D.I.Y LED Shift light

blackbusa03

Registered
Hi .new on forum,
I have a 2003 Busa and i want to make a 10LED shift light controlled by the flapper valve solenoid.
Can anyone tell me if i connect the Led parallel (all the + and all the - together), how many Ohm resistor i have to use?
 
Generally it's better to connect LEDs in series rather than parallel for power consumption and circuit simplicity. For instance, a red LED typically has a voltage drop of 1.5 volts. This means that you can use 8 of these in series to have a total drop of 12 volts and not need any resistance (thought it's still good to have a small amount in there). If you did them all in parallel, then you would have to account for 10.5 volts that the LED isn't dissipating, so you would need a 525 ohm resistor for each LED you use to keep the current at 20 mA.

For reference, series is connecting the LEDs in a line (the negative of one connected to the positive of another) and parallel is with all the positives and negatives connected together.

To get the resistor you need for your LED project, take the voltage drop of your circuit (for series add up the individual voltage drops), subtract this from 12 (the voltage of your system) then divide the remaining voltage by 0.018 A (this puts in a buffer in case voltages aren't what you expect) and that will give your resistance you need in series with your LEDs.

Ex. Two red LEDs with a voltage drop of 1.5V in series will have a total voltage drop of 3.0V. 12.0V - 3.0V = 9.0 V
9.0V/0.018A = 500 ohms.
 
Generally it's better to connect LEDs in series rather than parallel for power consumption and circuit simplicity. For instance, a red LED typically has a voltage drop of 1.5 volts. This means that you can use 8 of these in series to have a total drop of 12 volts and not need any resistance (thought it's still good to have a small amount in there). If you did them all in parallel, then you would have to account for 10.5 volts that the LED isn't dissipating, so you would need a 525 ohm resistor for each LED you use to keep the current at 20 mA.

For reference, series is connecting the LEDs in a line (the negative of one connected to the positive of another) and parallel is with all the positives and negatives connected together.

To get the resistor you need for your LED project, take the voltage drop of your circuit (for series add up the individual voltage drops), subtract this from 12 (the voltage of your system) then divide the remaining voltage by 0.018 A (this puts in a buffer in case voltages aren't what you expect) and that will give your resistance you need in series with your LEDs.

Ex. Two red LEDs with a voltage drop of 1.5V in series will have a total voltage drop of 3.0V. 12.0V - 3.0V = 9.0 V
9.0V/0.018A = 500 ohms.
Thank you man for replying..was very helpful.
 
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