I am running into problems creating a circuit that once a button is pressed ( clicked ) will latch and run for a few seconds and then turn it self off. This would be trivial with a MCU but now I am trying to do it with a monostable 555 circuit.

The idea is that I have a high-side mosfet acting as a switch, this is closed by the button. This should start the 555 timer circuit and now the switch would be kept closed by the timer output. And when the timer output is low again, the circuit will be inactive. This is the circuit I am trying to simulate using LTspice:

The timer is working if simulated, but now the simulation hangs as soon as the voltage controlled switch is triggered. Do anyone have any better idea how to achieve this functionality or how to get the simulation working?

Edit: I am trying to keep the power consumption to a minimum when the circuit is inactive, minimum meaning nano-Amps. The current draw when active is also good if it is low. But mostly the circuit will be off.

  • \$\begingroup\$ At a quick glance it looks to me that you are powering down the 555 timer with the top MOSFET. i.e., You are sawing off the branch you are sitting on. Is this your intention? \$\endgroup\$
    – Transistor
    Commented Nov 28, 2017 at 11:47
  • \$\begingroup\$ My intention is to shutdown the entire circuit after a delay after activated. So yes, I want the top pmos to shut down the circuit, but only after the 555 timer has put its output low. \$\endgroup\$
    – Kristoffer
    Commented Nov 28, 2017 at 11:57

2 Answers 2


As you say, the obvious and simple way is to do this with a small microcontroller, like the PIC 10F200.

However, if you really want to do this with analog electronics, it would be simpler to use a transistor rather than trying to somehow fit the evil 666 555 timer into this role:

R4 and Q2 is basically the same thing you already have for switching the power. Q1 is similar to your M1 in that it activates Q2. The difference is that Q1 is activated directly from the switch, and turns off according to the decay time of the voltage on C1.

When the switch is pressed, C1 is charged up quickly thru R2. R2 is only there to avoid excessive current thru the switch when it would otherwise short a discharged capacitor across power.

R1 causes the voltage on C1 to decay exponentially towards 0. While that voltage is about 600 mV or more, Q1 is kept on enough to pull its collector low, and thereby turn on the power switch, Q2.

This circuit will turn on quickly, but fade off over a few 10s to 100s of milliseconds. If that is acceptable, then there is nothing further you need to do. If you need snap-action, then a little hysteresis is in order. That could take the form of some AC feedback from the drain of Q2 to the base of Q1.

Current drain

The issue of the current this circuit would use was raised in a comment.

Look at the circuit carefully, and you will see that it uses very little current, especially compared to a 666 timer. When the switch (Q2) is on, the dominant current drain is thru R4. This should be obvious just from the values of the resistors. If V+ is 5 V, for example, then the current thru R4 will be less than 50 µA.

The current to keep Q1 on comes from the one-time pulse of current thru the pushbutton to charge up the timing cap. After the pushbutton is opened, no more supply current is used to keep the power switch on. Other than the initial inrush to charge C1, the steady state current with the pushbutton closed would be another 50 µA thru R1, and less than 1/10 of that thru R3.

Now compare that to the original proposal. The power voltage is kept across both R4 and R5 while the power switch is on, and this is before even considering the current to run the 666 timer.

In short, it should be quite obvious from even a cursory inspection that the circuit above draws considerably less current to keep the power switch on than the original circuit.

  • \$\begingroup\$ I was thinking in the same before, and i like the idea since it has less components. However, this requires me to bias the bjt. Wouldnt this draw some amount of current. My idea with the 555 timer was that the current consumption would be in nA when not active, the consumption when active would also be low, the 555 is available in 100uA versions now. \$\endgroup\$
    – Kristoffer
    Commented Nov 28, 2017 at 13:00
  • \$\begingroup\$ @Kris: Seriously!? See addition to my answer. \$\endgroup\$ Commented Nov 28, 2017 at 13:30
  • \$\begingroup\$ Yes, The worry is the current consumption when not active, which in this case would mean ~50uA. Way to much. The load is going to consume ~2mA when active. I can however use your circuit behind a high-side switch to reduce current when inactive, in the same manner I do with the 555 circuit. \$\endgroup\$
    – Kristoffer
    Commented Nov 28, 2017 at 14:15
  • 1
    \$\begingroup\$ @Kristoffer the circuit uses zero current when off. it uses 50uA when on. \$\endgroup\$
    – Trevor_G
    Commented Nov 28, 2017 at 14:17
  • \$\begingroup\$ When simulating this the bjt constantly draws 14uA \$\endgroup\$
    – Kristoffer
    Commented Nov 28, 2017 at 15:30

Here is your circuit, modified a bit, seems to work. Sorry, it's a bit messy.

enter image description here

Simulation says 20nA when off and 1.8mA when on (basically the leakage of the MOSFETs, especially M2 in this case).

I used the pulse generator V3 and M3+R4 to simulate the pushbutton, you would just need to connect the pushbutton in place of M3. If you hold the pushbutton down longer than the set time it would remain on until the switch is released then immediately turn off when it is released.


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