One experiment is worth 1000 opinions… so just do it. Find a decent LED strip (i.e. positive and negative conductors are well matched… you’ll need a good milliohm meter with kelvin connections to verify that) say 5 meters long.
Drive the strip from each end (+ to one end, - to the other). Then go down the strip with a volt meter and measure the voltage at each LED group (it’s ok to get lazy and measure it every foot or so). Report your results.
My strip has 150 ohm resistors in series with each set of 3 LED’s which are in series. Each of those set of 3 are in parallel. I think this is typical of how these strips are made. That relatively high input resistance of each set of three LED’s effectively simulates each set of 3 LED’s having its own constant current source. Current limiting resistors not only mute the effect of raising the voltage, but also mute the effect of lowering it and the voltage at each node will be almost identical no matter what. That’s why these strips are wired this way and it is also why you would need a very high resolution volt meter to detect any difference.
But you know that. 
If you would want to be able to run longer lengths of these LED strips, then the buss voltage could be increased along with the proper increase in the current limiting resistors.
I believe Eugene is right when he explained this, there is an extra I*R drop on the middle LED. And it seemed that it would make less and less of a difference percentage wise as the number of LED’s in the strip increased.
Just to clarify what I wrote here.
Certainly as the string of LED’s gets longer, the buss will load and the LED’s will get dimmer. What I am referring to is the difference, if at all perceptible, between the center most LED’s and the outer ones. As the string gets longer, the difference percentage wise will become smaller and smaller.
More and more. Let’s N LED segments in the strip sinking equal current.
U2 = U1 - (N-1)*I*R + I*R = U1 - (N-2)*I*R
U3 = U2 - (N-2)*I*R + 2I*R = U2 - (N-4)*I*R = U1 - (2N-6)*I*R
and so on.
For long strip it’s easy to write a differential equation for linearized network and get answer what LEDs current (i.e. brightness) is proportional to hyperbolic cosine of LED position as I explained before. Hyperbolic Cosine -- from Wolfram MathWorld
Only any significant current source, an ampere-meter and a voltmeter.
Here is a diagram that explains what is happening. The resistors represent the resistance of the buss wires on the strip
This is a real gotcha, until I drew this I assumed that all LED’s would get the same current when a strip is wired this was (positive and negative connected at opposite ends). I would guess this got Eugene at sometime, that is why he was up on it. We will be too, now that we know.
Dchomak and Eugene, you just blew my mind!
The hilarious part of it all is that we’re debating the best way to drive LED strips that cost less than $10 a spool
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I love this forum 
Dchomak, great work!
The other interesting feature of this network is that if we drive it from both sides by 4 wires and adjust the driving voltage accordingly then for strip wires of equal resistances the LED voltage distribution across the network will be the same as on your drawing. And the same as for LED srtip of half length driven by the same voltage from one side. For wires of different resistances the voltage distribution will be more uniform if driving voltages are equal.
PS Yes, I did the same mistake originally.
Let’s not loose sight of how superior a connection “texaspyros method” is vs connecting both leads at one end. Here is the diagram for that.
Notice that the variance in the voltage drops to each LED is much greater.
With the “tp method” a strip with 100 LEDs the first and last LED would have a drop of 5050R and the center most LED would have a drop of 5050R + (2*100)–6 = 5244R. Percentage wise this is next to zero, so all the LEDs would appear to be the same brightness.
dchomak, I do not understand the difference between 4, 3, 2, 1. What do those numbers actually represent and why are they not all the same (eg 4, 4, 4, 4)? Wait, I suddenly get it. Interesting. As we travel along the V+ bus and pass each LED, there is progressively less current flowing along each section of that bus. With less current flowing, voltage drop is reduced. A similar thing happens on the GND bus. You’ve drawn up the interaction between those things.
FYI the way you wrote a simplification of (x+x…)R on the same line really threw me off. If that’s a standard practice I don’t know about it - I’m just letting you know that it confused me! Not that I wasn’t still confused after I figured out what you were doing there. It still took me a minute to suss out what you and Eugene are talking about.
This variance is absolutely the same as if to drive from opposite sides LED strip of double length.
In opposite side driving method (i. e. “tp method”) LED brightness in a long strip is proportional to cosh(x/a), -L/2<=x<=L/2, a = const in linearized LED current model I = s*(V-Vth). The more L the more is relative difference between the center and side LED currents. The relative difference rise is exponential. For very long LED strips center LEDs will be dark as their currents will be near zero.
Please note also that in opposite driving method we should set VA = 12V by adjusting driving voltage. And for long strips where driving voltage greatly exceeds LED threshold voltage Vth but we set VA=12V, voltage drop on wires for all LEDs will be about the same indeed and approximately equal to the driving voltage, and their ratio will be near unity indeed. But not LED currents and not their brightness we speak about.
Also your wire voltage drop calculation for 100 LEDs is incorrect. For center LED the wire voltage drop is approximately 2*(N+N/2)/2*N/2 = 3*N*N/4= 7500 as there is an arithmetic progression of currents in wires connecting the center LED with sides. And for side LEDs accordingly the voltage drop is approximately N*N/2. So ratio of wire voltage drop for center LED to wire voltage drop for side LEDs has limit equal to 3/2. But this conclusion is valid for simple LED model only when we model LEDs by unit current sources.
