Time microcontroller can work without supply voltage

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Dknot--

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Hello guys,

I have relatively exotic question, and I did not find answer for that from internet. So, If one has some small gaps in supply, for example 20us gaps where supply voltage goes from 3.3-->0V, is functioning of micro controller disturbed? What is maximum time gap microcontroller, for example ATMEGA128, can handle without problems occuring? How could this kind of maximum gap be calculated, at least in principle?

THanks already for aswers,

-Dknot--
 
It is really so exotic a question that you aren't going to get a straight answer from me.

My attitude is that if you want the microcontroller to continue to function, assume that you have to keep the power on all the time. A small and cheap capacitor, which should be there anyhow, will keep the microcontroller running.

If you want it to reset on very short interruptions, the reset line will do that for you. The data sheet will show you the minimum reset time required to guarantee a reset.
 
as you can see from page 320 operation frequency is function of Vcc. if you work on 8MHZ clock you
can feed cpu with 2.7V, below this voltage it will not work properly.
If you know that you your power supply falls down to some time (few uS or mS) you can put capacitor that will supply current/voltage to cpu in this time.
 
Hello,

I have to agree with the other two replies in that if you want any controller to work when the main power source goes away for very short periods like that then you have to at least add a capacitor that will hold the voltage up for at least that long of a period. The min size of the capacitor can be estimated from:
dv=i*dt/C
where
dv is the change in voltage (decrease in this case) in volts, and
i is the average current drawn from the controller chip and associated parts,
dt is the time duration of the missing voltage,
C is the size of the capacitor.
Solving this equation for C we get:
C=i*dt/dv

Example:
The power goes out for 20us while it is at 5vdc and the minimum voltage the processor can take is 3vdc and it draws 5ma average while running normally.
Since the change in voltage can only be 2v that means dv=2, and the time is 20us so dt=20e-6, and the current is 5ma so i=0.005, so we have so far using the equation above:
C=0.005*20e-6/2

and doing the math on the right side we get:
C=0.00000005 farad

which is the same as 0.05uf, and double that for some margin of error in the cap value we get 0.1uf as the required capacitance.

Usually the drop out period would be longer though, like 20ms rather than 20us, and for that we would need:
C=I*dt/dv
C=0.005*0.020/2=0.00005 farad, and double that we get 100uf capacitance required.

Also, if you have a voltage regulator in the circuit it helps a lot if you put the capacitor on the input side of the regulator so that the output of the regulator maintains the correct voltage for the whole time. With the cap on the output side, the voltage will still sag a little bit which may through some of the AD conversions off depending on what you are using for a reference voltage for the controller chip.
 
Thanks for answers!

I'm quite disappointed to read that I should put capacitor in front of the micro controller.

My problem is following: I'm making analog low power consumption (30uW) switch circuit, which gives power with certain (complex) conditions to microcontroller after source voltage has died. So, voltage source has to go to the zero before this switch circuit is triggered.
Because of the strict power consumption demand, I can't put capacitor as a battery to the microcontroller, because I should use diode etc. to separate microcontroller from switch and voltage source (switch would not trigger!). Because of the voltage drop over diodes consumes power, I can't use it!

Would you have any guesses how to solve problem of having some extra source of power for 20us, which is the turning ON-time for my switch? I can't turn my microcontroller off for that time, because it is making calculations then.
Only demand for this is, that is does not consume almost any power when it is not used. It is used only about 20us.

I know this sounds like a bad and complex design, but I have to do my circuit this way.

Hopefully you understood what I mean,

-Dknot--
 

The diode doesn't consume any power, it simply makes a voltage drop across it - a diode and capacitor is the standard (and sensible) solution, it won't increase power consumption (and may even reduce it slightly).
 
Dknot,


You should be able to use a Schottky diode and capacitor to separate the main supply from the micro controller. The diode will drop a voltage yes, and that will take power yes, but that diode will also REDUCE the voltage to the micro controller thus reducing it's power consumption. In the end the power requirement should be the same or even a little less if the micro controller voltage/current curve current goes down with a decrease in voltage.
The only concern is that the diode will introduce a temperature variation in voltage to the controller chip so you'll have to evaluate that and see if it works out ok or not.
Unless of course you meant to say all along that the voltage drop itself is a problem, in which case i agree it can be a problem so you'll have to decide. There might be another way to do it too.
 
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You may be able to use a ideal diode circuit such as this to isolate the input from the microprocessor and capacitor.
 
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Hello again,


Yes, if he could get that to work with his app. Also look for a low voltage (10v or under) Schottky as they can have very low Vf too 0.2v or less.

If we could see more of the real circuit i think we could help more too.
 
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Hello,

thanks for your answers. Actually, your propositions helped me a lot, and I was able to figure out how to do my design! When my design is ready, I'll show the circuit to you ^^
 
Hello again Dknot,


That's great...it's always nice to see a circuit that someone else is doing and compare notes. Will be looking forward to seeing it..
 
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