leds as drls at 6/12v

No parallel resistor, just series with each LED. The LED/resistor combo will be run in parallel on the string. Aaron

Yes, so the resistors are in series with each string of LEDs, and all strings are connected in parallel. The relay is not needed unless you want to independently turn it on and off as shown, but instead I'd consider the following options. (If using the diode, place it between 86 & 85.) I'd suggest a single switch to do the lot... For off you merely need to interrupt the 6V/12V line. To have on, off & 6V/12V I'd suggest a 3-postion on-off-on (SPDT centre off) toggle switch - common pin to converter; one side from 6V/12V & the other from whatever +12V supply you want. The same or a 2-way switch (SPDT) can be used for the on switch (ie, on with switch = on or 6V/12V powered). Or maybe simpler still, +12V to an SPST switch to the converter thru a diode, and a diode between 6V/12V and the converter (1N400x diodes).    If power is a consideration, you are probably burning 5W needlessly; ~3W for a relay, & 2W using 20 resistors. No relay and 4 resistors drops that to under 0.5W with 1/5th the converter output current. (EG - 5 series LEDs with one resistor; converter voltage set to five times LED forward voltage drop plus 5.4V using 270R resistors, or plus a volt or 3 & recalculate the resistor - eg, calculated appropriately ie, 56 or 82 Ohms etc.) Having LEDs in series strings may also reduce any difference in brightness between the LEDs.

The purpose of the relay is to have the DRL circuit (6/12V high-beam) run the LEDs independently of the headlight (12V low-beam) circuit and not feed one into the other. All connections will be made inside the headlight housing so there are no external controls. I also thought of using diodes, but with the regulator, the less voltage boost that is needed, the greater current it can supply. So if I placed diodes on the inputs, the DRL circuit could potentially be feeding the regulator only 4.5-5V, then it may struggle to deliver 400mA. Once I have everything in hand, I will certainly follow your suggestions and try running the LEDs in series. My goal was to strike a balance between regulated voltage (8V) and current-limiting resistors since the regulator can't manage both. If it does work out best to step up the voltage more to something like 15V and have 3 LEDs ins series with (1) 270ohm resistor, does this make the circuit more efficient? Is it the number of resistors that is effecting the efficiency?Aaron

Not that you need use a resistor as large as 270R, but using that as an example... P=iiR = 0.02 x 0.02 x 270 = 108mW per 270R resistor. As to the diode voltage drop, IMO that's negligible assuming you use multi-LED strings. Keep in mind that with your 270R single LEDs off 8V, you are wasting (V=iR = 0.02 x 270 = 5.4V) 5.4V/8V = 68% of your power in the resistors. (That figure should be more like 10% or less.) Drop that by 50% (ie, to 18%) and you have halved your input current requirement which outweighs the input max current drop from ~500mA to 400mA.

So as to not discard the resistors that I've ordered, could I also theoretically run as mentioned (3) LEDs in series with (1) 270ohm resistor @ 15V and be wasting 1/3 less power? When running LEDs in series, does the forward current remain the same or is it the sum of the total? ie (3) 20mA LEDs in series = 20mA or 60mA?Aaron

In a series circuit, the current is the same thru all components. And the voltage across each component totals the "string" voltage. (The voltage is constant across all paralleled components, or strings etc.) Hence 3 or 5 LEDs in series takes the same current as a single LED. But the voltage will be 3x or 5x the single LED voltage. Hence most vehicle LED strings comprise 3 to 5 series LEDs (depending on color) usually to total ~10V or 12V or 14V etc.

Great, so I can test a series string of (3) LEDs @ 15V with the 270ohm resistors, or (2) @ 12V. Is it resistors that I'm trying to reduce the number of?Aaron

What is your (rated) LED voltage - ie, 2.7V @ 20mA?

3.3V @ 20mAAaron

So using 2 sets of 5 LEDs per side... 5 x 3.3V = 16.5V + 270 x .02A = 5.4 16.5 + 5.4 = 21.9V. So set the output to 21.9V = 21V to 22V. Wastage = 5.4/21 = 24%. Or with two 270R in parallel for a 2.6V drop = 12% waste. Of 4 for a 1.3V drop = 6% wasted (4 parallel 270R = 67.5R). Or get a 82R (Ohm) resistor, hence 82 x 0.2 = 1.64V & set the converter output to 16.5 + 1.6 = 18.1 = 18V. Hence each converter supplies 18V (or 21-22V) to two 20mA strings -= 40mA, hence 18 x 0.04 = 0.72W per side (or .88W if 22V). Hence input at 5V = .72W/5V = 0.144A (or .88/5 = .176A if output is 22V) - both well within converter specs. A 1-2V overhead (the resistor voltage) should be enough to compensate for LED variances.