Will this short-circuit protection circuit work?

Question

Here is a circuit designed to protect a load (an incandescent lamp) from short circuits. It is supposed to work as follows: VT2 is initially conducting — its gate is pulled to ground through R1. A voltage close to 12 V is applied to the base of VT1 (marked “out”) and keeps it off. Capacitor C1 charges through R3.

In case of a short circuit (the resistance of R5 suddenly drops close to zero), the potential at “out” sharply decreases, which turns on VT1, and a potential of about 12 V is applied to the gate of VT2, turning it off.

Please let me know if this circuit will work. I feel like I'm missing something important.

And of course, I would be happy to hear any suggestions for improving the circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You're on the right track, but as soon as drain voltage drops a bit, the BJT can't really pull off the PMOS anymore, can it?

Put it "in front" instead. This makes the traditional "ring of two" current source. PMOS must be large and well heatsinked, since it'll be dropping a good 6A -- at 12V, that's 72W short-circuit!

The capacitor and other pull-down resistor aren't necessary (R3 is simply in parallel with the load, and C1 doesn't do much paralleled by R4, or more to the point, its value has to be quite large to do much of anything), however you can make a divider between current sense, base and output (Rfb into R3) that helps turn off VT2 as its voltage drop increases -- this is called foldback current limiting.

Since you show an inductive load, then if V2 will be switched, or VT2 by additional wiring to this circuit, consider adding D to protect against drain voltage avalanche.

Note that foldback gives the output a negative-resistance characteristic, i.e. current falls as voltage (across VT2) rises. This will oscillate with an LC load (the load is shown as inductive, but wiring capacitance, or the transistor itself, can serve as the capacitance). It would be prudent to place an R+C series snubber, say where D is (in parallel with it), with C > 3 times the expected transistor, diode and wiring capacitance (or just make it fairly large because who cares, 100nF say), and R < -negative resistance (probably 1-2 ohms would suffice).

Answer 2

The circuit you show is a current regulator, not a fuse. Once the output is shorted, VT2 is controlled to limit the current at about 5-6A, and it will dissipate accordingly (>60W), as long as the short circuit is present. It will quickly overheat and get destroyed.

Instead, you'd probably want a circuit that latches into the OFF state. To reset it, the power must be removed then reapplied, after the short circuit was removed of course.

^{simulate this circuit – Schematic created using CircuitLab}

R4 is the current sense resistor, R3-VT1 detect the excessive current, VT3 latches VT1 once it starts turning on. R2 limits the SCR ON current. C1-R5 add rudimentary immunity against transients. D1 protects against negative spikes generated by the load inductance.

Instead of a permanent latch, we can turn the load off for some fixed duration, then turn it back on - and off again if the short persists. This can be done with a monostable.

^{simulate this circuit}

VT3 and VT4 are inverters. The 2nd inverter's output is buffered from C2 and VT2's gate capacitance by the buffer VT5 and VT6.

C2 adds positive feedback that speeds up the output transitions of the latch.

To prolong the off time, increase C1 and C2 by the same factor, e.g. both by 5x, etc.

Answer 3

If you continue your analysis you will see, that during a shortcut of the output, closed SW1 here, the circuit enters a half conducting state of M1, where more than 90W will be dissipated.

When Q1 starts conducting, indeed the gate voltage of M1 and the current is reduced, but this also reduces Vbe of Q1. This is a metastable situation until M1 burns away, except this high power loss is intended and proper cooling is established.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 4

To confirm what @Jens was saying, here is the "utility" curve of the "protection"...

You can see why the "protection" is not "very effective".

you should add a pair of BJT (Q1 & Q2 = SCR) like these, more effective.
Q4 is not very "useful" (only affected by temperature).

Answer 5

Depending on \$R_{DS}\$ of VT2 when fully switched on, and the robustness of VT1's base-emitter junction, under an output short-circuit condition you might find the system in a state like this:

^{simulate this circuit – Schematic created using CircuitLab}

The potential at VT2's drain, node D, is very low. Transistor VT1, which is responsible for raising VT2's gate potential, can never raise it above D. It would need to apply closer to +12V to VT2's gate (so that \$V_{GS}\approx 0V\$) to switch it off, but it can't.

Transistor VT1 and sense resistor R1 should be on the source side of VT2, so that it's always possible to apply the fully supply potential to VT2's gate:

^{simulate this circuit}

Answer 6

Below is the LTspice sim of a simple electronic fuse circuit using just 1 PNP BJT and 1 P-MOSFET.

For R1 = 100mΩ, it triggers at about 6A current through the load, with the load current then dropping to zero (yellow trace). At no time is there significant steady-state power dissipated in the MOSFET, other than from the load current times the MOSFET on-resistance.

It is reset by removing the short.