I am planning on using a 3.3V GPS module which draws 22mA. The module has a requirement that a shutdown sequence be initiated 1s before power is removed from the IC. If power is removed without allowing it to go through the shutdown sequence, the program flash memory can become corrupted and the module bricked. This seems like a really poor design to me, but that's the way it is.
The circuit I'm designing is battery powered, so the battery could die or be removed at any time. I considered using a separate battery as a backup, but it seems like capacitor would be cheaper and easier.
So, I need a capacitor to power it for 1s when the power is disconnected. I'm planning on putting a Schottky diode before the capacitor to ensure other ICs in the circuit don't draw power from the capacitor. I need to determine the capacitance to use and am unsure if my calculations are correct.
I know \$Q = I \times t\$. Therefore, the charge needed is \$Q = 22\text{mA} \times 1\text{s} = 22\text{mC}\$.
Since \$C = Q/V\$, the capacitance needed is $$C = \frac{22\text{mC}}{3.3\text{V}} = 6.67\text{mF}$$
Is that correct? Do you think using a capacitor like this is a viable alternative to using a backup battery?
Answer
The device can work from 3.6 to 3.0 V according to the datasheet. I assume you will use a 3.6 V supply. Also I simplify the device as an impedance of 150 Ohms. So, the problem can be translated into that: In a RC circuit, what is the capacitance needed for a 150 Ohm resistance, having 3.6 V on t=0, having 3.0 V on t=1.
v(t) = V(0) * e^(-t/(R*C))
v(1) = v(0) * e^(-1/(150*C))
3 = 3.6 * e^(-1/(150*C))
C = 0.0365F = 36500 uF
It is bulky for a conventional electrolytic cap. Use a supercapacitor.
note: This formula I used is a solution of differential equation for natural response of RC circuit. For more information look at: https://en.wikipedia.org/wiki/RC_circuit
note2: It may not be bulky for electrolytic if you find a capacitor a bit higher than 3.6V volts, but I guess it is difficult to find.
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