Device load is 1 Ampere at 11 Volts. Capacitor bank should sustain this power for 4 seconds, before dropping voltage below acceptable (unspecified) threshold voltage. Let's say a drop down to 10 Volts is acceptable.
- 1 Farad = 1 Ampere-Second / Volt
For 1 Volt of drop, over 4 seconds, at 1 Ampere,
Thus, required capacitors must total 4 Farads, with a recommended voltage rating of ~ 24 Volts or higher: Car electrical systems can see voltage spikes of over twice the nominal voltage in certain situations.
For charging the capacitor at start-up, the battery must support the inrush current the capacitor bank will draw. As car wiring and batteries are typically very low resistance, given that 100 Amperes or more may be drawn for starting the car. A discharged capacitor behaves pretty much like a short circuit initially when connected to a charging source, so 100 or even up to 1000 Amperes could well be drawn by it.
Use a bank of car audio capacitors that incorporate soft-charge circuitry into the capacitor units. These are in the $100 to $200 range. Keep in mind, though, that car audio capacitors are typically rated far higher than their actual capacitance:
For instance, this 10 Farad rated capacitor was found to actually be a mere 0.227 Farads.
The typical naïve car mod customer never figures it out, so the manufacturers continue to sell them.
Like I have implied in comments, this capacitor bank approach is non-ideal, could be dangerous, and is actually strongly not recommended. I've put down the numbers simply because that's what the question asks for.
Both the answers posted at this time other than mine, actually point out the more practical approaches to the problem:
- Either turn off the lights at crank time, or
- use a different power supply, i.e. a SEPIC or buck-boost.
Both those much more practical solutions have been rejected by the OP.