Arduino in car: capacitor for extra 3 sec of power

Question

I want to install an Arduino Uno in a car, powered by a consumer 12V->5V car voltage regulator plugged into the lighter socket. The socket is switched, i.e, there's no power when the motor's off. When I turn off the engine I'd like to keep the Uno powered for extra ~3 secs. Can I use a capacitor in parallel to the Uno to get the extra 3 secs of power once the motor's off? How would i determine the capacitance? Should the cap be placed before the volt. regulator (i.e. directly on car's 12V) or after volt. regulation (on the regulated 5V)? Would I need some diodes to go with it? I don't want to put the Uno on the car's unswitched circuit, because it seems wasteful to run the Uno 24/7 off the battery just so it can be used for extra 3 secs when the motor's off. Thanks.

Answer 1

Don't use the 12V\$\rightarrow\$5V regulator, the Arduino needs at least 7V in. Use the battery's 12V directly instead.

The capacitor's value will depend on the Arduino's power consumption. The Arduino webpage doesn't say what the Uno consumes, so you can't say right away what capacitor value it needs. In any case it's not designed for low power. I checked the datasheet for the voltage regulator, and that alone already uses 6mA. On the schematic I can see two microntrollers: an ATMega16U2 running at 16MHz, and an AtMega328P, also at 16MHz. The former may consume up to 21mA, the latter says 9mA at 8MHz, so it's safe to say 18mA at 16MHz. We already have 45mA, let's round it to 50mA for the other components.

If a capacitor is discharged at a constant current, then

\$ \Delta V = \dfrac{I \cdot t}{C} \$

You start at 12V, and the Arduino needs a minimum of 7V, so \$\Delta V\$ = 5V, I was 50mA and t = 3s. Then

\$ C = \dfrac{I \cdot t}{\Delta V} = \dfrac{50mA \cdot 3s}{5V} = 30 000\mu F \$

That's the minimum, I would pick a 47 000\$\mu\$F/25V capacitor. Add detection for power off, so that you can switch off all unnecessary outputs which also may consume current, a relay for instance.

If you want to know exact what the power consumption is, add a 1\$\Omega\$ resistor in series with the power supply and measure the voltage drop. A 50mV drop means 50mA consumption.

Also add a TVS (Transient Voltage Suppressor) at the Arduino's power input; a car's 12V is extremely dirty.

Add the diode clabacchio mentions. A series resistor of 10\$\Omega\$/5W will charge the capacitor in 1.5s when applying power.

Answer 2

An alternative to using a capacitor is to connect to the permanent supply but to use a timer to power down or disconnect after a suitable delay.

The circuit can be arranged to repower the Arduino via the switched circuit when power is next switched on.

Current drain when off can be essentially zero.

When power is switched on supply to the Arduino can be from switched or permanent supply as required.

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As Clabacchio notes, if a capacitor is used, the holdup time =

t = C x V / I or
C = t x I / V

where t = holdup time. V = allowed drop in Volts and C = capacitance in Farads.

eg for 3 seconds, 50 mA, 5 Volt allow droop

C = t x I / V = 3 x 0.05 / 5 = 0.03F = 30 mF = 30,000 uF.

Answer 3

You can use a capacitor, but you need quite a big one depending on how much your Arduino consumes. 3 seconds at - let's say - 25 mA is 75 mC (Q=I*t), that at 12 V are stored in a 6.25 mF capacitor.

$$\left( C = \dfrac{Q}{V} \right) $$

The problem is that the voltage will decrease linearly if you drain a constant current, and below a certain voltage your Arduino will turn off. If you put the capacitor before the voltage regulator it will store more charge for the same capacity value, and - more important - the regulator will allow a wider voltage range, so you'll be able to use better the capacitor.

Since the Arduino accepts 7-12 V supply, you have a 5 V range to have the capacitor discharge. Again, 75 mC over 5 V means 15 mF, so with a 20 mF capacitor you should be able to keep it alive.

Note: I don't know what your Arduino should do, so the power that it will consume; size your capacitor accordingly.

About how to connect it, I'd suggest a resistor and a diode in the lighter socket side, to prevent too fast charge of the capacitor and to avoid its discharge towards the lighter socket.

So, summarizing, if I is the average current absorbed by your Arduino, 7-12 V is its supply voltage range, the minimum capacitor size you need will be approximately:

$$ C = \frac{Q}{\Delta V} = \frac{I \cdot t}{\Delta V} = \frac{I \cdot 3s}{12V-7V} = \frac{3s}{5V}I $$