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The question is not that dumb as it sounds because the answer is 8. And what I really want to know if there are different 5050 LED's or I've got some particulary bad batch or am I making a measurement misstake.

I've bought gauss 14.4 LED strip which is marked "5050-60P-12V-10D-V1.1" So 60 5050 LED's per meter. This should consume 14.4W boths according to descriptions of such strips and according to calculations on 5050 datasheet. But my power supply shows 2.0A at 12V for 2.95 meter strip. That's 8W per meter. Voltage on the other end of the strip is 11.6V. The reason seems to be that voltage drop on a single LED is 2.9V when it should be 2.3V. Voltage drop on a single LED is 2.9V. For each 3 LED's there are two 39 ohm resistors (marked 390, and measurements confirm that). And I've checked my power supply (HY 3005) with resistor load, reading seems correct: 0.17A at 4V for 23.5 ohm (two 47 in parallel).

Update:

I've made a mistake by not describing the strip properly and not giving datasheets. This is a warm white strip and it's target voltage is really 12 volts. It's for home usage and is meant to be used with 12V power supply which you can buy in the same store. That supplies can't give 14.4 volts.

Unfortunately manufacturers site is very uninformative so there is no point to linking to it. There is also no datasheet available and I don't know LED's manufacturer. So I can only use datasheets for similar LED's. 3V drop for white LED seems OK, 2.3V from unedited question was for red LED's, which is a mistake.

Conclusion:

All components in the strip seems to be fine, it's just designed to drain less power than specified at 12V. Probably that's made for extending lifetime. Or, as @TonyStewart.EEsince'75 noted, extra room is left to allow usage of the same strips in cars at 14.4V. By increasing voltage I can make it brighter and still will be within allowed current limits for LED's, but stock power supplies for that strips can be maximally adjusted to 13V.

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    \$\begingroup\$ Depends if it's supposed no run from 12V, or from a "12V" car battery (which can be up to 14.4V if fully charged or if the car is running). Re-run your calcs for that higher voltage... \$\endgroup\$ – Brian Drummond Oct 29 '16 at 21:03
  • \$\begingroup\$ I don't know, but how much voltage shud a 12V LED take? can you supply the ref link or datasheet \$\endgroup\$ – Sunnyskyguy EE75 Oct 29 '16 at 21:04
  • \$\begingroup\$ @BrianDrummond 12V is 12V. This strip is for home usage. It can be bought in any store with electric devices for home. Comes with 220V-12V power supply. The same as for 12V halogen bulbs used in some furniture lamps or in bathrooms. \$\endgroup\$ – Alexey Guseynov Oct 29 '16 at 21:31
  • \$\begingroup\$ @TonyStewart.EEsince'75 That's a LED strip, not single LED. It is composed of blocks each one has 3 LED's and two resistors in sequence. So it is not current driven. It is designed for 12V supply and resistors take care ensuring proper current through LED's. And there are no datasheets in electricity stores unfortunately. \$\endgroup\$ – Alexey Guseynov Oct 29 '16 at 21:38
  • \$\begingroup\$ I was kidding ok? \$\endgroup\$ – Sunnyskyguy EE75 Oct 29 '16 at 22:03
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Let me try a simple analysis with no specs

  • my initial reading was you had two 39 in series = 78Ω, so I will show Rev B
  • let 14.4W / 60LEDs = 0.24W /LED including 1/3rd of Pd in 78Ω
  • let's assume they "cheated" ( or exaggerated marketing with no specs)
  • V+=14.4V ( car alternator = 14.2V nom) e.g. 14.4V*1A = 14.4W

    thus 60 LEDs in 20 strings = 50mA per string

    • and Vdrop on both R is 50mA*78Ω = 3.9V
    • with Vf = (14.4-3.9)/3led=3.5V (cheaper quality LEDs)

Conclusion

12V will never achieve rated power for these StripLeds, you need a battery and charger or a 15V or 14.4V power supply, because they are rated for Cars but still work on 12V since the LEDs add up ~9V at low current.

These will also be running near maximum temp so suitable for semi-trailer trucks with forced wind cooling. lol.

Rev B

Analysis with updates on Rs current limit using two 39Ω in parallel=20.5Ω

  • Given 60 5050 LEDs/m in parallel(P), series(S) array 20P(3S + 20.5Ω)
  • your results of 2.0A at 12V for 2.95 meter = 8W/m and not 14.4W/m
  • Now what would we expect at 14.2V with 2.9V/LED?

    • or a rise of 2.2V or almost 20% in power source to match opt. alternator voltage in cars
    • do we expect a linear incremental rise in power due to series R? no but,
    • a 20% rise in voltage, results in a 40% rise in power if it was linear
    • we need almost 6.4/8W=80% rise to achieve 14.4W, so it is quasi-linear
    • this is because the most of the voltage rise is across Rs=20.5Ω
    • You measured 2.9V/Led @ 8W/m thus If= 8W/(20S * 12V)= 33.3mA /string
    • thus V across 20.5Ω * 33.3mA= 0.68V
    • my trick is knowing LED ESR is due to incremental V/I above 2.7-2.8V
    • i.e. (2.9V-2.7V)/33mA=6.7Ω assuming your reading was accurate
  • now I can estimate power vs voltage for arrays in two ways, Can you?

