How to turn off an entire circuit when one LED fails (in the event of an open branch)?

Question

I'm wondering how I can have a circuit to turn off in the event that one (or even multiple) LEDs fails, causing an open parallel branch. I would like all the LEDS to turn off in that case. The example circuit is attached, however, there will be 18 LED branches rather than the 4 in the example.

LEDs: Vf: 3.0 V @ 30 mA

The LEDs will have a steady 12 V supply. I will be using only SMD components.

The problem I'm having is that when one branch fails, there is not much of a voltage difference to make a circuit with a diode and transistor turn off the circuit (which is how I usually design a circuit with less LEDs).

Ideally, I would want to use something like an LED driver with fault detection or comparators, but this is something I am trying to avoid, as that drives cost per piece up a lot.

Answer 1

Move the 100R resistors to the bottom.

At 30 mA, each 100R drops 3 V when its LEDs are conducting and 0 V if any of the 3 fail open-circuit.

Each 100R's voltage can drive a 3.3 V AND gate input through a 10K current-limiting resistor. A full AND tree can then be made from 4-input or 8-input AND gate ICs, producing the single output. If the final AND tree output is LOW, an LED has gone open-circuit.

If multiple LED failures take a gate pin voltage above the MCU's supply rail, its pin clamp diodes will conduct. The 10K series resistor will limit the clamp diode current to a safe level, even if driven with 12 V.

^{simulate this circuit – Schematic created using CircuitLab}

If a 3.3 V MCU with ADCs and enough channel inputs is used, each 100R's voltage can drive an ADC input through a 10K current-limiting resistor. Then the ADC can check for LEDs gone open-circuit (Vin near 0 V) and LEDs gone short-circuit (Vin above, say, 3.1 V).

For better voltage measurement of LEDs gone open-circuit, each 100R could be two 50R in series. The ADCs could then measure across a 50R, getting 1.5 V when all's well and measurable higher voltages from failures.

Answer 2

The series resistors can be used as current sensors. For each LED branch, a PNP transistor's E-B junction is monitoring the voltage across the series resistor. The PNPs are placed in series and are used to maintain a positive gate voltage on the NMOS M1 used as a primary switch.

The circuits below use easily available parts and are inexpensive.

Burn-Out Detector

SW1 is used to initially turn the LEDs ON.

If any LED goes open, the current sense voltage goes to 0, the associated PNP transistor opens, and the gate is pulled to 0V via R7.

If an "ON" pushbutton is not desired, it can be replaced with a 100nF capacitor. That way the circuit will start up with the LEDs turned ON.

^{simulate this circuit – Schematic created using CircuitLab}

The plot below shows the total LED current. SW1 is depressed from 0ms to 1ms. At 5ms, SW3 goes open, simulating a failure of the 3rd branch. The LEDs then turn off.

Voltage Regulator with Burn-Out Detector

It takes only a small modification to the circuit to use the pass element M1 to both turn the LEDs off when one burns out, but also to regulate the voltage.

^{simulate this circuit}

The voltage regulation is not too shabby for a three-transistor circuit. R10 should is used to adjust the voltage.

The gate- and LED-supply voltage throughout the lifecycle of the system are shown below, for supply voltages of 11.5, 12, 12.5 and 30V. From 0ms to 1ms, the ON button is depressed. At 5ms, SW4 opens, simulating a LED failure in the 3rd string.

One-Branch Current Regulator with Burn-Out Detector

This circuit, instead of regulating voltage, regulates current in one of the LED branches. The remaining branches are assumed to be at a similar temperature and will draw similar current. The current regulation is very good.

^{simulate this circuit}

The LED current for supply voltages of 11V, 12.5V and 30V is plotted below. The ON switch is depressed from 0ms to 1ms, and the 3rd LED branch opens at 5ms.

All-Branch Current Regulator with Burn-Out Detection

Instead of using just one branch as an "indicator" of current, the average current across all branches can be regulated. This will compensate better for unequal LED aging, etc.

^{simulate this circuit}

Again, the current regulation is decent:

Answer 3

This circuit tries to activate the LEDs during power on and turns off, if one ore more LED chains do not conduct.

^{simulate this circuit – Schematic created using CircuitLab}

1. C1 is initially not charged or has beed discharged via D1 and R7.
2. The P-FET M1 will conduct and turn on M2, if the supply voltage is above the threshold voltage of both FETs.
3. During the early startup time a small charge current flows to C1 via R1 and all the "wired or" diodes. This current is reduced if M2 starts conducting and all LED chains conduct as well.
4. The voltage at U_SUM will not be high enough to turn off M1 as long as all strings conduct.
5. An "open" failure of one string, simulated by SW2, allows to rise U_SUM up to 12 V - one diode drop. This will charge C1 and turn off M1 and M2.
6. This state is latched until the next power cycle.
7. The time constant of R1 and C1 can be shorter if the rising power slope ist fast. The initial test flash of all LEDs will be shorter as well.
8. D1 and R7 are obsolete if the power off period is long enough to discharge C1 via R1 and R2. If not, the circuit may not (re-)start.