# Crowbar Circuit for 35V OVP - What parts should I use?

Please excuse my confusion, because I really don't understand what the "150mA max continuous cathode current" means on the datasheet for the LM431. I'm also not sure it'll work in the crowbar circuit I intend to build. (I suspect it will, seeing as it's rated for 36V).

I have the following circuit diagram, based on a Wikipedia description of a crowbar circuit:

simulate this circuit – Schematic created using CircuitLab

D3 leads to a(n) LM3481 used as a SEPIC (Pg 26) to output 7V for a load that draws between 500mA to 1A (1.25A to be on the safe side).

This leaves me with a number of questions, assuming I want the crowbar to reject (short out) input voltage higher than 35V:

1. If the input voltage at D1 is higher than 35V (unlikely, but possibly as high as 40V), will the LM431 become damaged or will the fuse burn out first?
2. Will a polyfuse/PTC work for F1? What part type/specifications should I look for?
3. To what should I connect the wiper for R1, GND? Does R1 need to be a POT or can I use a fixed resistor?
4. What are the values for R1, R2 and R3?
5. DI1 is a TRIAC. What specifications should I look for?
6. Does the 150mA mentioned in the data sheet have anything to do with the load after D3? What is it actually telling me?
7. I know that I can use Vout ~= (1 + R1/R2)Vref to work out what R1 and R2 should be, but I don't know how I work out Vref. Is that Vin from D1 or from the LM431?

I have seen this as a proposed solution, but apparently the MOSFET doesn't short to ground (according to one of the comments).

• Asking "what part(s) should I use?" is a good way to get your question closed as off-topic. Try asking something more like "what specs should I look for?" instead. – Hearth Dec 1 '18 at 15:53
• Is your goal to blow the fuse in the case of overvoltage? Do you want it to auto-reset after some specified time? Or is it good enough to have to replace a fuse each time? – jonk Dec 1 '18 at 19:54
• @jonk Yes, the goal is to blow the fuse in the case of overvoltage (> 35V). Auto-resetting after a few seconds would be good, but I don't mind if the fuse has to be replaced once blown. – Agi Hammerthief Dec 3 '18 at 6:50
• @AgiHammerthief Added something that may be acceptable. – jonk Dec 3 '18 at 7:54

## 1 Answer

Something like this, then?

simulate this circuit – Schematic created using CircuitLab

$$\Q_2\$$ is a big BJT easily capable of causing the fuse to burn out in the case of an over-voltage situation. (You could try something smaller, of course. But this BJT is way more than you need to make sure.)

Here, $$\R_1\$$ and $$\R_2\$$ are used to set the threshold voltage to trigger on. I've set the threshold just a little high, just by way of an example. But you can adjust these, as needed, to hit the point you want.

$$\R_3\$$ is designed to provide about $$\7-8\:\text{mA}\$$ for the base of $$\Q_1\$$. Assuming $$\Q_1\$$ is driven into saturation, this should still mean at least $$\75\:\text{mA}\$$ into the base of $$\Q_2\$$ (and probably a lot more than that.) I might consider adding a base resistor for $$\Q_2\$$, but since the point is to blow the fuse I'm not sure it's really needed. If you do add it, then something like this would be fine:

simulate this circuit

That should blow the fuse.

The main thing here is to adjust the values of $$\R_1\$$ and $$\R_2\$$ to get the trigger voltage you want. Or, use a $$\10\:\text{k}\Omega\$$ potentiometer in there, as appropriate, to allow an adjustable setpoint.

The voltage across the LM431 shouldn't exceed your trip-point setting, less a diode drop or so. You could add more circuitry to make absolutely certain to protect the LM431. But this arrangement may be okay for your needs.

• Thank you; much appreciated. Any suggestions on what the specifications for F1 should be? – Agi Hammerthief Dec 3 '18 at 9:42
• @AgiHammerthief I think you'd suggested $1.25\:\text{A}$. That sounds fine to me. You could select anything from $1\:\text{A}$ to $2\:\text{A}$ (or more.) $Q_2$ in the circuit can sink a LOT of current on its own. And it will, if permitted. – jonk Dec 3 '18 at 17:53