LEDs and Power Supply

[falcon3497]

Hello all,

So I'm playing around with LEDs for the first time. I'm trying to power 3 LED diodes using a 12v 5A DC plug. I did some research online and calculated the forward voltage of each diode (two of them are 2 volts and one is 3.4 volts) and determined the resistors needed for each diode if I'm supplying 12v (the 2 volt diodes each have a 510 ohm resistors and the 3.4 volt has a 470 ohm resistor). I soldered them to a female 12V receiver, and they light up. Success so far.

My concern is, is that the resistors seem to be getting unreasonably hot. Each diode is supposed to run at 20 milliamps, and so given that the 12V power source is 5 Amps, I fear that may be the issue. Is the power source the issue here, or did I do my math wrong? If so, how can I rectify this? I'm doing a parallel circuit by the way.

 

[Harald Kapp]

given that the 12V power source is 5 Amps, I fear that may be the issue.

No. The power supply is capable of supplying 5A, but will supply only 20 mA, as calculated, to each diode.
The problem is the 10 V (or 8.6 V) dropping across the resistor which causes power dissipation in the form of heat.
Example: P = 10 V × 0,02 A = 200 mW

I'm doing a parallel circuit by the way.

Switch to a series circuit with only one resistor. The added voltage of the 3 diodes is 7.4 V. The resistor is (12 V - 7.4 V) / 0,02 A = 230 Ω (you can use 220 Ω).
Power dissipation in the resistor is P = 4.6 V × 0,02 A = 92 mW only.
If you want (or need) to stay with a series circuit your options are:
- use a low voltage power supply (e.g. 5 V)
- step down the 12 V from your power supply to a lower voltage (e.g. 5 V) using one of cheap step-down converter modules (also known as buck modules).
- Instead of using a single resistor for each LED use a series connection of multiple smaller resistors, e.g. 3 × 220 Ω instead of 470 Ω. The total power dissipation will be the same, but distributed across multiple resistors so each resistor will not get as hot as a single one.