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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asked Oct 7, 2010 at 17:57
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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\$– endolithOct 7, 2010 at 18:42
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\$\begingroup\$ Yes that is what I am looking for. \$\endgroup\$– PICyourBrainOct 7, 2010 at 19:04
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10\$\begingroup\$ I'd call that an "undervoltage lockout". "Regulate" isn't really the right word. \$\endgroup\$– endolithOct 7, 2010 at 19:31
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\$\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\$– user3662759Jun 19, 2014 at 16:40
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\$\begingroup\$ Im thinking just a Voltage divider and MOSFET. hard to find the right mosfet though. any help appreciated. thanks in advance \$\endgroup\$– user3662759Jun 19, 2014 at 16:58
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3 Answers
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According to your description, something like this should work:
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". :)
Similar:
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.
answered Oct 7, 2010 at 19:04
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5\$\begingroup\$ It's very unlikely you'll find something passive that does what you want. \$\endgroup\$– Jason SOct 7, 2010 at 19:22
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1\$\begingroup\$ the chip Endolith mentions has an internal vref generator. \$\endgroup\$– Thomas OOct 7, 2010 at 20:00
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1\$\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\$– JustJeffOct 7, 2010 at 22:10
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3\$\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, 2011 at 23:14
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1\$\begingroup\$ @Matt: The question doesn't say anything about batteries, but hysteresis can be added with a few resistors \$\endgroup\$– endolithJul 6, 2011 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.
(graph is actually from an MC34064 datasheet, but would look similar for the mAX809)
answered Jul 16, 2011 at 14:45
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You can connect a push/pull voltage detector (PMIC supervisor) such as the Microchip MCP112 or TC54 to a logic-level MOSFET, e.g. as described in the TC54 datasheet:
(page 6)
There are some popular thresholds versions of these parts, e.g. MCP112-300, MCP112-315, TC54VC30. So perhaps the nominal 3.15 V tripping point is sufficient for your application. In case it isn't you can look into getting a customized version or use one with V_trip=3V and voltage divide V_In as described in the TC54 datasheet:
(page 6)
For example with V_trip=3V and a target threshold of 3.2V you could chose R1=1780 Ohm and R2=26700 Ohm).
In comparison with the MAX809, the above voltage detectors are more efficient (e.g. I_DD_typ=1µA and I_DD_max<4µA instead of 17µA and 50µA) and implement hysteresis.
