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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.

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  • \$\begingroup\$ Is that a rechargable battery in the camera? Or is it one which is supposed to provide, like, 500 shutdowns and then be replaced? \$\endgroup\$ – JimmyB Feb 26 '15 at 10:34
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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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    \$\begingroup\$ If the camera's internal power supply is a switcher, as the capacitor/battery voltage drops it will draw more current, not less, to keep the device supplied as internally it will be drawing that 100mA at a set (regulated) voltage which the switcher will have to work harder and harder to keep up. \$\endgroup\$ – John U Feb 26 '15 at 11:55
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The main difference between a battery and a capacitor is that the battery maintains the voltage while it sources current, while the voltage of a capacitor will decrease linearly over time.

So, let's say your battery has 3V, and the cam would just switch off if the voltage drops below 2.5V. Also, 3.5V would be an over voltage and could damage your device (These are just random numbers, to find out the upper limit, you would need a few cams...). Further more, let's assume your cam sinks 100mA for 10s. You have to charge your capacitor to 3.5V, and after 10s delivering 100mA (Charge: Q=100mA*10s=1As), it still needs to have 2.5V. It is

$$ C=\frac{\Delta Q}{\Delta U}=\frac{1As}{1V}=1F$$

So you need a 1F capacitor, which is quite huge. Even if you find one, it has to be charged to 3.5V.

You did not say if your cam recharges the battery, but even if, it will for sure not try to charge it to 3.5V

And there is still plenty of charge in the capacitor at 2.5V.


To use a capacitor of reasonable capacitance, you would need a circuit which charges the cap to a high voltage like 12V from the car and then a voltage converter to 3V. There are voltage converters on the marked which give you 3V from lets say 0.8V to 12V with >80% efficiency. (Yes, they also boost a lower voltage to a higher.)

As you can see, it's not just plug-and-play. It now depends on how much effort you want to put into it.

Another way is to use the un-switched (i.e. always-on) 12V from your car, regulate it to 3V, and put this into your cam. But make sure you do not drain the battery and also use fuses, as there's a lot of power behind the 12V in a car.

(Again: The numbers were just exemplary, I have no idea what the real numbers are. I also doubt that 100mA could be a far to low guess.)

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  • \$\begingroup\$ "maintains the voltage while it sources current" I would love to have such a battery, mine always drop voltage when I attach a load, and open circuit voltage gets lower all the time too after use. \$\endgroup\$ – PlasmaHH Feb 26 '15 at 9:07
  • \$\begingroup\$ Of course. But looking at a cap, this still is somewhat constant, eh? \$\endgroup\$ – sweber Feb 26 '15 at 9:41
  • \$\begingroup\$ The curves have different shapes, and due to usually having lower power density than batteries, same sized caps deplete much faster, but for the total amount of stored energy there is a surprisingly large overlap between batteries and capacitors these days. \$\endgroup\$ – PlasmaHH Feb 26 '15 at 9:51
  • \$\begingroup\$ +1 for suggesting a 12V-to-3V DC converter; you can find those remarkably cheaply online (random example), or probably in your local auto supply store. \$\endgroup\$ – Ilmari Karonen Feb 26 '15 at 13:52
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What kind of measuring equipment do you have ?

Have you tried to create a charging circuit of your own , supplying enough power (amps !) bypassing the original circuit ?

Not being specific, but it seems to me, that the charging circuit of the battery (capacitor ?) was not well dimensioned from the start. Or is this a "new / modern" way of emptying your pockets ?

Any short circuits ? ( I don't think so, but I just have to ask ).

Kris Norway

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    \$\begingroup\$ Kris, this answer is better suited as a comment. It would be better to remove your answer and re-post it as comments to the question above. \$\endgroup\$ – Dan Laks Feb 26 '15 at 8:53
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This is a 2.5F Supercapacitor assembly, that is good for up to 5.4V, which is typically used as a direct replacement for internal prismatic lithium ion polymer cells: http://www.amazon.com/Mobius-Action-Camera-Super-Capacitor/dp/B00JG7CRYU

The unit is quite small, and will fit into a variety of dash cam units (not just the Mobius), and can be parallelled with additional units to increase the overall C. The major advantage gained, here, is that the Supercapacitor arrays will better withstand the typical high temps developed in a vehicle, during the Summertime months, than do the lithium ion polymer cells that you're having short service life issues with (LiPo cells hate heat).

Now, to give you an idea of the successful implementation of this particular supercapacitor, the Mobius utilizes a LiPo cell of 820mA/h capacity, with a nominal terminal voltage of 3.7V (full charge terminal voltage is 4.2V). The above indicated supercapacitor assembly, having a capacitance of 2.5F, is sufficient to soft power down the Mobius unit, allowing it to close the video file current at the time of power down, preserving the recording successfully, while maintaining sufficient charge to maintain the internal clock function and parameter settings for several days.

I believe that you will find this to be a satisfactory solution for your needs, in this instance.

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