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Apologies before I dive into the question. This is the best title I could think of for a more complex question.

Is it possible to have a timer circuit that consumes no power when quiescent? I mean that the circuit draws no power whatsoever, until a button is pressed momentarily. The button press might only last 0.1 seconds. Following the actuation, the timer outputs a 100 mA signal for say 5 seconds. The timer then reverts to it's quiescent state again.

This is for:-

I have a large posh antique door bell push. I don't want to replace it with some plastic thing from Taiwan. Unfortunately it's very prone to contact oxidation, so the electrical operation is very erratic. I want to operate it from batteries only. Hence I'm looking for zero current drain whilst not operating. Due to the electrically noisy contacts, I thought of using some form of monostable based pulse stretcher plus capacitor thingie. Upon pressing the button, the timer would operate a loud electro mechanical ringer, then stop.

This circuit arrangement seems possible to me, but it's realisation remains just out of reach. Perhaps something with relays? Is this actually possible, or would I be violating some fundamental law of logic?

Edit: I'm looking for a simple electrical solution to this with such as relays, capacitors, monostables, e.t.c. I don't have access to a micro controller development environment, nor any credible C skills.

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  • \$\begingroup\$ Replace the guts with the guts of a modern one? \$\endgroup\$ – Passerby Sep 26 '15 at 2:08
  • \$\begingroup\$ Bit of google suggests digikey.com/en/product-highlight/t/texas-instruments/… \$\endgroup\$ – pjc50 Oct 22 '15 at 21:18
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    \$\begingroup\$ The other approach is to have a relay that closes while the timer is running. then opens leaving the circuit entirely disconnected. \$\endgroup\$ – pjc50 Oct 22 '15 at 21:21
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It's possible to have a timer that has significantly less current draw than the self-discharge of a given battery, so I think we can it's possible to do what you want, though technically the draw may not be exactly zero. It's even possible to use a standard microcontroller as the timer and achieve that low a current draw.

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  • \$\begingroup\$ Even Lithium-SOCL2 have a shelf-life (10 years.) A simple micro that self-latches power when started up (from existing pushbutton) and turns a pin high for 1 second sounds plausible. Zero consumption while idle. \$\endgroup\$ – rdtsc Sep 26 '15 at 2:52
  • \$\begingroup\$ @Spehro-Pefhany So you just recommend a low power 555esque timer? \$\endgroup\$ – Paul Uszak Sep 26 '15 at 20:08
  • \$\begingroup\$ Probably not, depending on the battery. They're pretty power hungry, even the TLC555. There's not really enough info to make recommendations, just to say that it is possible. \$\endgroup\$ – Spehro Pefhany Sep 26 '15 at 21:19
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Here is kind of a sketch. Something like this might work. When you push the button, the Q1 turns on, and the cap C1 gets charged up. Q1 will stay on for some time after the button is released because C1 will be discharging through its base. As long as Q1 is on, it will keep the PMOS (M1) on, which will power the load. You can fine-tune the duration by changing C1 and R2.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Oh, and the standby power is pretty much zero, because everything is off. \$\endgroup\$ – mkeith Dec 27 '17 at 0:44
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My mircowave oven is old enough to have a schematic inside a service panel. It looks something like this (some things not shown for clarity):

schematic

simulate this circuit – Schematic created using CircuitLab

The Timer switch is attached loosely to the Motor shaft to allow the user to start simply by turning the knob. When the knob gets back to "home", it turns off.

This is probably the kind of architecture that you're looking for. Maybe a crystal- or RC-based timer instead of a synchronous motor, and maybe a relay or MOSFET in parallel with the flaky power button, something like this:

schematic

simulate this circuit

U1 is an open-drain comparator, which you can get two of in an 8-pin DIP package. Two rows of 4 pins on a 0.1-inch grid.

  • When you push the button, it provides power to the rest of the circuit. C1 is completely discharged at this point (0V) and takes a while to charge via R4, but R2/R3 jump immediately to 1/2 the supply. In this state, the comparator outputs low, which turns the FET on, which keeps the circuit running when you let go of the button.
  • Eventually, C1 will charge enough to pass the level of R2/R3, and the comparator will let go of its output (it doesn't output high because it's open-drain). R1 then pulls the FET off, which kills the entire circuit.
  • C1 then discharges through D1 and the load, but remains higher than R2/R3 so that the comparator never flips while the circuit powers down. D1 should be a shottkey type to avoid discharging C1 through the comparator's input protection diodes instead.

SOT means Select On Test. It basically means that I don't know the correct values and so you'll have to determine that by experimentation. Bigger R4 and bigger C1 both mean longer time. I would use an absolute minimum of 1k for R4, preferably more like 10k to 100k.

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  • \$\begingroup\$ Something like this eleccircuit.com/off-after-delay-switch-by-mosfet I think this might be what you mean, but I'm having trouble reading the English... \$\endgroup\$ – Paul Uszak Jul 10 '16 at 21:53
  • \$\begingroup\$ Close, but two problems: 1) As drawn, it won't start when it gets power, but needs a button to do that. I guess you could swap the button and the capacitor to make an auto-start, then use the button as an early stop. 2) The RC timer will go slowly through the switching region of the FET so that it's somewhere between on and off for a long time. That's going to be hard on both the FET and the load. \$\endgroup\$ – AaronD Jul 10 '16 at 23:41
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This 74LVC1G123 monostable has a typical quiescent current of 0.1 µA (that's 100 nA), and a maximum of 10 µA. 100 nA is pretty close to zero power.

With two AA batteries (~2000 mAh), even with the maximum quiescent current value, they would last something like 22 years, way beyond the life and self-discharge of the batteries. With the lowest power 555 I could find (maximum 250 µA quiescent current), the the batteries wouldn't last quite a year.

Neither of the figures above take into account the occasional current drain required when the timer is running; if the timer operated 12 times a day, this would eat up 1.6 mAh, or 584 mAh per year which would limit the batteries to about four years. If the average number of times the timer runs is only four times per day, then you would theoretically get three times as long (but once again, probably beyond the life of the batteries).

So you can see with this lower power device, the quiescent current is not worth considering, whereas with a low power 555 the quiescent current uses a significant fraction of the batteries' power.

The 74LVC1G123 is also much more accurate than the 555, so you don't have to experiment with different component values to get the timing right. The 74LVC1G123 can only source 24 mA, so you will need a BJT or N-channel MOSFET as a low-side switch to operate the 100 mA device.

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  • \$\begingroup\$ I agree with @tcrosley. I would add that the CD4538 has a typical quiescent current of 5nA and, if necessary, will run at higher supply voltages. \$\endgroup\$ – henros Oct 31 '18 at 10:19
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See this interesting design: https://makezine.com/projects/the-greenest-delay-timer/

It uses a relay that switches on both the load and the timer itself. When it times out, the timer circuit is disconnected from the battery.

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    \$\begingroup\$ Can you please add some more explanation of what's going on with the linked project? Links rot... \$\endgroup\$ – ThreePhaseEel Dec 26 '17 at 19:38

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