leds as drls at 6/12v

I would like to run a string of (20) LEDs off the high-beam DRL circuit (around 6V), but when the low-beams are on, it would switch over to the 12V low-beam circuit and stay on with the headlights. Then it would go back to the DRL (6V) circuit as normal when the headlights are switched off.

What I will do is run the LEDs in parallel with a 150 ohm resistor each, but when the relay kicks over to the 12V circuit there will be a single large resistor (?? ohm) in series with the string. I don't know how to calculate this single resistor, and I don't want the relay to change back over to the high-beam circuit when the high-beams are on as anything but DRLs because it will burn up the LEDs as their individual resistors are sized for 6V only.

I have found some 'low battery relay circuits' that will disconnect power from a battery when it's voltage drops below a setpoint (about 10V) that appears to be on the track of what I need to do with controlling the relay, but I'm not sure how to implement these features.

In a nutshell, there are 2 circuits feeding 1 string of LEDs, one circuit that is dual voltage (6V/12V), and the other 12V only.

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Aaron

You mean a resistor in series with the LEDs, not parallel.

If you can find the 12V source for your DRLs, the problem should be solved - there should be one 12V lamp terminal that is common to both DRL and hibeams; otherwise maybe from the DRL switch itself.

A DRL triggered shorting relay will work if it's a single resistor, but I don't recommend a single resistor unless you mind burning all LEDs if some start to fail (go open circuit), tho it should also work with the relay connected to all string resistor junctions via diodes. But that requires 6V sensing - unless you have the 12V source as per above & hence no relay needed.

Or you could use a buck-boost converter & set to (say) 12V, or a boost converter & set to say 15V or 18V and design the strings for those voltages (one resistor per string).
Or a buck (step down) converter if using strings less than 6V.

Otherwise I'd suggest current limiters for each 6V string. Can use one resistor with an LM317 voltage regulators, else 2 transistors & 2 resistors.
If the single "12V" resistor was acceptable for the lot, so too a single current limiter, or two for two groups of strings, etc.

I'd suggest 2 circuits if running off the 6V so you don't unbalance the current, tho if the DRL current is much larger than the total LED current, that wont matter.

I agree that a regulator is the way to go. Hadn't thought of a buck-boost converter so I will have to look in to how they work. I'm leaning toward the use of a fixed voltage converter like the LM1117, 3.3V model. Since the dropout is 1.2V, it should work well with the DRL circuit running at 5.7-6V and it only requires 2 capacitors for stabilization.

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Aaron

I'd prefer more than one LED in a string...

But check out (switching) converters on eBay etc. They are price comparable with the 317 and have much less heat (the 317 has to dissipate over 2W for ten 20mA LEDs)

I've ordered 2 of these - Auto DC-DC Boost Buck Converter Solar Voltage Regulator 3-30V to 0.5-30V T9

Will do extensive testing prior to installation.

The intention with the buck/boost converter is to run 2 separate strings of LEDs (1 per side). All the LEDs will run in parallel on the string at 8V constant voltage w/current limiting through 270ohm resistors in series with each LED.

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Aaron

Why single parallels? It's such a waste of power & wiring. Plus if each LED is above 30mA, the converter won't handle it.

Since these will be installed in sealed headlights, in my mind 1 can fail and not take out 2 or 3 in series. I can also assemble and test the strings without touching the headlight assemblies, and if I place 1 LED every 1 to 1-1/2", I can simply trim the string to the length I need when the time comes. I was figuring 20 LEDs as worst-case.

The LEDs are run at 20mA, the resistor will dissipate about 11.7mA. Though I don't have the components in hand right now, by my calculations 20 LEDs will draw 634mA. If these controllers are not sized right, then I can certainly order some that are and use these elsewhere.

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Aaron

20x 20mA LEDs in single-LED strings draw 400mA. There is no additional current draw from the resistors since they are in SERIES. (Again - they are NOT in parallel with the LEDs!!)

I hope you're right, because the resistor has to convert the excess power into heat, which calculates as 94mW or 11.7mA @ 8v.

https://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator

If this is not the case, then 400mA should be easily doable by the buck/boost regulator.

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Aaron

None of the pics in that link show a parallel resistor - they are all IN SERIES with the LED or LEDs.

The function of the resistor is to drop the current (voltage) to an acceptable level for the LED(s).
Putting anything in parallel with a load (LED) has no effect on the current or voltage going to or thru that load (LED) - unless it's a short circuit and somehow I suspect that is not what you want.
Putting something in parallel merely increases the total load (current) - it has no effect on the other parallel component (excluding current limited situations, but that's not relevant here).


