3
\$\begingroup\$

LED Option

1 led strip = 3a, 36w, 12v, 16.5ft ..............6 strips= 18a, 216w, 12v, 99ft

Will this transformer/ power supply (20A, 240w) be capable of handling the 6 strips with the specs mentioned above? Each strip has on its end an external female power adapter that can be plugged with any 3a male power adapter.

1- Which gauge of wire would be sufficient to use for power supply to the female adapter, I have 100ft of 12AWG and will option 1 of having the transformer in middle and run 50ft to left and right be best or with a 30A, if that.

2- Will running two 20a transformers linked to that ONE power feed cause issues?

// Each strip are connected to one another and power on but dim thus the need for external power. I think of it more like a water hose being pushed by two outlets from each end to maintain pressure :-D

\$\endgroup\$
  • \$\begingroup\$ How are the strips to be arranged, physically? That is, are they going to be a single long array or something else? \$\endgroup\$ – EM Fields Sep 30 '16 at 8:35
2
\$\begingroup\$

The power supply seems to be capable of driving the whole strip chain.

If you run the supply from only one end of the chain, there will be voltage drops. Because, first strip draws 3A and the remaining 15A flows through its copper tracks. Then second strip draws 3A and the remaining 12A flows through its copper tracks. Then third strip draws 3A and the remaining 9A flows through its copper tracks ...and so forth. This means, there will be a couple of voltage drops on the strips which results in brightness decrease from one end to another. That's why the supply should be applied from both ends.

1) The current need of cables for each option is about 9A. According to the table on this page (http://www.engineeringtoolbox.com/wire-gauges-d_419.html) at least AWG #16 will be sufficient.

OPTION 1: If you use AWG #12, the cable resistance of 50ft will be 0.08ohms and the voltage drop on each cable will be about 9A x 0.08 = 0.75V. So, input voltage on the strings will be about 12 - (2 x 0.75) = 10.5VDC. This will cause a decrease of brightness, clearly.

OPTION 2: If you use two separate power supplies and make the connection cables shorter, this will give you a better result (i.e. nearly no voltage drops on the cables).

Note that there will be a little and equal amount of brightness decrease on LED3 and LED4.

2) If you run the power supply from both ends of the strip chain (OPTION 2 with one supply or two), strip 3 and 4 shouldn't be connected to each other. Because you apply the supply voltage from both ends.

\$\endgroup\$
  • \$\begingroup\$ AWG #12 copper wire has a resistance of about 1.6588 ohms per thousand feet, \$\endgroup\$ – EM Fields Sep 30 '16 at 12:18
  • \$\begingroup\$ @EMFields According to the link I shared in my answer above, AWG #12 cables have a resistance of 5.4 Ohms / km. And 50ft = 15.2m, thus that long cable will have a resistance of 5.4/1000*15.2 = 82mOhms. Am I wrong? \$\endgroup\$ – Rohat Kılıç Sep 30 '16 at 12:24
  • \$\begingroup\$ AWG #12 copper wire has a resistance of about 1.6 ohms per thousand feet, so 50 feet of it would be about 80 milliohms. For 50 of cable, though, that's 80 milliohms out and 80 milliohms back for a total of 160 milliohms for the 50 foot run, making the drop 1.5 volts with a 9 ampere load. \$\endgroup\$ – EM Fields Sep 30 '16 at 12:27
  • \$\begingroup\$ Ah, yes! You're right. Input current does drop voltage on the return path as well. Now I'm editing my answer. \$\endgroup\$ – Rohat Kılıç Sep 30 '16 at 12:32
0
\$\begingroup\$

If your lamps are going to form a single long array, then you can save about 33 feet of cable by using a single supply and hooking the lamps up as shown below.

enter image description here

12 gauge (AWG) copper wire has a resistance of about 1.6 ohms per thousand feet, so a 16.5 foot run of that wire would have a resistance of about 26 milliohms.

That's 26 milliohms out and 26 milliohms back, which would be a total of 52 milliohms for 16.5 feet of cable.

With that in mind, if we give the LED strings a resistance of \$R =\frac{E}{I}=\frac{12V}{3A} = 4\text { ohms} \$

and draw a schematic of the circuit, we come up with this:

enter image description here

And this DC operating point analysis from LTspice.

       --- Operating Point ---

V(n002):     11.7654     voltage
V(n001):     0.234551    voltage
V(n004):     11.8433     voltage
V(n003):     0.156718    voltage
V(n005):     12  voltage
V(n008):     11.7654     voltage
V(n009):     0.234551    voltage
V(n006):     11.8433     voltage
V(n007):     0.156718    voltage
I(R12):  5.80437     device_current
I(R11):  -5.80437    device_current
I(R14):  2.88272     device_current
I(R13):  -2.88272    device_current
I(R10):  -5.80437    device_current
I(R9):   5.80437     device_current
I(R8):   -2.88272    device_current
I(R4):   3   device_current
I(R5):   2.92164     device_current
I(R6):   2.88272     device_current
I(R7):   2.88272     device_current
I(R3):   3   device_current
I(R2):   2.92164     device_current
I(R1):   2.88272     device_current
I(V1):   -17.6087    device_current
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.