I want to replace a light bulb in a Peugeot 308 SW with a LED. The bulb works (lights) normally, but the LED blinks. I thought they are controlling the bulb in alternative current but my digital voltmeter shows 9V on DC scale.

And no. It is not a blinking LED (I also thought it was). I have tried it with a battery and works perfectly. The blink must come from the dimming system - yes, the car dims the lights instead of suddenly turning them off.

I have tried also to use a diode and a capacitor to convert from AC to DC but it is not working. Maybe they are suing an open collector circuit?

My guess is that they control the bulb in DC pulses (PWM), and the circuit is an open collector.

  • \$\begingroup\$ "Blinks" or "flickers"? \$\endgroup\$ Commented Sep 27, 2014 at 21:59
  • \$\begingroup\$ @IgnacioVazquez-Abrams-What is the difference? I think it stays on for 0.5 seconds and then off for 0.5 \$\endgroup\$
    – IceCold
    Commented Sep 27, 2014 at 22:17
  • \$\begingroup\$ "Blinking" would be indicative of a circuit shutoff. "Flickering" would be indicative of PWM control. \$\endgroup\$ Commented Sep 27, 2014 at 22:22
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    \$\begingroup\$ Easy test: Connect the original lamp and the LED lamp in parallel. The original lamp acts as the "wasteful power resistor" and shows whether the supply voltage is steady. \$\endgroup\$
    – gbarry
    Commented Sep 28, 2014 at 17:08
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    \$\begingroup\$ @IgnacioVazquez-Abrams True and I have bought some of these for use in my dash but the fact that it is blinking at about 1Hz strongly suggests to me it could be a blinking LED. I'm simply working on probability in the absence of hard information. Its up to the OP to confirm or eliminate this. \$\endgroup\$ Commented Sep 28, 2014 at 17:10

1 Answer 1


It sounds like the system is using PWM dimming at a slow rate (1Hz, 50:50 duty cycle, and that this is fast enough that the decay in an incandescent bulb is not very noticeable (thermal time constants are long). If the car has a 12V electrical system, then the fact you measure 9V at this bulb is odd... Perhaps the ratio is 75:25?

You could try adding a (large) capacitor in parallel with the LED to create a lowpass filter with the R. The problem is that when the supply is off, all the LED current must be sourced by the capacitor. This will require a big capacitor. From a quick LTSPICE simulation, to bias a 2V LED 20mA average current from a 75:25 duty cycle 12V supply, this takes a 330 ohm series resistor and a shunt capacitor on the order of 100,000uF (in parallel with the LED). It's a little better with a blocking diode in series with the resistor, but not much. There is still a variation in the LED current of about 20%.

Perhaps you can trace this back and figure out what is providing the PWM to the light bulb and run the LED circuit from 12V with something else switching it on. For example, with a few transistors, you can build a nearly constant current source for the LED and turn it on with the PWM'd signal filtered with a series R and smaller C. I can post a diagram of this approach if you are interested.

Here's the circuit with no added capacitor, a 12V supply switched at 75% duty cycle, and about 20mA through a Vf=2V LED with R=360 ohm. The LED current just shuts off when the supply is off.

enter image description here enter image description here

With a 100,000uF cap you can still see droop in the current, but the average is better. This would probably still have noticeable dimming. Note the current scale is different. Partly the current drops here because the LED is effectively regulating the voltage on vled when the supply is on. So the current drops rapidly when the supply is off.

enter image description here enter image description here

Based on that observation, you really want an RC filter first and then the LED and its series resistor. This has much less ripple, and the cap can be a little smaller. Here it's 33,000uF.

enter image description here enter image description here

  • \$\begingroup\$ Thanks. The car is new and I don't want to break it apart to look for a wire that has constant 12V on it. I am interested in the first solution even if it requires a large capacitor. But still... why so large? 100000uF is a lot for a small LED. Can you post please a diagram? \$\endgroup\$
    – IceCold
    Commented Sep 29, 2014 at 9:34
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    \$\begingroup\$ @Altar- The problem is by trying to filter the supply by just adding a cap, the cap value is determined by the resistor used to bias the LED. For 20mA at 12V, R=330 ohms. I've updated the answer with some images. \$\endgroup\$ Commented Sep 30, 2014 at 2:57
  • \$\begingroup\$ Wow. Great answer. I will do this. 33000uF is more acceptable! I hope this will trigger the Open Collector circuit. Can you please tell me the power for R1 and R2. I guess they don't have to have to be high power. I see a maximum of 31mA in the circuit. \$\endgroup\$
    – IceCold
    Commented Sep 30, 2014 at 9:18
  • \$\begingroup\$ @Altar- R1 and R2 have the same average current, equal to the LED current. Power = I^2 * R. For R1, P=0.02A^2 * 68 = 27mW, and for R2 P=0.02A^2 * 260 = 104mW. A 1/4 watt resistor is fine for both. \$\endgroup\$ Commented Oct 9, 2014 at 0:59

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