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Zener over voltage protection

I have a board powered by a hand crank generator. Under normal conditions the voltage peaks at 18-20 volts. However spinning the handle violently will peak higher at 30+ volts. The generator runs through a schottky bridge rectifier and into a 5 volt regulator rated at a max of 26 volts. The circuit as a whole pulls less than an amp.

In trying to keep the simple circuit simple I was thinking about a 20 volt zener (D5) and a tiny 10 ohm series resistor(R1). When it comes to protecting a voltage regulator is this the most ideal method or should I be looking at something different?

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  • \$\begingroup\$ "peak at 30+ V ... max of 26 V". So don't do that. \$\endgroup\$ – Olin Lathrop Jun 29 '17 at 22:55
  • \$\begingroup\$ I'd be thinking of a slipping clutch on the crank, frankly. Sometimes mechanics beats electrics. \$\endgroup\$ – Ian Bland Jun 30 '17 at 0:10
  • \$\begingroup\$ This is a hand crank. \$\endgroup\$ – Ryan Mills Jun 30 '17 at 21:10
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Answer to the question in the comment below -

If you know the current and voltage you will expect, and want to dump the energy into a resistor instead of the zener, build you can build an emitter follower or source-follower-type circuit running off a small zener with a load resistor. I don't think I can draw a schematic in the comments so I will put in another answer.

Follower Circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Won't work. Unless you intend to remove the wire sticking out to the right and make R2 the load (ie using the MOSFET and zener as a linear regulator). \$\endgroup\$ – Oskar Skog Jun 30 '17 at 8:47
  • \$\begingroup\$ You are correct Oskar \$\endgroup\$ – John Birckhead Jun 30 '17 at 13:09
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    \$\begingroup\$ See my edit - I had a short between the headset. \$\endgroup\$ – John Birckhead Jun 30 '17 at 13:15
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A linear regulator is inappropriate here. Use one of the many many buck regulator chips out there. Pick one that can handle a bit more than the max possible output after the full wave bridge, and you don't have to worry about that part at all.

A buck switcher will be smaller and cheaper than a linear regulator after you include the cost and space of getting rid of the heat. Even at only 500 mA at 5 V out and 20 V in, a linear regulator will dissipate 7.5 W of heat. That's not trivial to get rid of, and it's not the worst case either.

Use a buck switcher rated for 40 V, and you're done.

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  • \$\begingroup\$ I agree the buck regulator is likely the more ideal candidate. I looked at it first but I need the circuit to run at 1.5 volts to 5 volts. I found one or two but they were cost prohibitive as a single package. \$\endgroup\$ – Ryan Mills Jun 30 '17 at 21:36
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Chances are you can put out more than an amp from your generator, but even if it is only 1 amp, your zener will get hot. You will have 20 volts times one amp or 20 W. This will blow any board mount zener. If you want to use a zener, it will have to be a chassis mount with a heat sink. It should work fine, however.

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  • \$\begingroup\$ Hitting the peak would be a very short duration not sustained (in the millisecond range). I had a 5 watt zener in mind. I will check the real world amps. Is there a reason you don't add a series resistor inline with the zener to take some of the load? Did not see that in any of the examples I found. \$\endgroup\$ – Ryan Mills Jun 29 '17 at 22:39
  • \$\begingroup\$ If you know the current and voltage you will expect, and want to dump the energy into a resistor instead of the zener, build you can build an emitter follower or source-follower-type circuit running off a small zener with a load resistor. I don't think I can draw a schematic in the comments so I will put in another answer. \$\endgroup\$ – John Birckhead Jun 29 '17 at 23:57
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I would replace the passive regulator with a simple buck converter. It will make more power from the generator available to your load and it will reduce the thermal dissipation. You will find that the component count will approach that of your linear regulator design.

If this is indeed powered by a generator, you can also eliminate the full wave Schottky bridge.

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  • \$\begingroup\$ It's plan B, the VR turns on at 1.5 volts, something that's very important. The VR works perfectly as is. But in very rare circumstances its getting peaked too high. Its not so much of a step up/down issue as avoiding the peak. \$\endgroup\$ – Ryan Mills Jun 29 '17 at 22:32
  • \$\begingroup\$ Why is there a bridge rectifier in the circuit? \$\endgroup\$ – Glenn W9IQ Jun 30 '17 at 2:06
  • \$\begingroup\$ @GlennW9IQ: Because the generator produces AC, not DC. Unless you know that the proper term for this is "alternator". OP might not know that. \$\endgroup\$ – Oskar Skog Jun 30 '17 at 8:45
  • \$\begingroup\$ For clarity this is a DC motor, however the direction you spin the handle changes polarity. It is hand cranked so there is possibility of it going both ways but I need to maintain polarity on the board. \$\endgroup\$ – Ryan Mills Jun 30 '17 at 21:14

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