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).

  • \$\begingroup\$ 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. \$\endgroup\$ – Hearth Dec 1 '18 at 15:53
  • \$\begingroup\$ 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? \$\endgroup\$ – jonk Dec 1 '18 at 19:54
  • \$\begingroup\$ @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. \$\endgroup\$ – Agi Hammerthief Dec 3 '18 at 6:50
  • \$\begingroup\$ @AgiHammerthief Added something that may be acceptable. \$\endgroup\$ – jonk Dec 3 '18 at 7:54

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.

| improve this answer | |
  • \$\begingroup\$ Thank you; much appreciated. Any suggestions on what the specifications for F1 should be? \$\endgroup\$ – Agi Hammerthief Dec 3 '18 at 9:42
  • \$\begingroup\$ @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. \$\endgroup\$ – jonk Dec 3 '18 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.