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I was just thinking theoretically on how long I would be able to run a circuit on a capacitor.
I'm wondering - is my approach for this approximate calculation correct?
Let's say the circuit consits of one Arudino (5V) and one other sensor (5V). Let's say that the total current draw of the Arduino + Other Sensor is 150mA at 5V (asssume constant draw)
Now, I model this cirucuit as a resistor of resistance R = (5V)/150mA = 33ohms.
Next, I choose a capacitor of capacitance 1F with a rating of 10V. Now, I can model my entire circuit as an RC circuit with a 33ohm resistor and a 1F 10V capacitor. The time constant of this circuit is thus (33)(1) == 33. Thus, in a discharging configuration, the capacitor will deplete by 63% of it's total charge in one time constant (33s).
Next, I consider the fact that my circuit needs 5V to run and thus I cannot utilise the full capacity of my capacitor. I assume that I will be using some sort of LDO that provides stable 5V to my circuit for any voltage of the capacitor above 5.3V (5V + LDO drop out).
By the equation Q = CV - if it takes 33s to deplete charge on the capacitor to (100-63)=36%, then it will also take 33s to reduce the voltage across the capacitor from 10V to 36% of 10V which is 3.6V.
By a rough estimate then, I assume I'll get somewhere between 10-20s of usable time.
Is my thought process correct (I know it's very rough calculations) - particularly about modelling the circuit as a resistor?