Sizing power supplies for constant voltage led strings, What about the resistors?

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
A simple question. Say I have a 24V power supply, a series string of 18 leds with a V(f) of 3V @ 500mA.
A popular calculator gives me this:

  • 6 series.3 parallel strings
  • each 15 ohm resistor dissipates 3750 mW
  • the wizard thinks the power dissipated in your resistor is a concern
  • together, all resistors dissipate 18000 mW
  • together, the diodes dissipate 27000 mW
  • total power dissipated by the array is 45000 mW
  • the array draws current of 1500 mA from the source.
So do you just use the mA and voltage to size it?
24V , 36W, 1.5A
Plus a standard efficiency cheat of adding 20%.
Do the resistor watts not err count?
Like use a 24V 45W 1.875A power supply?
(add 20% later)

What simple thing am I missing?

An online quote..
Take the length of your LED strip in feet and multiply it by the watts consumed per foot, then multiply by 1.2. This will give you the minimum sized power supply you'll need to run your strips.
 

WBahn

Joined Mar 31, 2012
30,295
I don't know where your "popular calculator" learned to do math.

If the total current draw from a 24 V supply is 1.5 A, then the total power dissipated is 36 W (not 45 W).

If your LEDs take 500 mA at 3 V each, then six in series will require 18 V. To drop the remaining 6 V at 500 mA requires a 12 Ω resistor (not 15 Ω).

The diodes in each string would dissipate 9 W (for a total of 27 W, which agrees with the calculator).

Each resistor would dissipate 3 W, for a total of 9 W.

The total of these is 36 W.

A 15 Ω resistor with 500 mA would drop 7.5 V, making the needed supply voltage 25.5 W, not 24 V.

While this would make each resistor dissipate 3.75 W, the total dissipation would only increase to 38.25 W (still not 45 W).

They are claiming that the total dissipation by the resistors is 18 W, which makes no sense. They themselves claim that each resistor is dissipating 3750 mW. To get a total of 18 W, that means that they have 4.8 resistors.

Huh?

If the Vf of the LEDs really is 3 V, then 15 Ω resistors would limit the current to 400 mA and make the total power draw 28.8 W.


If you go with three parallel strings of six LEDs each, I would recommend that you use a supply that is rated for more than 1.5 A. It's always nice to have some margin -- and the more margin you have, the longer your supply will last. Using at least 20% is a start. Unless there's a good reason not to, go with 50% or even 100% margin. Often the size and cost of the larger supply is quite minimal, but if that's not the case, then rethink it.

Keep in mind that your 12 Ω resistors are likely to not be 12 Ω, especially when they are dissipating 3 W of heat. So you need to allow for that. Also, the voltage and current on those LEDs is not engraved in stone. Look at the data sheet (if you can get it) to see how much the voltage and current can vary and still be in spec. Use the worst case combination to size your components. Whatever heat they must dissipate, make sure that they can dissipate at least twice that (if there's a damn good reason why that's not an option, pare it back to a 50% margin, but thing long and hard before going any lower).
 

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
Hmm ..
Maybe just start with one string
6 diodes.
Power supply is 24v.
Chosen voltage per diodes is 3v to achieve a part. current draw, 500mA.
Or one could say I want to apply 500mA which creates a voltage drop of 3 V at the diode.
So you want a resistor to drop the 24v to 18v.
15 ohm, 3.75 watts
V= IR
Diode string draws 500mA at 3V.
6= .5*x
12 Ohms.
Calculator uses "closest popular fit" so I assume 15 is easier to get than 12?

Diode wattage is 9 watts.
Soo a 24v 9W 375mA power supply is " calculated minimum" if one ignores the resistor.
Resistor wattage is......???
15 Ohms, 500mA, 7.5v ( deviation from 12)
3.75 watts.

Do you use it in calculating power supply?
Sooo .3.75 + 9 = 12.75
So 24v 12.75 w 531 mA power supply
To START ....
:)

I appreciate the more err precision you refer to but I'm looking for a concept and I do recognize all the " fudge factors"you can add ..
 

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
Thanks but I know how to do it with a constant current driver.
I assume most of those resistors were an attempt to balance the current spread out over the 3 parallel chains.

