Hello
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!
Maybe a monostable (one
Maybe a monostable (one shot)?
Analogue Delay
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
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
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.
http://datasheets.maxim-ic.com/en/ds/MAX6467-MAX6468.pdf
… 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)
Analog Delay
Hello Lucas (I presume),<o:p></o:p>
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).<o:p></o:p>
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.<o:p></o:p>
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.
ON#-----R2---------------------------------------R1-----------+5<o:p></o:p>
I I<o:p></o:p>
Sw1 C1<o:p></o:p>
I I<o:p></o:p>
--------------------------------------------------Gnd
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,
will give this a shot, thanks guys
