3
\$\begingroup\$

I have a series of LEDs I would like to have display at 2 brightness levels and am confused how to properly wire this. This is for additional running lights/brake lights on my bike.

LEDs - Typical Voltage 2.2V, Max Voltage 2.6V, Forward Current 150ma

schematic

simulate this circuit – Schematic created using CircuitLab

Source voltage 1 (running lights): 11.5V

Source voltage 2 (brake lights): 0.5V at all times, 11.7V when brake is engaged


When using the series LED calculator I believe for the Source voltage 1 I should set the forward voltage to 1.1V and 75ma (half of typical voltage and half of forward current) and then set the Source Voltage 2 to 1.1V and 75ma (the other half of voltage and forward current). That in my eyes would result 2.2V and 150ma. http://led.linear1.org/led.wiz

Is this right or should I be approaching this wiring different?

\$\endgroup\$
8
\$\begingroup\$

Note: The Voltage sources are indicated the wrong way around in the question's schematic, going by the LED direction shown (negative ground circuit).


First the always-on running light case. For the LED resistance calculator, use:

  • supply voltage = 11.5 Volts
  • Vf = 2.2 Volts ( x number of LEDs)
  • current = 75 mA

The entire forward voltage of all the LEDs would appear at the supply leads (not half), but you would want only about half the rated current to flow.


Now, the brake lights:

In order to prevent current from flowing from the running light lead to the 0.5 Volt brake light line when the brakes are off, you would need a diode on the brake light line, connected so as to be forward biased, same direction as the LEDs. A 1n4001 diode should do fine.

schematic

simulate this circuit – Schematic created using CircuitLab

For the LED calculator, use:

  • Supply voltage = 11.7 - 0.7 = 11 Volts (the diode drops around 0.7 Volts)
  • Vf = 2.2 Volts ( x number of LEDs)
  • current = 75 mA

The reason for doing this is, the currents from the two sources add up in going through the LEDs. Hence, when both supplies are high, 150 mA will flow. When just the running lights are on, 75 mA will flow. The voltages do not add up between leads.


As an added twist, if the LEDs need to light up only at one-third intensity for running lights, and full intensity for braking, this is easy: Just take 50 mA for the running lights calculation, and the remaining 100 mA calculation for the brake lights calculation, in the bullet points above.

\$\endgroup\$
9
  • \$\begingroup\$ Thanks so much for your answer. This is my real electronics project since high school electronics class (9 years ago). This was my last holdup for finishing the project and if i would of done it my way it looks like i would of had issues. I like your suggestion of doing 50mA for running lights and 100mA for brake lights. With that being said here are the ratings on the LEDS: Forward Current IF 150 mA Pulse Forward Current IFP 300 mA Allowable Reverse Current IR 15 uA Power Dissipation PD 0.5 W \$\endgroup\$ May 28 '13 at 15:03
  • \$\begingroup\$ Do you think 100mA running and 250mA brake lights would be better? The Pulse Forward Current max is 300mA and brake lights arent on long ever. powerledworld.com/en/… \$\endgroup\$ May 28 '13 at 15:04
  • \$\begingroup\$ @JustinKissinger Pulse forward current ratings are usually accompanied by a maximum duration per pulse, and further, that duration is usually much much smaller than typical braking behavior. Leave us not forget, also, about those people who tend to ride their brakes at a traffic signal, bad habit though that may be. So, no - do not use the pulse current rating for this purpose. \$\endgroup\$ May 28 '13 at 16:02
  • \$\begingroup\$ Well i just did a lot of reading on diodes. I cant find any info on 1n2001. I have some 1n4001 diodes at my place of work i can use, from what i can tell that should work just fine. Sorry for all the questions, starting to feel needy, but dont know where to go to learn yet. \$\endgroup\$ May 29 '13 at 6:53
  • \$\begingroup\$ @JustinKissinger Sorry, that was a typo, it's the 1n4001, a really inexpensive diode. Edited answer with datasheet link. \$\endgroup\$ May 29 '13 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.