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.