Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.
Sign up to join this community
I'm trying to make a simple ups for a raspberry pi dash cam. The goal is to provide power to the pi for about 10 seconds after the car has stopped running. When the power supply turns off, the gpio pin will read low and the pi will begin to shutdown, using the capacitor for power. I haven't decided on R3 or D1 yet, they will be determined by what the 5v supply can output.
Will the circuit below work for this purpose?
The relevant formula for this sort of circuit is below:
$$C_{min} = \frac{t_{HOLD}\cdot I_{OUT}}{\Delta V}$$
In your case, neglecting the voltage drop of the current-limiting resistor, and assuming you can run down to 3.3V, this becomes
$$C_{min} = \frac{10\text{ s}\cdot I_{OUT}}{5\text{V} - V_{diode} - 3.3\text{V}}$$
Ballpark numbers:
Use a Schottky diode, and you'll maybe have 300 mV for the diode drop. Let's say the Pi draws 0.4A.
With these numbers, the equation above gives 2.9F. So your capacitor is close, but might not be quite enough.
A 12V battery, charger, and buck converter might be a better way to do this. You'd at least be able to avoid the giant capacitor.
The best solution I think in this case is to avoid a supercap altogether and instead draw directly from the car battery. You can easily get a direct lead to the battery from the car radio, or in a pinch run a wire from the fuse box. Use the switched power to start a timer and shutdown 10 seconds later. When the switched power comes on assert power to the board.
In this circuit, the large capacitor will start at ~4.7V (5V minus V-D1, here I assume 0.3V but be careful if you have a generic 0.7V diode, and also beware of the voltage divider effect of R3) and decay from there, so it will be a game between the 4.7V sagging down and when the RPI browns out. If this was a production design, then you are spending way too much on the huge capacitor. Also, if the capacitor is fully discharged, then be aware of the inrush current that will occur when it is first powered up. With such a huge capacitor, D1 and R3 will see a ton of amperage for a bit of time.
So if you make D1 and R3 big enough (in terms of wattage), and the brown-out level for the RPI is well below 4.7V, and R3 isn't too high of impedance, then it will work. But I wouldn't do this in a production product.
It would be more efficient to connect your 12->5V converter to the battery (use a fuse). That way, the RPI isn't racing against time. You could put in a hold-up circuit so that when the RPI detects the car is off, it can do its tasks and then flip an output that turns off its own power, thereby conserving your car battery. No worries about brown-out behavior, minimal inrush current, no rush against time for the RPI, minimal drain on battery, and much cheaper in component cost.
