate circuit could use both parallel and serial connection as shown:
upload_2021-8-12_6-42-25.png
Total current here is 40 mA (compared to 60 mA in the fully parallel configuration).

A note on the side: A typical alkaline 9 V block has a capacity of ~600 mAh. I don't know how much current your amplifier/distortion circuit draws but drawing 40 mA from a 600 mAh battery will drain the battery in ~15 hours. Faster when you add the current drawn by the amplifier/distortion circuit.
 

Martaine2005

May 12, 2015
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You may not require 20mA. Some LEDs are still very bright at 5mA or less.

Martin
 

SparkyCal

Mar 11, 2020
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Thanks guys. The additional leds are just for aesthetics. I can revert to only one, if it will cause a voltage drain challenge. Having three is not important.
 

SparkyCal

Mar 11, 2020
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I agree, it is either all anodes together and all cathodes togehter, which would make it a parallel connection. Or it is a serial connection, then it is anode to cathode.

I disagree: LEDs are current driven. In a parallel circuit as proposed by you the difference in turn-on voltage may lead to different currents through each LED and thus different brightness. Also power dissipation is 3 times that of a series connection.
If, on the contrary, the LEDs are put in series, current is the same for each LED and thus brightness matches much better. Plus power dissipation is reduced.
But: in this case the voltage drops across the LEDs sum up to more than 9 V and a series connection will not light them up at all (or only very dimly) on a 9 V power source. This is no issue with the proposed parallel circuit.
A power saving intermediate circuit could use both parallel and serial connection as shown:
View attachment 52625
Total current here is 40 mA (compared to 60 mA in the fully parallel configuration).

A note on the side: A typical alkaline 9 V block has a capacity of ~600 mAh. I don't know how much current your amplifier/distortion circuit draws but drawing 40 mA from a 600 mAh battery will drain the battery in ~15 hours. Faster when you add the current drawn by the amplifier/distortion circuit.
Harald- I may go with your schematic. I am not planning to use a battery at all. It will be powered through a DC barrel.
 

DBingaman

Jun 27, 2021
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Harald- I may go with your schematic. I am not planning to use a battery at all. It will be powered through a DC barrel.
If this is not a battery operated device, I still think 3 separate resistors is better. It allows you to adjust up/down current to each one for the best effect you are looking for.
 
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