24v 9W 375mA
Or
24v 12.75 w 531 mA.

Which is the more " correct" power supply for 6 LEDs in series with a terminating resistor as described above?
 

LowQCab

Joined Nov 6, 2012
4,290
Actually, I've never seen anyone make this procedure so complex.

The problem is not with the Resistors,
the problem is the Forward-Voltage shift of the LEDs with Temperature.

I assume that You want 500mA X 3-Strings,
this equals 1.5-Amps total Current.

If your Power-Supply is well Regulated at 24-Volts,
and capable of producing ~2 to ~3-Amps of Current without loosing Regulation,
then You can skip any Power-Supply considerations.

24-Volts / 6-LEDs = 4-Volts per LED.
This would appear to be a good match for the LEDs since they are specified at ~3-Volts.

6-LEDs X ~3-Volts = 18-Volts .......... at what Temperature ???

What is the Forward-Voltage of the LEDs at, let's say, ~70C ???
Maybe ~2.8-Volts ???
6-LEDs X ~2.8-Volts = ~16.8-Volts.
So, which Voltage do You use to make your Resistor calculations, ~18-Volts, or ~16.8-Volts ?

What Temperature will your LEDs be averaging after maybe ~10-minutes of operation ?
Will the Current go up, or down, as the LEDs get warmer ???
You should check the Graph on the Spec-Sheet.

Using just plain-ole Resistors is not the best plan for High-Power-LEDs.
A better plan is to use a proper Current-Regulator.
A Switching-Current-Regulator is far more efficient than Resistors,
and automatically compensates for changing Forward-Voltages due to Temperature-Swings,
it will also stabilize the Light-Output, and Color of the LEDs,
and protect the LEDs from accidental over-Voltages, or extreme Temperatures.
It will also allow You to more safely "push-your-luck" with
the amount of Current You can use on a continuous basis,
instead of having to "err on the side of safety".
.
.
.
 

WBahn

Joined Mar 31, 2012
30,295
Hmm ..
Maybe just start with one string
6 diodes.
Power supply is 24v.
Chosen voltage per diodes is 3v to achieve a part. current draw, 500mA.
Or one could say I want to apply 500mA which creates a voltage drop of 3 V at the diode.
So you want a resistor to drop the 24v to 18v.
15 ohm, 3.75 watts
V= IR
Diode string draws 500mA at 3V.
6= .5*x
12 Ohms.
Calculator uses "closest popular fit" so I assume 15 is easier to get than 12?

Diode wattage is 9 watts.
Soo a 24v 9W 375mA power supply is " calculated minimum" if one ignores the resistor.
Resistor wattage is......???
15 Ohms, 500mA, 7.5v ( deviation from 12)
3.75 watts.

Do you use it in calculating power supply?
Sooo .3.75 + 9 = 12.75
So 24v 12.75 w 531 mA power supply
To START ....
:)

I appreciate the more err precision you refer to but I'm looking for a concept and I do recognize all the " fudge factors"you can add ..
If it chooses 15 Ω over 12 Ω based on assumed availability, that's fine.

But doing so means that the resistor doesn't have 500 mA flowing in it and it makes no sense to assume that the current draw hasn't changed as a result. .

If it is dropping 6 V, then it only has 400 mA flowing in it. So it is not dissipating 3.75 W, but rather 2.4 W. The drop from 500 mA to 400 mA also means that the LEDs (in one string) are only dissipating 7.2 W instead of 9 W and the total power needed from the supply is 9.6 W. You need a supply that can deliver 400 mA plus whatever margin you want. If it's rated at 500 mA, that's about the minimum I would recommend.
 

LowQCab

Joined Nov 6, 2012
4,290
I forgot to mention that the LT-1074 Switcher also has
an "Enable-Pin" which will allow PWM-Dimming,
and/or, On/Off control via a Micro-Controller, or other Low-Voltage-Circuit.
And, all 4 of the Resistors combined, only dissipate around ~3 to ~4 Watts of Heat.
.
.
.
 

Dodgydave

Joined Jun 22, 2012
11,348
If you use a Constant Current supply then all the leds go in series take the same current, the driver automatically adjusts the output voltage depending on how many leds are in series.
So choose how many leds you want in series and then sort out the current driver you need.
 

