Or maybe a different way of doing it?
Problem:
Electrical circuit (12vDC) on my boat is supplied from the battery but on engine start there is a brief voltage drop. This doesnt affect any equipment other than the chartplotter (marine satnav) which shuts down. When re-started it takes a minute to boot up and find itself which can be a touch nervewracking at times!
Some boats deal with this by having the instrumentation powered by a separate battery to the one that is used exclusively for starting the battery, but that isn't really economically viable to retrofit
A friend has suggested that a capacitor and diode will provide adequate temporary power to maintain the voltage until the engine start is finished. The plotter draws maximum 13 watts at minimum operating volatge of 10.8 V.
Is this viable – and if so, what size Cap?
I am not too worried about load dump as there is little else to shed!
Alternatively (it has been suggested) a circuit with a small lead acid gel 12v battery that will cut in when the voltage drops and otherwise charge up when the main circuit voltage is at normal 13+V with engine running etc.
Best Answer
The voltage on a one farad cap will drop one volt per second given a uniform current draw of one ampere. Given three of the four values (capacitance C, acceptable voltage drop V, required hold time T, and current I) one can determine the value of the fourth using the equation CV=IT.
With regard to the capacitor and diode idea, you need to be aware that if one connects a large high-quality initially-discharged capacitor to an "already-on" power supply such as a battery, the cap will try to draw as much current as the supply will put out. It's conceivable a diode alone may work, but it would more likely turn into a "fryode" when the battery is connected; such a direct connection may also seriously damage the battery or the capacitor. Thus, rather than simply using a diode, you should use some kind of circuit which can safely handle that condition (charging a 1 farad cap to 12 volts will require 1200 amps for 10ms, 120 amps for 0.1 second, 12 amps for one second, or 1 amp for twelve seconds).
From an efficiency standpoint, you may be best off using a pair of switching-power-supply circuits, one of which would step your capacitor voltage up to some higher voltage, and one of which would step that higher voltage down to whatever your device needs. If your cap is charged to 12 volts, and your circuit will fail when it drops to 10, you'll only be able to use about 30% of your caps energy storage ability (the circuit will fail when the cap still has has about 70% of the energy that was put in). By contrast, if you were to charge a cap up to 35 volts and could operate down to 11 volts, and if you used a switching supply to extract energy from the cap, you could probably utilize more than 80% of the cap's energy storage capability. Given that large caps are expensive, the reduction in cap size may more than make up for the cost of the electronics.