Eugene, I have been on the road all day and am using my iPhone. Because of that, I may have made a mistake in my calculations for a 100 led strip, but I think not.
Are you getting you arithmetic progression by applying Tellegens Theorm, because if you are I don’t know that it can be applied here. The LEDs are diodes and there can be no counter currents flowing in any of the loops. I took my calculations from my diagram.
Voltage drop at the ends is the sum of the digits 1-100 =5050 (remember Gauss?)
Voltage drop at the mid point would be 5050 + (2N-6) = 5050+194 = 5244.
Not a big difference.
I calculate sum of arithmetic progression by multiplying number of elements in sequence by mean value of sequence elements. It is very easy rule to remember. It is valid for any sequence but for arithmetic progression {n, n+d, n+2d … n+md} the mean value of elements is equal to simple mean of first and last elements of progression (n+(n+md))/2. So sum of such progression is (m+1))/2. And I through off small constants like ’1’ added to big numbers as they are not important for approximate calculations. So sum of long enough arithmetic progression with step 1 from N to M is approximately equal to (M-N)(N+M)/2 and sum of arithmetic progression from 1 to N is approximately equal to N*N/2.
Let be N LEDs in strip. Current in positive wire drops in sequence from (N-1) down to 1 with step 1 (yes, from N-1, not from N!), and current in negative wire increases accordingly from 1 to N-1. The same are voltage drops on unit wire resistors in sequence. First LED sees wire voltage drop on negative wire only. So it sees wire voltage drop as sum of natural numbers from 1 to N-1 what is precisely equal to (N-1)*N/2 or approximately equal to N*N/2. So if N=100 the first LED sees wire voltage drop 4950 exactly or 5000 approximately.
LED in the center sees equal voltage drops on positive and negative wires, each starting from N/2 approximately near the LED up to N approximately on sides with N/2 elements in sequence. So it sees wire voltage drop approximately equal to 2 * (N/2)*(N/2 + N)/2 = 3N*N/4. So for N=100 the wire voltage drop for the center LEDs = 7500 approximately.
Yes you are absolutely right. I used the wrong formula in calculating the voltage drop on the middle LED. It is apparent to me now that you have done excellent work here and have nailed it. It has taken me awhile to grasp all that you have done, but now I do. I hope you will stick around on this forum, I think you will enjoy it here and I can see that you would be able to contribute to this community.
dchomak,
thank you. In community everybody does his job.
Now I see that such high-power LED strips rated for 14.4 W/m no doubt will have problems with overheating also. 1 cm wide copper line generating 14 W/m of heat glued from bottom to wood panel of kitchen cabinet must become very hot. Power rating of 7 W/m is much more adequate as have half the temperature rise. And water protection plastic tubes of some LED strip types will rise the LED temperature even more. May be such race for Chinese LED strips power is blind. Minimum electrical power with maximum luminosity is what must be goal.
BTW I see that less powerful LED strips have less of copper also. So the characteristic length I spoke about before is about the same 2.5-3 meters for them also. In fact the best driving strategy is to use small cuts of such LED strips (0.5-1 m length) attached to thick 12V copper cable. It doesn’t matter how to drive them: from both sides with double length cuts or from one side with single length cuts. Also voltage drop in driving line must be carefully controlled to achieve LED brightness uniformity as current setting resistors have 2-2.5 voltage drop only.
The temperature rise of the strips in open air is actually not too bad. Don’t drive them while still wrapped on the reel… you’ll melt the reel… don’t ask me how I know… :party:
Good idea. It is the same as to drive two strips of half length independently. Or drive the same strip from both sides.
Use two 12V power supplies with isolated secondary circuits connected in sequence to get 24V.
Are they small scraps? Can you post up pics of what you’ve got or a link to pics of the configuration of these strips?
Not extremly bad but if you glue 12V 14.4W rated LED strip to bottom side of half inch plywood and drive it by maximum allowed 1.2 A/m, the temperature rise of LED package will be about 50°C above ambient.
Your oblation saved a lot of other reels, Im sure.
BTW I’ve found site of Chinese LED factory producing both LEDs and LED strips. They have English datasheets available http://www.snowdragonledhk.com/smd-5050-flexible-led-strip-light.htm The maximum LED strip rated power they produce is 7.2 W/m. Also they have water-protected LED strips. Hope they use non-aggressive electronics-grade compound.
Number of LEDs connected in sequence and value of current-setting resistor. For 14.4W-rated 12V LED strip with three 20mA LEDs in package it is 3 LEDs in sequence and 150 Ohm resistor.
My dining room table uses LED strips to mimic Lumiline light bulbs ( Replacing Lumiline light bulbs with LEDs) Each “bulb” has 1.6 meters of strip on them driven at 1 amp. They just get pleasantly warm. I’ve never measured it because their temperature is not high enough to be anywhere near a concern. If that 1.6 meters was spread out over the length of a shelf or counter the temp rise would be quite a bit less.