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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:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ @IgnacioVazquez-Abrams I'm not aware of any statistics to support my experiences, but I've seen a fair share of shorts in power BJT's in my life. \$\endgroup\$ – jippie Jan 4 '14 at 12:15
  • \$\begingroup\$ Add a second opamp OA2 with Vin2 slightly greater than Vin. If OA2 ever has positive output the regulator circuit must have malfunctioned. Use OA2 output high to disconnect the supply. One method is to use a clamp transistor across the supply to blow a fuse. As Vclamp ~=0 the dissipation will be relatively low until the fuse blows. Or add a second series pass element which is usually hard on and is switched off by OA2 high. \$\endgroup\$ – Russell McMahon Jan 4 '14 at 12:25
  • \$\begingroup\$ FYI:" As shown dissipation in R3 at 4A = 16 Watts which is OK but high. You could increase R3 until the lamp can receive only slightly more than 4A with Q1 hardon and Q1 will then be unlikely to fail if properly heatsunk. That of course makes R3 dissipation even higher. Lower R3 reduces dissipation at cost of heat in Q2. \$\endgroup\$ – Russell McMahon Jan 4 '14 at 12:28
  • \$\begingroup\$ Follow Russell's suggestion, increase R3 until I=4A at Q2 saturation. Then go one step further : eliminate Q2, eliminating the problem. Most R failures are high impedance : pick R such that this is its failure condition. \$\endgroup\$ – Brian Drummond Jan 4 '14 at 12:33
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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: enter image description here

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.

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  • \$\begingroup\$ what is this circuit editor program? \$\endgroup\$ – Standard Sandun Jan 4 '14 at 12:05
  • \$\begingroup\$ The schematic editor is Altium Designer. \$\endgroup\$ – Martin Jan 4 '14 at 12:13
  • \$\begingroup\$ Does Altium Designer have a button you can press to fill in the "?" part designators? ;-) \$\endgroup\$ – RedGrittyBrick Jan 4 '14 at 15:37
  • \$\begingroup\$ It doesn't have a single button per-se... You need to annotate the schematic sheets. "T", "A" will bring up the annotation options where you can configure the indices for your component designators. Alternatively you can select the "Annotate quietly" option after pressing "T" (tools) and it will use the default annotation options which are normally configured in the annotation options screen. \$\endgroup\$ – Martin Jan 5 '14 at 1:47

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