Okay, here is the scoop:
forget the op amp circuit, way too complicated and what was described wouldn't do what you want anyway. Same with the DC-DC converter.
The basic problem is that LEDs need a certain amount of current. The LEDs are very non linear with regards to the voltage, you just can't predict what the voltage will be to get the desired current. The "rated maximum voltage" is just an indication of how high the voltage might have to go to in order to get the desired current.
The real problem is that at a different temperature, or just a different LED, might required a substantially lower voltage to get the same current. If you give it too much voltage you will burn out the LEDs.
What you really want is a circuit that regulates the current, not the voltage. Such a circuit will require a voltage drop in order to function, so you will need a voltage source at least a volt or two higher than the LED maximum voltage.
So, to answer your first question, NO, three 4.5 volt LEDs wired in series will probably produce no light at 12 volts. They *might* work fine when the engine was running and the alternator was providing ~13.5 volts. But then, they might burn out when the engine was turning fast and the voltage increased to 14.2 volts.
The easisst circuit for an LED is to have a series load resistor. You could hook two LEDs in series, which would have a combined maximum voltage of 9 volts. Your boat's electrical system can go as high as 14.2 volts (max charge voltage) so you need to protect the LEDs from this voltage. The required voltage drop is 14.2 - 9 = 5.2 volts.
Now you need to know the current that the LEDs will need. A rough guess would be around 25 milliamps (0.025 amps) but check the specifications for the LEDs. The basic equation for electrical work is known as Ohms Law, and goes E=IR
Where E is the voltage
I is the current (in amps)
R is the resitance (in ohms)
Back to your circuit: Forget the battery voltage we need the "drop" voltage, which is 5.2 volts. Given the desired current (let's call it 25 milliamps) we can solve for R
E=IR
5.2 = 0.025R
R = 5.2/0.025
R= 208 ohms.
A 208 ohm resistor will be hard to find, so use the next size larger (or you could use the next size smaller if you wanted to drive it a bit hard).
When you buy the resistor you will need to know the power rating (physical size). The power the resistor will see is the voltage times the current, or 5.2 * 0.025 = 0.13 watts. Use a 1/4 watt resistor.
You will need to duplicate this cicuit as many times as necessary to get the necessary brightness. DO NOT attempt to hook additional diode pairs to the same resistor.
A disavantge of this is that as the voltage drops (engine slows down or shuts off) the LED brightness will drop considerably.
There are other options that require a bit more knowledge of electronics. The ideal approach is to use a current regulator. Similar to a voltage regulator, this circuit attempts to hold a specific current, regardless of the voltage. It will end up with the same voltage drops so the efficiency will be the same, but the LED brightness will remain constant. A "linear" current regulator can be built using a voltage regulator and a couple of resistors, just hooked up a little differently.
An ideal circuit would use "switching" technology to create the current regulation. The switcher would increase the efficiency and reduce unwanted heat.
Or, you can just buy this and be done with it:
http://www.lopolight.dk/
Linear Mode
