Electronic – 2 switching power supplies…and diodes

An alternative strategy would be to power the Raspberry Pi continuously and use the ignition line to initiate the power down sequence. I've done that in the past but with systems where the exact solution wouldn't apply to a Pi but in general:

Use a DC-DC converter for the best efficiency, there are many examples around but the following is one example of something that would be convenient to use and it can supply 1A at 5V from a 6.5V to 32V input:

http://www.digikey.com/product-detail/en/V7805-1000/102-1715-ND/1828608

A car supply can be quite harsh, so you might want to use a 30V TVS diode across the input to protect against spikes with a chunky Schottky diode with the anode at ground and the cathode at the 12V input to protect against negative voltages along with either a normal fuse or a PTC resettable fuse in series with the connection between the car's power and your system. Otherwise you may be able to 'hack' a car to USB charger that should already have all that in place.

I'm not sure what a Raspberry Pi draws in normal idle mode, but presumably well under 500mA which is the maximum USB can supply and more likely 100mA. Say it's using 100mA at 5V that will be under 50mA at 12V using that circuit, a car battery is normally in the order of 50Ah so that would be around 20 days to drain the battery to 50%. If the car is in regular use there's probably no need to go any further, and you may just be able to leave it running and just turn off any peripherals you're not using.

Otherwise for detecting the ignition change either way and both informing the Pi it needs to shut down followed by removing power a minute later the most practical way is probably to use an external microcontroller that drives a FET. It could be done with discrete logic but you also need to make sure power is re-applied when the ignition goes high, so it's not an entirely trivial excercise but parts costs will be lower than using a large cap.

The problem is that with the LM7805 (or any linear regulator) the output voltage must be less than the input voltage by a certain amount, which varies from a fraction of a volt (for so-called LDO = Low Drop Out regulators) to several volts. LDO regulators bring with them some disadvantages (stability can be an issue) so they're not a panacea.

In the case of the 7805 it's at least a couple of volts. You can find the number in the "dropout voltage" specification in the datasheet.

Since you're only giving it 12 volts, it can't output higher than about 10V (and if truth be told, it's best to keep it a bit less than that, maybe 9V).

Options if you need more output voltage than input voltage include boost converters (which store pulses of energy in an inductor then spit them out at higher voltage and lower current) or a DC-DC converter (which chops up the input voltage into AC and passes it through a transformer or does something similar to the boost converter, then converts it back to DC.

One of the cheapest boost converters is the (old) MC34063- it requires a fairly large inductor but it's cheap. Newer ones from Linear Technology and TI can work with smaller inductors due to high operating frequency.