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
The question was a "simple" power supply question in general.

All these tangents, though important in the real world is distracting from the orig. question.. do resistors count in the sizing estimate for power supplies.

As to constant voltage led "real world" useage bring that up with the millions of miles of led ribbons/boards produced.................

I don't think "I" am making this complicated that is for sure....

The "actual" power supply chosen based on a "napkin" estimate of minimal conditions would be well over those said conditions because of all the above mentioned factors..
Take the length of your LED strip in feet and multiply it by the watts consumed per foot, then multiply by 1.2
. This will give you the minimum sized power supply you'll need to run your strips.
The above statement seems to include ALL "watts consumed" thus implying the resistors are included in the calculation.
When one looks at the "general recommendations and formulas" it seems to imply ONLY the led wattages matter.. resistors are ignored..
Thus, calculated right or wrong isn't exactly important for the answer
  • together, the diodes dissipate 27000 mW
    [*]

    [*]total power dissipated by the array is 45000 mW

Which would you use to size the power supply? No not the REAL power supply but one just based on the above as the only data that matters (YES there are other factors)

Maybe my question is soo basic as to not being able to be seen..
I build aquarium lights for a hobby and only use constant current in any arrays SO I understand the benefits. Still like I mentioned there are millions of constant voltage arrays in the wild.
There are "system watts" and "led watts".

OK so here is a real world example:
xhopower.JPG
And why my question is important to me..
According to the above the diodes at say 24" XHO model has a wattage listed of 27W.
System watts or diode watts?
The resistor (yea it uses constant voltage) is chosen to drive the leds at like 1.3 watts due to the voltage decrease from the resistor. These are "3W class" diodes so capable of taking more power (with PROPER thermal design).
Guesstimating 433-500mA of current to the leds.

So say I want to run 4 in parallel off one power supply.
Using the above I'd START around 27x4 =108W capable power supply @ 24V (YES add in x,y,z for real world sizing).
I don't believe that would be correct.
Full power to consider is 24V
YES I ass-ume 3V at the diode voltage drop @ 433mA current.
I used the 433 based on such an assumption and the "diode wattage" in the above chart.
Point is in one calculator the Wattage of the resistor is ignored..the other it is not.
The 2.81 watts..

ledcalc.JPG



Soo considering I want to run 4 24" XHO's off one power supply what is THE BARE minimum 'IN THEORY"
Based on diode watts or "system" watts?


OK that "mess" written let's get back to the real world...
FEEL free to ignore all the above and go to "scenario 2" below.

Regarding 4 of the 24" XHO lights.. Someone was sent a 24V 5A (120W) power supply and enough Y cables to bring 4 lights down to one power supply.
I advised them that the above power supply is,if anything, the bare minimum of what he needs to run 4 "27W"
lights with the light total wattage would be 108 Watts.
129.6 Watts is needed if one applies a "standard" 20% fudge factor.

Using the resistors total system wattage was calculated at:
10.6 x 3 x 4 = 127.2 W
So the power supply gets WORSE.

Either calculation shows the power supply, when considering "EVERYTHING" would be too small but convincing someone that the included one is, though "appearing" large enough (120W ps vs lights listed at 27x4 = 108W) is in fact not large enough if one uses a different calculation of wattage shows it is DEFINITELY too small.

Thus the "do you include resistor wattage" in the calculation.
For the sake of argument assume the data and calculations are correct.

My suggestion was to upgrade to something like this:
https://www.amazon.com/dp/B082TTQZ4...&linkCode=w00&linkId=&creativeASIN=B082TTQZ4H

See, I understand the concept of "oversupply".. ;) though with good power supplies going close to 100% utilization often gets the ps into the higher efficiency range
360 Watts is overkill with a decent power supply in my mind.
And I'd definitely prefer something like a Meanwell supply.
On the plus side at 360 Watts one has room to add another light or 2.
But I digress.

There now I included both real world and calculated world..which is it?

Look I appreciate the constant current advice but as you can see the "pieces" in the real world are already SET to constant voltage. Just a matter of using the cards I was dealt with.
 
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WBahn

Joined Mar 31, 2012
30,295
Why do you keep using these online calculators that are clearly giving bogus results?!

