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The schematic beneath is a new project of me.

The theory behind this is: My car has a very fool proof bulb failure detection and warning system. CANbus proof LED bulbs can't fool the car and just a parallel resistor will make a faulty LED bulb undetectable. So when the parallel resistor (wich imitates as a incandescent bulb) can be "switched" off when a short or open circuit LED bulb is "detected", the car will see a faulty bulb.

But... another common fault in LED bulbs is that the LEDs are burned, but without a full open or short circuit. They just take less current/voltage when this happens. So there should be a third "detection": the voltage/current drop. How can I managa that, without heavily increasing the number of parts and most important, the costs.

At this moment, it is only for the rear brake and indicator bulbs and every bulb needs a circuit as the example below, with 2 circuits combined in 1 box and 1 box for each side.

Thanks in advance.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Figure 2 by @Transistor. A more conventional drawing with voltage decreasing from to to bottom, chassis symbol used for the vehicle groun and ground symbols all pointing down (towards the ground).

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  • \$\begingroup\$ Would you like to make a system which tells you when the lights should be changed before they burn out? \$\endgroup\$ Commented Feb 8, 2020 at 12:54
  • \$\begingroup\$ Does this really waste 50W in normal operation? \$\endgroup\$
    – user16324
    Commented Feb 8, 2020 at 13:11
  • \$\begingroup\$ @Horror Vacui: Yes and no. With an incandescent bulb, it is faulty or it is not faulty (filament is whole or broken). With an LED bulb, some components can fail, so the light output can be very dim, without a burn out. The circuit as designed, will only detect an open circuit or short circuit, not something in between, like a current/voltage decrease, without a full drop out. \$\endgroup\$
    – htevents
    Commented Feb 8, 2020 at 13:12
  • \$\begingroup\$ @Brian: no the original incandescent bulb is 21W. But when using a 50W rated resistor, less heat is generated. A 6.8 ohm 21W resistor will get extremely hot. \$\endgroup\$
    – htevents
    Commented Feb 8, 2020 at 13:14
  • \$\begingroup\$ Why is R1 6.8Ohm? Why not kOhms? Isn't it a digital signal? It has to be , since the current through Q1 will vary a lot with different Q1 instances or even with different transistors. \$\endgroup\$ Commented Feb 8, 2020 at 15:16

2 Answers 2

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A product I designed has a diagnostic test for the LED that simply measures the voltage across it and compares it to high and low limits (it would be possible to temperature-compensate those limits).

That's done as part of the supervisory functions of an MCU that is there anyway, so it occupies one GPIO pin and requires an extra resistor or two (one or both of which may have been otherwise unused in networks). The software overhead is negligible, in program size and speed, even with a lower-end MCU.

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The voltage of diodes and transistors might vary significantly with temperature, not to mention the variation between different pieces and types, and therefore I do not think that you can do anything better than measuring both the current and voltage, save them and compare to their previous values. You have to do these all in a relatively noisy environment. Not an easy task!

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  • \$\begingroup\$ So how about this: A 5W LED bulb is drawing 360mA at 14V, the voltage in idle. With R2 being 1.8 ohm, it will take 350mA (the threshold) to turn on Q1. So if I change R2 so it has a threshold of 320mA to open Q1 for example, it will turn Q1 off when the bulb is drawing less then 320mA to open. Is it correct that an LED bulb will draw less current when more and more LEDs burn out? Correct or is this thought too simply? \$\endgroup\$
    – htevents
    Commented Feb 8, 2020 at 14:21
  • \$\begingroup\$ An LED is basically a diode. There is already 5W single package/die LED, but multiple smaller power LED units might be used as well. In the latter case it is logical to connect some LEDs in series due to the available 12-14V supply. BAU15s is not just a LED. It has CAN interface and I am pretty sure some current generation for the LEDs inside. It can be even a series resistor. The I(V) characteristic of both BAU15s and Q1 will vary with temperature, and you need a different circuit, or at least trimming if you change the lamp. The current of the lamp will depend on the voltage as well... \$\endgroup\$ Commented Feb 8, 2020 at 15:12
  • \$\begingroup\$ @htevents whether your assumptions are correct depends on the internals of the lamp, what I am not familiar with. BTW: the transistor is not a switch in this operational mode. It is an analog device. \$\endgroup\$ Commented Feb 8, 2020 at 15:14
  • \$\begingroup\$ I think I need to clear things up. The LED in this scheme is a 5W BAU15s replcament bulb, which replaces the 21W incandescent bulb. R1 will imitate that bulb and Q1 will cut off the power to R1 if the LEDA lamp fails by either open or short circuit. Everything is on the 12V line coming from the light switch. \$\endgroup\$
    – htevents
    Commented Feb 8, 2020 at 17:19

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