    • such as 20S(3P+Rs) using LED ESD & threshold Vth or from specs

http://lianhaolighting.en.made-in-china.com/product/kbCEjFuMQZVg/China-LED-Strip-Light-LH-5050-60P-12V-.html

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  • \$\begingroup\$ Thank you. I think you are right: manufacturers make the same strips both for car market and for home lighting and specify maximal power consumption rather than actual consumption which can be achieved with stock power supply. \$\endgroup\$ – Alexey Guseynov Oct 30 '16 at 9:06
  • \$\begingroup\$ I'm glad you agree. I can tell many things from specs like ESR of LED and hence Vf and quality. 200mW in 78Ω @50mA and 720mW/3led&R means each 5050 LED uses 520mW/3 = 173mW in their 14.4W spec. I know ESR ~1/Pd or less thus 5050Led = 5.7 Ω +2.8V then at 50mA, Vf=3.1V and the fact they must have used 3.5V LED to get a bigger power value @ 1A means they used high ESR LEDs which are the cheapest. So power isn't everything, in fact lower ESR is better for cooler operation at same current. \$\endgroup\$ – Sunnyskyguy EE75 Oct 30 '16 at 18:39
  • \$\begingroup\$ I just read your revised question and realized my assumption or understanding was wrong . I thought you said initially 2x39 in series and now parallel makes more sense with lower Vf as a result.. see what happens with no stripleds specs? \$\endgroup\$ – Sunnyskyguy EE75 Oct 30 '16 at 18:51
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After a cursory look at the datasheets (Way Jun and Cree), it appears you have a power supply compliance issue.

White 5050 LEDs have a forward voltage comprised between 2.8V and 3.6V at 60 mA. Cree's datasheet for TR5050 says 3.3V typ for 120 mA. The fact that you are seeing 2.9V seems to point out to underpowering the strip. And the fact that you are measuring 2.0 A in the strip makes me think that perhaps you have set current limiting in your bench power supply to 2 amps.

BTW: 14.4 W / 12 V = 1.2 A; since (IIANM) every meter of strip has 60 LEDs in three-led series, the current in each of the 60/3 = 20 series is 1.2 A / 20 = 60 mA. Expect a voltage between 2.8 and 3.6 V per LED, then (according to Way Jun - but you might post the datasheet you used for your computation if you do not agree).

EDIT: correct typo, 60/2 in 60/3, as noted in comment.

EDIT: After reading the other answer I realized that I had underestimated the effect of the two resistors. One 39 ohm at 60 mA would drop 2.34 V leaving room for more than 3V per diode. But 78 ohms would drop more than 4.5V, thus bringing the voltage under 2.5V per diode with a supposedly required current of 60 mA, too far from typical spec for any datasheet I've scoured. So it still is underpowering, but it appears you cannot rule out voltage as a culprit.

Indeed, 2 amps for 2.95 meter (I take it you cut a 3 led series to experiment with?) corresponding to 59 3-LED series, would amount to almost 34 mA per series, 2.64V cumulative drop on the two resistors leaving 3.1 V per diode. Which is in line with what you observe (the difference is probably due to the metallic strip resistance and or tolerances in components or instruments). You need the I-V curve of the diode to verify if that's what your device is supposed to do at that supply voltage. (Incidentally, 2.9V at 35 mA is more in line with CREE's devices than with the other chinese datasheet).

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  • \$\begingroup\$ Cree does not make standard 5050 leds, you cannot take their specs for the average led strip. \$\endgroup\$ – Passerby Oct 29 '16 at 23:21
  • \$\begingroup\$ What about Way Jun? Moreover: do you know of white LEDs with 2.3 V forward voltage (we need a new quantum physics in that case)? \$\endgroup\$ – Sredni Vashtar Oct 29 '16 at 23:22
  • \$\begingroup\$ Plenty, when run at less than full current. But no, thats likely a red diode, or its a type of 3.3, much like your typo of "60 / 2 = 20"... \$\endgroup\$ – Passerby Oct 29 '16 at 23:25
  • \$\begingroup\$ Ah ok, so I have this 4.7 V zener but I will call it a 3.3 zener, since I am running it at very low current... Thanks for pointing out the typo, though. \$\endgroup\$ – Sredni Vashtar Oct 29 '16 at 23:31
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Hi, it is no big relation with the LEDs; The voltage dropping problem is because of the PCB board, the resister value of PCB board is big enough to reducing the voltage as the lenght of strip going on; the solution is increase the thickness of PCB board or width of PCB board; or add constant IC on the LED strip to save the voltage.

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  • \$\begingroup\$ That's a common problem but it's not the reason for this particular case. I've measured voltage on the other end of the strip and it's 11.6V. So 0.4V drop on the whole strip. \$\endgroup\$ – Alexey Guseynov Nov 16 '16 at 17:42

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