How to turn off simple LED circuit when a 'short' is detected?

Question

I have a question about how to control this simple LED circuit that is being driven by a battery. I am not using a microcontroller in the set up so the control will have to be something else. The conditions are that sometimes the D1 LED can be shorted (from an external condition) so that the only load will be R1. I would like to be able to detect when D1 is shorted and shut down the circuit and then reenable the circuit when the short is cleared.

I was thinking of using a comparator to sense the voltage between R1 and D1 to make sure it is the forward voltage of D1 but that wont work because when the mosfet turns off this just floats to Vbatt and you couldnt detect the short to know when to turn it back on. Im trying to come up with a simple solution because this seems easy but Im stuck. There is a control signal I will use out of another circuit but it isnt a microcontroller so I was thinking of using an and gate with some other logic signal to detect the presence of a short across D1. The LED has a Vth of about 2.7V. If you think there is an easier way to drive this LED thats helpful too, maybe using the nFET isnt the best way.

Thanks for your help

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You might be able to do this with just MOSFETs, see the below image.
You can play with this circuit here
In this circuit, I'm using the switch as a stand-in for the short.
Under normal conditions, the forward voltage across the diode is used to create a negative V_GS for the P-channel MOSFET, opening it. This, in turn, creates a positive voltage for the V_GS of the n-channel, opening that.
When the short happens, it also shorts V_GS, causing it to close.

I say might, because you need all your values to line up nicely for this circuit to work. You for starters need the forward voltage of the diode to be more than the V_GS of the p-channel MOSFET.
You also need the circuit to be able to start. Which I'm not actually 100% on. You need the leakage through the n-channel, to create enough forward voltage of the diode to start opening the p-channel, to in turn open the n-channel fully. If things didn't line up well, you could get a circuit that doesn't start at all. There might actually be a way to do something similar with BJTs in a more reliable fashion.

Answer 2

This constant current circuit provides nearly the same collector current with open and closed switch. This way the shorted LED will not be detected, but it has no impact on the rest of the circuit except more power loss in the transistor. This may be important if your circuit feeds a power LED.

The value of R1 defines the current as long as the base voltage is above, say, 0.8 V. The regulation is better and less temperature dependent at higher base voltages.

If you replace R2 by two diodes in series, the current is less dependent on the supply voltage.

^{simulate this circuit – Schematic created using CircuitLab}