How to make transistor fail protection for light bulb

Question

In the circuit below I choose an optimal current for the lamp set to 4A (the lamp specs are 12V/5A).

I need to make a safety circuit for the lamp in case transistor Q2 fails.
I add a fuse in the circuit but I'm looking for a better solution, what is the best way to protect the lamp?

This is current regulator circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Firstly, I would consider a different approach.

I would consider Pulse width modulation to the lamp similar to the function of an SCR in an AC dimmer circuit.

You could also increase the 1R sense resistor such that it achieves a 4A current limit. You would need to know the hot resistance of the lamp.

If however you cannot use a PWM drive circuit, then I would propose a high side FET current limiter such as this:

The way it works, is that when 4A current is exceeded, the voltage across the 0R15 resistor is sufficient to turn the PNP resistor on, lifting the P-Channel Gate up to 15V. THis clamps it off to maaintain a constant current limit which is determined exclusively by the value of the 0R15 sense resistor. The 100k/10k form the bias network which keeps the P-CH fet in conduction until the PNP is enabled. THe Mosfet should be rated to handle the required power. Ordinatily, it will be saturated but at fault condition it could dissipate up to 60 Watts.

I hope this helps.