The results they come up with aren't even self-consistent.

How much power would a 15 Ω resistor dissipate if it has 433 mA flowing through it?

More than 2.8 W. Yet that calculator recommends using 2 W resistors.

What is the voltage across a 15 Ω resistor that has 433 mA flowing through it?

6.5 V. Add that to the 18 V across the six LEDs and you are at 24.5 V, which is more than the supply voltage they are claiming (but at least it's closer that the one that was claiming 500 mA).

You don't size a power supply based on wattage, you size it based on how much current you need it to deliver. The wattage needed is then the voltage output multiplied by the current.

If you are driving LEDs with a constant voltage source, then of course the power in the resistors has to be taken into account, but only indirectly.

Let's say that all I could get were 100 VDC supplies and I needed to drive one of these strings-of-six LEDs. If I want 433 mA in the LEDs, then I need a 100 V supply that can deliver at least 433 mA. That's what I'm going to be looking for. If they are marked in terms of power output, then I'm looking for something that is at least 43.3 W (leaving all the margins and fudge factors aside). I don't need to know the wattage of the LEDs or the wattage of the resistors to determine this, I just need to know that I need a supply that can deliver at least 18 V (again, ignoring real-world issues) and at least 433 mA.

Where the wattage comes in is in making sure that I'm not going to burn anything up when I drive those LEDs. Each LED is going to be consuming 1.3 W, so I need to be sure it can handle it with whatever heatsinking I provide. I have to be sure that that heatsinking can also handle the total of 7.8 W from all six of them. Then I have to choose resistors (or an array of resistors) that can handle the 35.5 W of power that they are going to have to dissipate.
 

LowQCab

Joined Nov 6, 2012
4,290
There is no such thing as "Constant-Voltage" LEDs.
The manufacturer has added either Resistors,
or some other form of Current-Limiting or Current-Regulation to the "assembly".

"Wattage-Ratings" are generally bogus advertising numbers.
You supply the Voltage that the manufacturer provides,
then the manufacturer has hopefully designed the system to
perform as stated when supplied with XX-Voltage.

You can't depend upon manufacturer "Wattage-Ratings" for ANY Power-Supply calculation.
This "might" get You a "Ball-Park" number, but don't believe it until You measure.
You must measure the normal operating Voltage vs Current to determine the actual Wattage.

A larger problem, with long Strips of LEDs, is Printed-Circuit-Voltage-Drop.
--------------------------------------------------------------------------------------------------
Resistors,
and whether to "count them" in the "Wattage" calculations.

In a series Circuit,
all devices in the Circuit,
always have the exact same Current flowing through them.
Therefore, if You measure the Voltage across a Resistor of known value,
which is part of any series Circuit,
then You can calculate the amount of Current flowing in every other device in the Circuit.

If there are multiple series-Circuits wired in parallel,
simply add the measured Current-values of each of the individual series-Circuits together.

The Current flowing in an LED cannot be measured by measuring the Voltage across the LED,
there must be some sort of Resistance in series with the LED in order to determine the Current.
( not counting Hall-Effect Current-Sensors )

The Wattage that must be dissipated by the Resistor is
simply calculated from the Ohms-Law Formula .........

V / R = A, then, A X V = Watts,
or,
V / R X V = Watts.

As for whether or not the manufacturer includes
the Wattage dissipated by any Resistors they may be using,
vs the supposed "Wattage" dissipated by the LEDs ..................
Advertised "Wattage" of LEDs is not a reliable number for any calculation.

Perceived LED-"Wattage", as a "brightness comparison" is a highly subjective thing,
and many assorted LED "performance-factors" can have a profound influence,
even when comparing different LEDs with the same, or very similar, supposed "Wattage-Ratings".
This makes any "Wattage-Rating" numbers of questionable value.
.
.
.
 

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
You don't size a power supply based on wattage, you size it based on how much current you need it to deliver. The wattage needed is then the voltage output multiplied by the current.

If you are driving LEDs with a constant voltage source, then of course the power in the resistors has to be taken into account, but only indirectly.
That was the kind of statement was asking for.
Also what do you mean by " indirectly"?