I still don't see why you cannot use longer strings. I fail to see how mounting a LED & a resistor is easier or takes less space than a few LEDs and a resistor.
Given your argument, I'd have 2 strings of 5 LEDs with "alternating" LED placement. That means 1/5th the converter output current compared to your single-LED string solution.
Sure - a failed string is leaves only half the LEDs, but IMO that'd look better than a single missing LED.

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



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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?

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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?

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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?

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Aaron

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

3.3V @ 20mA

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Aaron

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.   

Will test. Thank you. How will I know when I am overdriving the regulator or with a most efficient V/I combo? I have a DMM, so do I just try difference combos and test the current draw?

The primary driving voltage will be about 5.7V because these will run off the DRL circuit 90% of the time. But I don't know how it will run when trying to achieve 18-22V on a 5.7V input. What is the best way to test this?

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Aaron

If you overdrive the reg it will fail - maybe a self-protection shut down or an alarm using smoke signals, else pops or black trails.    


What "most efficient"? If it works, it works. I've already saved 5W of energy/power and it should now be using ~1.5W total (excluding converter inefficiency).

The specs say nothing about it not being able to supply 30V out with a 3V input.
To test, input (say) 5V, attach your LEDs & see if it adjusts to 18V or 22V or whatever you decide & design.
Check too that the output remains the same even if the input rises to 12V or 15V etc. (That should not be an issue.)

Since you have pointed me on how to reduce the 'wastage', I was assuming that less wastage = more efficiency as long as the regulator can deliver the necessary current. And the way I'm understanding it, if we are driving more LEDs in series, then the current draw goes down as the voltage (and efficiency) goes up. It is my challenge to keep the series of (X) LEDs so that when installation time comes, I'm not cutting off many to fit them in the allocated space. If I can fit more in a series, then I hope to increase the density as well. MORE=BETTER.

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Aaron

You need to specify which currents etc you are talking about.
Each LED takes (say) 3.3V @ 20mA.
For 10 LEDs in parallel, that's 3.3V @ 200mA.
10 in series means 33V @ 20mA.

There is no efficiency there - the power consumed is the same (ignoring higher iiR power losses thru component and wire resistances - ie, 100x greater in parallel than if in series.

The saving is in the number of resistors - 4 versus 20.
Plus the selection of voltage. I'd probably chose 3.3V with no resistor if paralleled, else (5x3.3=) 16.5V if 5 in series with no resistor, tho I'd probably increase that to 17 or 18V and add a resistor.

I'd expect that with one resistor per 3 or 4 or 5 (etc) LEDs you can fit more - not that resistors usually change the packing density since they sit behind a LED lead - not beside; else beside instead of between.
Plus of course much quicker assembly (16 less resistors) and less prone to failure (less component count).


But if you can adjust the voltage reasonably accurately, you could consider omitting the resistors. Resistors are used mainly to drop the supply voltage to one that is suitable for LEDs tho there is also a stabilising effect if individual LED voltages vary (for a given current) or if a LED fails in short circuit mode etc.

I've read somewhere that even when designing with exact LED/supply voltage parameters, a current limiting resistor is still advisable because a small increase/fluctuation in voltage can yield a large increase in current. I simply chose 270ohm resistors to have safe overhead. Per our conversation and the installation intent, as of now my goal is to run (3) LEDs in series @ 15V with (1) 270ohm resistor. By my calculations, this scenario would draw 137mA with 7 sets of LEDs.

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Aaron

That's consistent with what I've said. However many don't use resistors. EG - if 2.2V LEDs they might use 14.4/2.2 = 6.5 hence 7 series LEDs. Or maybe 6 => 6x2.2= 13.2V depending on the charging system (older 13.8V, or 14.2 or 14.4V, or maybe a drop to 13.4V (battery float voltage etc).

It's not as if LEDs will instantaneously burn out if they only slightly exceed their ratings - eg, 22mA instead of 20mA etc - and many usually design for under 20mA (hence 7 x 2.2V LEDs etc).

But yes, because LEDs are a non-linear load/resistance {ie, your 3.3V LEDs might be like a 3.0V zenor diode and a (15 Ohm) resistance; hence negligible operation below its voltage knee (3.0V) and then a big increase in current above that} the added series resistor acts like a "variable voltage" compensator.

However the extra voltage need not be too extreme - usually 2V if not 1V (or less) is more than enough.
If I had have been using 8V I would definitely have run 2 LEDs, hence 6.6V with a "1.4V" resistor (ie, 1.4V/.02mA = 70 Ohm; hence 68R or 82R).
If running singe LEDs I might ditch the resistors and set the converter to 3.2V, or maybe 3.3V. Only if brightness varied significantly would I increase the voltage and add a resistor (per LED - ie, per string).


