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I'd like to build a low-battery circuit. It should default on when first connected, but if a particular signal goes high for a small amount of time (say, 10 microseconds,) then it should turn off, and stay off. It's the "default on" that gives me trouble. The closest I can get is something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

From null, "OFF" is floating. Supply 15V, and the capacitor starts out acting like a bit of wire, which pulls the M1 N-channel gate high, which makes the N-channel conduct (I'm likely to use a BS170, not the IRF on the schematic.) That pulls the gate of the P-channel M2 low, which makes it conduct. That, in turn provides voltage into R3/R2 divider that keeps the circuit latched open.

Now, when OFF goes low-impedance to ground, it will turn off M1, and thus R1 will pull the M2 gate high, and M2 will turn off. Because C1 already has a charge, the system will now stay with M2 turned off even if the OFF signal goes high impedance again (which it will, because it's a MCU output eventually powered by the SWITCHED voltage.)

At this point, the system is unlikely to turn on again unless I short "OFF" to "+15V." In fact, I may add a pushbutton to do just that for manual turn-on.

While off, the leakage through C1 and the leakage through R1/M1 (and, I guess, leakage through M2 and the entire powered subsystem) will still draw a little bit on the battery, so if it's a LiPo, I shouldn't leave it in that state forever or it will die, but as a safeguard for a hobby robot, I think it might work.

Is this correct, or am I getting something wrong? Will the system oscillate when the "off" is triggered? Will the system automatically turn on from "zero" as intended?

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    \$\begingroup\$ It's vulnerable to any other sources of voltage on "Switched" (such as decoupling capacitors or leakage paths from other logic) turning M1 back on via R3. \$\endgroup\$ – Brian Drummond Apr 28 '13 at 8:34
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    \$\begingroup\$ Is that the correct symbol for M2? It's P channel and shouldn't the arrow be pointing away from the channel? Isn't it also a source follower which means it won't switch on very well? \$\endgroup\$ – Andy aka Apr 28 '13 at 13:35
  • \$\begingroup\$ I think M2 may be flipped; you're right. CircuitLab uses the reverse symboks from what I'm used to. The source and drain should be swapped (unless I'm using it as reverse voltage protection :-) I think the R3 + R2 impedance (and assumed load impedance) is high enough to not turn M2 into a source follower, but if I'm wrong, please show me the math! \$\endgroup\$ – Jon Watte Apr 30 '13 at 22:13
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schematic

simulate this circuit – Schematic created using CircuitLab

Q1 and Q2 together are actually an SCR, but CircuitLab doesn't have an SCR symbol. This is essentially a crowbar circuit, with the input left unconnected.

F1 should be some sort of current limiting device. You said "stay off" but didn't specify for how long. A fuse will stay off for a long time. Replace this with a resistor and it will stay off until V1 decreases enough for the SCR to unlatch.

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  • \$\begingroup\$ "Stay off" means in the conventional sense, where no current is drawn from the voltage source. Crow-baring it down by burning up the battery is not within the solution space :-) \$\endgroup\$ – Jon Watte Apr 30 '13 at 22:12
  • \$\begingroup\$ @JonWatte this doesn't burn up the battery: it just blows the fuse. If this is a protection thing, and you don't expect the battery to run low enough for this to happen normally, maybe this is a good thing. With the fuse blown, there's no current drawn from the voltage source: not even a small leakage bit. \$\endgroup\$ – Phil Frost May 1 '13 at 2:37
  • \$\begingroup\$ Fair point. I think in this use case, getting to bottom would be reasonably common. Thanks for the suggestion, though! \$\endgroup\$ – Jon Watte May 2 '13 at 6:18

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