In an assembly, I'm using spring loaded pins to connect a battery to a circuit board. It's possible from dropping the assembly that power may become disconnected for milliseconds at a time from the shock of dropping it. I have a capacitor bank that holds a charge for 30+ms when the positive node of the battery is disconnected. The problem is I don't think the capacitors will have any affect if the negative battery terminal is disconnected, or both at the same time because there will not be a complete circuit. Is my logic flawed? Is there a better way to do this than using a capacitor bank?
How to keep circuit powered when battery disconnected for short time
battery-operatedcapacitorvoltage
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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.
C1 was first connected to the dry cell to charge till V volts. Then C1 is connected to C2. So C1 should be connected to the 'fixed' terminals of DPDT switch and others must be connected to the 'throwable' terminals.
The schematic of the circuit with switch thrown to initial condition (connected to dry cell) is shown below.
The capacitor C1 is charged to V volts. So total charge stored in capacitor C1 will be $$Q = C_1\times V$$
When C1 is connected to C2, C1 and C2 are parallel and hence they will share the charge to make the voltage across them the same. So the voltage across C1 decrases and that across C2 increases. Let the final voltage be \$V_f\$ then by theorem conservation of charge,
$$C_1V = (C_1+C_2)V_f$$ $$V_f = \frac{C_1}{C_1+C_2}\times V$$
So \$V_f< V\$.
Best Answer
The battery and the capacitor should be connected in parallel. It is better to solder the capacitor to the main circuit board if possible. Considering the capacitor connection is always okay it will work just fine.
The beauty of parallel connection is that a damaged or disconnected or open branch does not affect other branches in parallel of the network.