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What is the simplest way to regulate the min DC voltage in a circuit? Is it possible to do with zener diodes?

Desired performance:

Input > 3.3 VDC     Output = Input
Input = 3.3 VDC     Output = Input
Input < 3.3 VDC     Output = 0.
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  • \$\begingroup\$ So if the voltage is >= 3.3 V, it should pass through, but less than 3.3 V, it should be disconnected? \$\endgroup\$ – endolith Oct 7 '10 at 18:42
  • \$\begingroup\$ Yes that is what I am looking for. \$\endgroup\$ – PICyourBrain Oct 7 '10 at 19:04
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    \$\begingroup\$ I'd call that an "undervoltage lockout". "Regulate" isn't really the right word. \$\endgroup\$ – endolith Oct 7 '10 at 19:31
  • \$\begingroup\$ Could this be done with out the comparator or any other chip. I dont need a sharp cutoff and trying to make the circuit as simple as possible \$\endgroup\$ – user3662759 Jun 19 '14 at 16:40
  • \$\begingroup\$ Im thinking just a Voltage divider and MOSFET. hard to find the right mosfet though. any help appreciated. thanks in advance \$\endgroup\$ – user3662759 Jun 19 '14 at 16:58
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According to your description, something like this should work:

undervoltage cutoff

The TLV3012 has the reference voltage built-in, so that's 1 6-pin IC, 1 FET, and 2 resistors. Not sure if you consider that "simple". :)

TLV3012 pinout

Similar:

Similar circuit based on this IC

Using a pMOS to switch a load is summarized here.

You could probably use a Zener as the reference for the comparator, too, or maybe rig something up to drive the MOSFET from the Zener more directly, but I don't know how precise that would be.

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    \$\begingroup\$ It's very unlikely you'll find something passive that does what you want. \$\endgroup\$ – Jason S Oct 7 '10 at 19:22
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    \$\begingroup\$ the chip Endolith mentions has an internal vref generator. \$\endgroup\$ – Thomas O Oct 7 '10 at 20:00
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    \$\begingroup\$ check out the '1381 voltage trigger' - it's a tiny (TO-92) 3 terminal package that does everything to the left of the PMOS depicted in the 1st schematic \$\endgroup\$ – JustJeff Oct 7 '10 at 22:10
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    \$\begingroup\$ It doesn't sound like this circuit is to protect a battery from overdischarge, but if it was, the circuit in the diagram wouldn't work. When the battery gets disconnected from the load, its voltage will rise somewhat as it "recovers". The comparator would then turn the load on again. The effect will be that the load oscillates on and off until the battery is very thoroughly discharged to that voltage level. If left long enough this can destroy the battery even though you were trying to protect it. The circuit needs some kind of hysteresis or latching to prevent this problem. \$\endgroup\$ – Matt B. Jul 5 '11 at 23:14
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    \$\begingroup\$ @Matt: The question doesn't say anything about batteries, but hysteresis can be added with a few resistors \$\endgroup\$ – endolith Jul 6 '11 at 1:56
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If the voltage doesn't have to be quite 3.3V you could use a MAX809 reset circuit. Exists in several voltage versions, closest is 3.08V, custom voltages are possible.
For low power applications (< 20mA) you can use the output directly to power your circuit, otherwise use the output to drive a low \$R_{DS(ON)}\$ MOSFET.

enter image description here

(graph is actually from an MC34064 datasheet, but would look similar for the mAX809)

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