As to the "2w" resistor yea one calc said it was ok, other would imply a 5w class resistor in the "discussion"
Round 2.8 up to the next class ..at least.
I wasn't happy w/ the 2 w recommendation either
Was more interested in the similarities between the 2 than the differences.

So let's go real world again..

https://www.digikey.com/en/resource...ors/conversion-calculator-led-series-resistor
What we have:

24v ps
Cree diode run at
3V
.150A
4 lights
18 diodes in each , 3 resistors, 6 diodes per series strings

Each light:
Diodes:
.150A * 6 * 3v * 3 = 8.1W
(.5 * 6 * 3 * 3 = 27 rated power of light above)
Resistors:
24v - 18 = 6 volts
6/.150 = 40 Ohms .
P = IV = .9 watts

Power dissipated .9 watts
.9 * 3 = 2.7 watts
Total watts for one light:
8.1 + 2.7 = 10.8

-----------------
Total led watts for 4 lights:
32.4 W
Total system watts
43.2 W
Calculating amps from this is 2 different numbers .
1.35A
Or
1.80 A (.150 * 3 * 4)
So...yes resistors are ( indirectly) accounted for.
Sooo, not adding all the corrections (20% add on or more)
24V ,1.8A, 43.2 W ps

24V 100 Watt ( 4.17A )
ps would be a "real world" fit

Trick is determining the actual amp draw of the diodes.

So using that mA estimate and the 4 lights the amp draw is.....
.5 *3*4 = 6A
@ 24v = 144 watts.

BEST ps would be like 24v 8.33A+(200W or better)

Thanks, think I got it now .

@LowQCab thanks for the Synopsys
All good stuff.

One note... Yea you can't trust manuf specs.
I'll give you a Great example.
There is a reef aquarium light, Viparspectra and a number of like " black boxes" that advertise wattage as 165 "W" based on using 55 what I like to refer to as "3 watt class" diodes.
55 * 3 = 165 right?
Yet inside are constant current drivers with typically 550 mW setpoints.
The actual diode string voltage is unknown.
Using the best guesstimates ( and some who put kill a watt meters in line)
is the lights actually run at about 110 watts.
No resistors.. :)
Now that does not include diodes efficiency in converting electricity to photons.
So yes .... most can't be trusted which is why many have gone to using quantum meters to actually measuring " real world" photon counts.
That said sometimes " guesstimates" based on things like wattage do have value...IF one understands the error factors.
 

WBahn

Joined Mar 31, 2012
30,295
That was the kind of statement was asking for.
Also what do you mean by " indirectly"?


As to the "2w" resistor yea one calc said it was ok, other would imply a 5w class resistor in the "discussion"
Round 2.8 up to the next class ..at least.
I wasn't happy w/ the 2 w recommendation either
Was more interested in the similarities between the 2 than the differences.

So let's go real world again..

https://www.digikey.com/en/resource...ors/conversion-calculator-led-series-resistor
What we have:

24v ps
Cree diode run at
3V
.150A
4 lights
18 diodes in each , 3 resistors, 6 diodes per series strings

Each light:
Diodes:
.150A * 6 * 3v * 3 = 8.1W
(.5 * 6 * 3 * 3 = 27 rated power of light above)
Resistors:
24v - 18 = 6 volts
6/.150 = 40 Ohms .
P = IV = .9 watts

Power dissipated .9 watts
.9 * 3 = 2.7 watts
Total watts for one light:
8.1 + 2.7 = 10.8

-----------------
Total led watts for 4 lights:
32.4 W
Total system watts
43.2 W
Calculating amps from this is 2 different numbers .
1.35A
Or
1.80 A (.150 * 3 * 4)
So...yes resistors are ( indirectly) accounted for.
Sooo, not adding all the corrections (20% add on or more)
24V ,1.8A, 43.2 W ps

24V 100 Watt ( 4.17A )
ps would be a "real world" fit

Trick is determining the actual amp draw of the diodes.

So using that mA estimate and the 4 lights the amp draw is.....
.5 *3*4 = 6A
@ 24v = 144 watts.

BEST ps would be like 24v 8.33A+(200W or better)
It's hard to keep track of things because they seem to constantly be changing.

So now you are talking about 72 LEDs?

And I can't tell what current you shooting for in each LED. You start out talking about 150 mA, but then end up talking about 500 mA. Which is it?