Actually I am being reminded of how fussy LEDs can be. I added a red LED bar as a 3rd stop light. I decided to also use it as a 3rd tail light but went against my normal choice of PWM for the dimming and opted for a voltage regulator (which is set somewhere around 7-8V).
However LED drop outs are a problem; sometimes just one LED but usually a full string of 3.
At first I thought I had a burnt string, but after replacement, the original tested fine. I found a suspect SMD resistor (dry solder joint?) but also wondered about moisture etc.   
I conformal-coated the replacement LED bar far more thoroughly than the first since I thought moisture may have caused the burning of the strings(s). However new bar has also developed an occasional dim or dark LED or string which seem more noticeable as the dimmer tail rather the the bright stop light.
Anyhow, I've decided to ditch the voltage regulation for good old PWM. If part or all the problem is a moisture or solder/track issue, or if it is the dynamic behaviour of LEDs where a weak(??) string may extinguish, I expect the full voltage of the PWM to punch thru or help solve the problem.

Great info. Thank you for all your time and attention.

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Aaron

No probs. And sorry of I do tend to ramble, tho IMO my main crime is not planning before I respond.

Tho I said how LEDs can be tricky, generally they are far less critical wrt design than many seem to think.
The usual 'rule' is to string as many in series as you can have with a bit of extra voltage at the lowest voltage you want them to fully operate at. EG, tho a battery may be 11V etc, taillights etc might be designed for 12.5V to 14.4V since this is the normal range for voltage. If it dips below 12.5V so what if the LEDs are dimmer?
And a few on this site omit resistors - they just use enugh LEDs to match (or slightly exceed) the highest voltage they "know" they will get.   

But if using say High Eff Red LEDs (2.0V) and a resistor, then 6 LEDs = 12V with 0.5V to spare at our 'design minimum'. Max charging voltage of 14.4V means the resistor must drop (14.4 - 12.0 = ) 2.4V.
Hence at 20mA, R=V/i = 2.4/0.02 = 120 Ohms. (What a coincidence - an exact preferred resistor value. If it were 130R I'd probably round UP to the next preferred value of 150R.)
Resistor power check: P = iiR = .02 x .02 x 120 = .048 or ~0.05W way below even a 1/8W resistor, tho I'd chose the common 1/2W resistor. (Alternatively P = VI = 2.4V x 0.2 = .048W.)
If the voltage drops to 11.3V, the LEDs drop to from 2.0V 20mA to 1.8V @ 4mA and the resistor voltage drop = .004 x 120 = .48V or ~0.5V - ie, 6x1.8V + .5V = 11.3V. Note that the LEDs were 20mA with a 14.4V supply but dropping to 11.3V which is 0.8V less than our 'low' of 12.5V, the LEDs have only dropped 0.2V each (10%) whist the resistor has dropped from 2.4V to 0.5V (~80%). That demonstrated the 'elastic absorbency' of the resistor wrt 'regulating' the supply voltage for the LEDs.
However, since the LEDs are 4mA (11.3V supply) as opposed to 20mA (with 14.4V supply), they will only be 4/20 = 1/5th the brightness. (LED intensity is linearly proportional to current, or average/RMS current if PWM is used).
Likewise I can tell you that the 2.0V red LEDs will be 1.5V @ 0mA, hence with a supply of (6 x 1.5V =) 9V they will be fully off.
That demonstrates the non-linearity of LED dimming to voltage - ie, from full 20mA brightness at 14.4V to off (0mA) at 9V. And hence why if you dim LEDs in parallel with traditional bulbs using a linear dimmer or variable resistor (a pot aka rheostat), the LEDs will dim & extinguish much faster than the bulbs. The bulbs will still be at over half brightness at 9V when the LEDs turn off.
Luckily rheostats and their linear equivalents for vehicle dimmers are things of the past. Japs went to PWM during the 1980s tho I have seen other vehicles using linears as late as the late 1990s


FYI - an example of LED current and voltage for a given supply voltage and resistor is shown at (arduino's) Electrical Engineering Stack Exchange's Voltage drops and current for LED? - see the bottom graph/diagram. (We used to call that "linear programming" back in the days when maths involved logarithm tables (sliderules had been superseded). It's handy to know when you haven't got batteries or the package for modern digital programming toys.)

I guess I could do a similar analysis using 5 LEDs and whatever resistor would be required...

Alas I rambled again, but it's been a while since I thought about detailed the actual basics...