Hey Electrical Engineerers!

I am currently working on making an LED tail light for an older car.

Overview of the Taillight

Each tail light has 4 inputs connected to it: Park, Brake, Turn, and Reverse.
The 4 inputs are switched 12V. For example, if you engage your turn signal, the turn input will switch 12V on and off each time the blinker or hazard lights are activated. The most current this circuit could draw is ~2.5A.

When I initially created this circuit, I had a regulator on each of the inputs. Soon after, I realized that four regulators was much too expensive. So with that, I created this circuit which aggregates all the power to one buck regulator (LM2596) and uses MOSFET transistors to determine which light should be on.

NOTE: The LEDs for each light are actually an array of LEDs, but I simplified it to just one for this schematic.


Tail Light Circuit


  1. Is this an effective circuit, specifically the one regulator being supplied from multiple sources and the MOSFETs being used to switch on the lights? Is something like this available in an IC?
  2. I would like to add over-voltage and current protection to this circuit, how could I effectively do that?

Thanks in advance, I really appreciate any feedback.

  • 1
    \$\begingroup\$ The basic idea seems OK. What is the total forward voltage for each of your diode strings when the LED is on? Is 9V the right choice for output voltage? Also, if you want to use PMOS, you will need to invert the control signals, I think. You could use NMOS between resistors and ground of each string. OVP could be achieved using a PTC and power Zener diode. Brutal but effective. I believe automotive circuits need to be designed to survive the alternator load-dump waveform (you can google that). \$\endgroup\$
    – user57037
    Commented May 21, 2016 at 22:42
  • \$\begingroup\$ The forward voltage for each LED is 2.5 V. There are 3 LEDs in each string. I have tried using a zener diode and PTC for over voltage protection, but blew my zener diode. But I think that is because I used the wrong zener diode, it could not dissapate the amount of power I was throwing at it. Are their zener diodes made for handling a large amount of power? \$\endgroup\$
    – Matt
    Commented May 23, 2016 at 4:06
  • \$\begingroup\$ yes. Try googling "transorb." Or search digikey and use max power dissipation as one of the parameters. By itself, the zener doesn't have a chance. The idea is to use a PTC in series, and then after the PTC, put the zener to GND. During over-voltage, the Zener must dissipate enough power for long enough to trip the PTC. Then it will just sit there dissipating power until the voltage comes down again. If Vforward is 7.5V, in total, then 9V seems about right for the regulator output. Make sure you test over temperature. The brightness will vary somewhat due to Vforward shift. \$\endgroup\$
    – user57037
    Commented May 23, 2016 at 4:12

2 Answers 2


As Dave Tweed pointed out, the logic will be wrong the way you drew it. Suggest adding a TVS and fuse on the input to protect the regulator and wiring. Use 200V or higher ordinary diodes (eg. 1N4004) for your input 'or' gate. Consider adding a small series fusible resistor before the TVS to help it clamp. This is 'low side' switching so you need to provide a return wire for the lights rather than the chassis. To make high side switches requires another transistor for each LED bank.

You do need to protect the MOSFET gates and probably to provide some kind of pull-down. Something like this:


simulate this circuit – Schematic created using CircuitLab

Resistors can be 1/4-W 5%.

If the lights are also connected to a computer or something to indicate bulb failure you may have strange things happening (this stuff started in the 1980s so the car could still be plenty old).

  • \$\begingroup\$ You missed the fact that PMOS transistors invert the logic of turning the lights on. NMOS used as low-side switches would make a lot more sense, and would be easier to protect. \$\endgroup\$
    – Dave Tweed
    Commented May 21, 2016 at 23:20
  • \$\begingroup\$ @DaveTweed Good point! I think it's fixed- same to protect. Unless the OP wants to maintain the chassis ground return in which case, it will be more complex to provide high-side switching. \$\endgroup\$ Commented May 22, 2016 at 0:04
  • \$\begingroup\$ Would I want to add a fuse and TVS diode to each input? Also @mkeith mentioned just using a zener diode and fuse. I guess I don't understand why I would use one over the other (TVS vs Zener). Lastly, for the MOSFET protection, how exactly does that work? I understand the zener diode is pulling the voltage at that node up to 12 V, but what exactly are the two resistors doing for the circuit? Also thanks for the help so far! \$\endgroup\$
    – Matt
    Commented May 24, 2016 at 2:11
  • \$\begingroup\$ TVS is like a zener but designed to absorb big pulses. The zener is fine for the gate where there is series resistance R2. R1 is not gate protection- it's for when the (not shown) switch opens and you need some way to pull the gate down to ground. Fusing depends on what else is there, probably you only need at most one since cars will have the wiring fused to begin with. \$\endgroup\$ Commented May 24, 2016 at 9:49

Your mosfets should be N channel and logic level so when the drive signal is active 12V logic high the relevant lamp is on .Gate protection like Spehro stated is essential.R2 can be 10K if R1 is kept at 1K .Remember that the incoming logic is active high not active low .Your off the shelf IC circuit makes a fixed output voltage which has to be greater than the LED drop .Say if one of your lamps in the LED cluster is White and has a terminal voltage of say 3V and you set up your chip to 4V DC then your overall efficiency is only 75% of what the IC is .Now lets say that one of the LEDs is red and drops 2V then your overall efficiency wont get to 50% .If you set up a switchmode buck current source then the power would not be wasted in the resistors.The current setting can be steered to different values for different coloured lamps with the mosfets.

  • \$\begingroup\$ Can you explain the efficiency rating in more depth? I don't quite follow what you mean. I have three different color of LED arrays (White, Red, Amber). The red and amber have forward voltage of 2.5 V and the white has a forward voltage of 3.65 V. \$\endgroup\$
    – Matt
    Commented May 24, 2016 at 2:17
  • \$\begingroup\$ Your DCDC buck wastes power .Simple switcher is not the most efficient switcher but it is simple .You may be on say 80% here but the led dropper resistors waste more reducing your overall efficiency even more. \$\endgroup\$
    – Autistic
    Commented May 24, 2016 at 4:30

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