Why are you calculating a current using 32.4 W and 24 V (to get 1.35 A).

Power is not just some random voltage multiplied by some unrelated current. It is a voltage multiplied by the current flowing through THAT voltage!

The 32.4 W is for a current flowing through an 18 V drop (the voltage across the LEDs), not a 24 V drop. So you need to divide 32.4 W by 18 V, for which you get 1.8 A.

The 43.2 W is for a current flowing through a 24 V drop. Using these you get the same 1.8 A. The difference of 10.8 W is for a current flowing through a 6 V drop (i.e., the power in the resistors) and that gives you, again 1.8 A.


If you use a 40 Ω resistor per string of six LEDs, there is no way you are going to get 500 mA of current from a 24 V supply. If you shorted out all of the LEDs you would only get 600 mA.

The CREE datasheet, doesn't give minimum Vf values (which is not uncommon). At 350 mA, the Vf is typically 3.3 V but can be as much as 3.9 V. Based on that VERY limited data, you can probably expect the Vf to have a max of about 3.6 V at 150 mA. The minimum is a lot harder to estimate, but based on looking at the very few datasheets where a min Vf is given, you're probably looking at around 2.7 V at 150 mA.

So figure that the voltage across the string of six will be somewhere between 16.2 V (let's call it 16 V) and 21.6 V (let's call it 22 V). That means that, with a 40 Ω resistor and a 24 V supply, you're looking at a current somewhere between 50 mA and 200 mA. That's quite a spread, but the most likely currents are much narrower, perhaps somewhere between 125 mA and 175 mA.

If you put fewer LEDs in each string, or use a higher voltage supply, you can significantly reduce that uncertainty. Let's see what happens with just three LEDs in a string. Now you would have a nominal voltage of 9 V at 150 mA, leaving you to drop 15 V across the resistor, making it's nominal value 100 Ω. If we use the same range of Vf, namely 2.7 V to 3.6 V, that would give you a current that could be anywhere between 130 mA and 160 mA with the most likely current range looking more like to 145 mA to 155 mA. But the price is that you will be consuming twice the power, since you have twice as many strings, each with the same current as before. So it's a tradeoff that depends on what is important for your application.

You can also see why constant-current drivers are so strongly preferred.
 

LowQCab

Joined Nov 6, 2012
4,290
You've got to get the terminology right so that You can be understood.
"" Trick is determining the actual amp draw of the diodes. ""

1)
Diodes don't "Draw-Amps". Nothing "Draws-Amps".
2)
Diodes have a "Forward-Voltage-Drop", or a "Threshold-Voltage-range", which varies with Temperature.
The "Forward-Voltage-Drop" must always be measured at a specific Current and Temperature.
3)
Current, ( or Amps ), flow in a Circuit depending upon
the Voltage-Source-Level, ( measured in Volts ), and
the total Resistance of every component in the Circuit added together.
Diodes don't have a "Resistance", and therefore they are not measured in "Ohms".
Diodes are "Semi-Conductors",
which means that they "partially-Conduct", depending upon several factors.

Diodes can "withstand" a certain level of Heat,
the amount of Heat that will be generated in a Diode is directly related to
the amount of Current flowing through the Diode.
.
.
.
 

Thread Starter

jeffkrol

Joined Dec 8, 2015
56
Sorry yes there is a "theory" fixture (what if at 150mA) and an actual real life fixture (assume 500mA based on manuf statement of running at 1/2 err "normal" (3W) and assuming about 3V V(f) at 500mA ).
I know the thread is really messy....

the Real life lights are designed for constant voltage.. can't change that.
Bottom line that I STARTED with was given 4 lights supposedly rated at 27W and putting them in parallel with a single 24V 5A (thought it was 6 looks to be 5 rated.. hard to see image)
Each light has 2 err"parts" in my mind.
1) LED wattage
2) Resistor wattage
And as I assumed both play into sizing a power supply.
I think I got the whole picture straight now..
See orig. you have led wattage (SUM V(f) @ x current * current * 3 strings)
For the REAL light, assumptions based on info is 3V V(f) * 6 in series * 500mA * 3 = 27 watts
Of course each light 3 series strings in parallel

I was looking at concepts not hard numbers..
I'll get back to this

Ok back..
and "how I see it"
Resistors play the part of changing the voltage across the array.
IF I took 6 diodes with a V(f) of 3 volts and ran with a constant current driver of 500mA :
My system voltage would be 18V
My wattage would be 9W

Use constant voltage and you have 24V across the string. Still at 500mA (CHOSEN value based on resistor choice, all other parameters the same 3V drop across each of 6 diodes but now 6 V dropped across the resistor)
System voltage is 24V @ 500mA current through the circuit
My wattage would be 12W

See, the difference threw me for a moment.
Current through both is identical.. wattage isn't..because voltage changes..

