Electronic – replace a car dash cam battery with a capacitor

batteriescapacitor

My car dash cam (G1W) has a weak battery. When it dies the camera will not function at all. I have already replaced it once. The original battery was 200ma, but I found a 320ma one that would fit. It lasted about the same as the original – a year. The newer models (G1W-C) of the camera have a capacitor as original equipment, but I can't get any information from the Chinese manufacturer(s) except that it's a super capacitor. How do I calculate the necessary capacity? It only powers the camera for about 10 seconds so that it can save the recording before the camera shuts down.

Best Answer

There are two electrical relationships that are useful for approaching this problem. First, current is the amount of charge per time (one Amp is one Coulomb per second): $$I=\frac{Q}{t}$$ Charge is measured in Coulombs and literally represents a quantity of electrons. \$6.241x10^{18}\$ electrons, to be exact. So if you were controlling current flow through a wire and could magically see the electrons flowing by, you would achieve exactly 1 Amp when exactly \$6.241x10^{18}\$ electrons were flowing through the wire every second.

Second, capacitance is the amount of charge per volt (one Farad is one Coulomb per Volt): $$C=\frac{Q}{V}$$ The capacity of a capacitor is measured by the amount of charge (Coulombs) it takes to change the voltage across the capacitor by 1V.

It's easy to see the relationship between the two equations: $$Q=I*t=C*V$$ Rearranging, we get: $$C=\frac{I*t}{V}$$ Now we have an equation that tells us the capacitance necessary to support a given current flow for a given time, given a desired change in voltage across that capacitor.

For example, let's say the capacitor was initially charged to 5V while the camera had power. Let's assume the camera can operate down to 3V, so we can afford to lose 2V during those 10 seconds. We'll also assume the camera draws a constant 100mA during that time. $$C=\frac{I*\Delta t}{\Delta V}=\frac{100mA*10s}{2V}=0.5F$$ So you'd need a 0.5F capacitor in this example. Do not assume any of these numbers actually apply to your specific camera.

An important thing to note. In this example, I assumed a constant current flow which makes everything nice and linear. In reality, as the voltage across the capacitor drops, the camera will draw less current, which will reduce the rate at which the voltage drops, etc. To properly solve the equation, you need to form it into a differential equation. The end result is actually an exponential decay curve. That said, if the change in voltage is relatively small, performing a linear approximation as above is "good enough" for a rough estimate.

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