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Analog Logic Circuit with Timer/Delay, need some help


I have a processor which needs to be powered on by pulling a pin LOW for 1 second.  I would like to set the system up so that whenever a battery is present, the processor turns on.  I have a programming background so I am having a little trouble turning ths into an analog design.


Here is what I have:

3.7V battery input (Li-ion), 900 - 1100mA, call it BAT

ON# pin, which needs to be pulled LOW for 1 second.

POWRMON pin, which is HIGH when the unit has been turned on OR ON# pin is pulled LOW.


So what I need is logic that says

if(BAT & !POWRMON) { ON# = LOW; pause 1; }

I could use a logic IC and a timer, but I feel like I should be able to do this with just a few analog components.


Can anyone help out?  thanks!


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Hello Lucasvicke,

So you have two pins, one which needs to be high (powermon) and one which needs to be low for 1 second (on#), ignoring Batt.

(I assum the unit is battery powered only).

So if I understand correctly the sequence you need is:

1.   Power on (batt connect....)

2.   PowerMon to High (power good)

3.   On# low for 1 second then High 

Is this correct? Does it matter if both the PowerMon and On# start at a low or do you need the ON# High/Low transition after power good? What processor is this?

What you want to do is similar to the treatment of reset pins and is normally done useing a resistor and a capacitor, 1 second is an unusually large time.

I assum the On# pin is designed to be connected to a push button. If you can clarify the items above we should be able to do something.

Hey Guys,

cwingnell:  You are close. It will only be battery powered yes. The POWERMON line will be low unless the unit is turned on.  What I expected is that I would pull the ON# pin LOW for 1 second, release, and then the PowerMon pin would go HIGH.  It turns out the second I pull the ON# pin LOW the POWERMON pin goes HIGH as well.  Still I need ON# low for a full second for the unit to turn on.

I figure I can use a simple CMOS NOR chip.  I wanted to do the logic analog but from what I read this is just easier.  Anyhow it will be true if I have BAT and no POWERMON.  It will also be true if I have POWERMON and no BAT, but since it's powere off BAT I don't have to worry about it.  I now have the trigger I need.  Next I just need the reset circuit that  OddBot is describing.  I can use that circuit to trigger a transstor which will pull the ON# pin low.

The auto reset circuit you described makes sense to me.  The only thing is the second it has charge enough to pull the  #ON LOW, the POWERMON will go high (even before system is 'on'), and then the capacitor will start to discharge, very shortly after the ON# pin will go high again, and the system will not have been restarted.

It's a cellular chip, I guess that's why it has such a slow on time (user would be pushing a holding a button)

this isn't the analog idea I was going for, but looks like it's a pretty low profile chip that does what I need.



... hmm no that won't work the MR pin needs a falling edge to trigger.  I can't get that at startup (POWERMON will just always be low)

Hello Lucas (I presume),

From your description you do not need to set POWERMON, it seems to be controlled by the chip, so all you need to do is to pull ON# low for 1 second, if this is the case the standard reset circuit will do the trick for you (that is what OddBod was describing).

Basically you have a capacitor (we will call this C1) between the ground and the ON# pin, this capacitor is charged by a resistor (R1) and a push button (Sw1) is wired between the ON# pin and the ground, so that when the button is pushed the C1 is discharged to deck.

When the button is released C1 will start charging via R1.  Now this particular arrangement has something called a Time Constant which is simply t = RC, which is the time in seconds the combination will need to charge to ~65%, this will be close to what you Processor will recognise as a High.  Now if I have done my maths correctly with a 10uf cap (C1) you are going to need a 100K resistor (R1). One last thing is good practice is normally to put a smallish resistor between C1 and the ON# pin 500 to 1k ohm is enough, this protects the processor from the transient currents and voltages.


                        I                                   I

                       Sw1                            C1

                        I                                   I


Hopefully that will do it for you, you may need to fiddle with the R1 value to get the exact time you want.

will give this a shot, thanks guys