There are companies that tell you the exact diode wattage. Those usually are run constant current.
All AFAICT constant voltage lights just list wattage with no explanation.. thus knowing the ACTUAL output of the diodes is dependent on the resistors used.
Most 12V led strips run (or were run) based on automotive specs so the strips needed voltage flexibility.. i.e won't burn out if voltage surges (common in auto circuitry) to say 15 volts..soo under-driven at the diodes..
 
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Thread Starter

jeffkrol

Joined Dec 8, 2015
56
You've got to get the terminology right so that You can be understood.
"" Trick is determining the actual amp draw of the diodes. ""

1)
Diodes don't "Draw-Amps". Nothing "Draws-Amps".
2)
Diodes have a "Forward-Voltage-Drop", or a "Threshold-Voltage-range", which varies with Temperature.
The "Forward-Voltage-Drop" must always be measured at a specific Current and Temperature.
3)
Current, ( or Amps ), flow in a Circuit depending upon
the Voltage-Source-Level, ( measured in Volts ), and
the total Resistance of every component in the Circuit added together.
Diodes don't have a "Resistance", and therefore they are not measured in "Ohms".
Diodes are "Semi-Conductors",
which means that they "partially-Conduct", depending upon several factors.

Diodes can "withstand" a certain level of Heat,
the amount of Heat that will be generated in a Diode is directly related to
the amount of Current flowing through the Diode.
.
.
.
LED "resistance"
http://lednique.com/current-voltage-relationships/resistance-of-an-led/
https://electronics.stackexchange.com/questions/76367/accounting-for-led-resistance


Yes, yes yes, LEDs don't "draw" amps..
Apply a voltage across a diode and based on it's "structure" it will "have" X amount of current
The 2 are intertwined to the point of it is semantics to separate it. At least in my mind.


Pick one axis... Apply or allow or set current (pick a term) Y get X voltage drop (yes at a specific temp/time/quantum vortex/black hole event horizon:))
I understand it is a passive concept with leds. IF you say apply 3.5V across the diode and a power supply restricted to an output of .1A you won't "pull" it till you burn out the ps. It just sits there.. under-powered..and dim unlike other things.
Or apply X voltage across the diode and get Y current
I know the description may not be technically correct but the application is..
In above a 700mA constant current driver will ramp up the voltage until it "sees" say 700mA of current flowing.
Power supply generally needs 2V over your projected voltage for common cc switching buck regulators. YMMV

Voltage across the diode will be approx 3.5V
Power supply should be >5.5V and capable of proving 700mA of current without stress.

Yes and diodes will become more efficient (more photons/watt i.e less heat) as you decrease the current but gross output decreases (less overall photons)

As diodes heat you get to a point where all "resistance" stops and the diode will draw as much current as the ps can put out and go into thermal runaway.. thus burning itself out..


It all starts with the profile of silicon. The electrical resistance of silicon increases with temperature up to about 160 Celsius, then starts decreasing. When resistance decreases, it allows more current to enter through the overheated areas, in turn causing yet more compounding heat, eventually leading to total failure of the device and hence the term Thermal Runaway.


With adequate thermal management and the use of a constant current driver, an LED will not exceed its maximum operating temperature and will never approach Thermal Runaway.
Call it Resistance or "resistance" your choice..

"Theoretically" one needs neither resistors nor current limiting circuitry if you EXACTLY match the power supply voltage to a desired string voltage at a desired current. Add desired temp to it if you must.
I do know of people who have done just that.
SOOO many things wrong with the concept though..

Haven't even touched methods of current balancing constant current controlled multiple parallel strings
 
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