0
\$\begingroup\$

I have the following LED array:

LED Layout

enter image description here

I need help calculating the power supply needed. All the LEDS are the same - 1W, 700mA and instead of just 5 columns my actual circuit has 14 columns (as in the second picture). Please can someone teach me how to do the calculations for this circuit. If it were a straight forward parallel string of series LEDs, I would not bother you, but the additional junctions throw me off a bit.

Please don't suggest alternative layouts as this is what I am reverse engineering.

Thanks

\$\endgroup\$
  • 1
    \$\begingroup\$ ”instead of just 5 columns my actual circuit has 14 columns” Draw what you have instead of something else! \$\endgroup\$ – winny Apr 9 '18 at 19:02
2
\$\begingroup\$

Your power supply will need to supply 2.8 amps (700 mA x 4 - current adds in parallel strings) and approximately 42 volts (~3V x 14 - Voltage adds in series strings).

\$\endgroup\$
  • \$\begingroup\$ Are you sure about this because it is not 4 parallel strings of 14 series LEDs? Or am I missing something? \$\endgroup\$ – Dylan144GT Apr 9 '18 at 19:59
  • \$\begingroup\$ @Dylan144GT: no, i think we're missing something, do you not have 4 parallel strings of 14? \$\endgroup\$ – dandavis Apr 9 '18 at 20:39
  • \$\begingroup\$ No, the PCB layout is exactly what you see in the left drawing, except for the extra 9 columns of LEDs on the right. \$\endgroup\$ – Dylan144GT Apr 9 '18 at 20:51
  • 1
    \$\begingroup\$ You need to be more clear on the configuration - what does "the extra 9 columns" mean? You need to draw out the COMPLETE circuit. And are all the LEDs the same? \$\endgroup\$ – Norm Apr 9 '18 at 21:33
  • \$\begingroup\$ @Norm As requested, I have now added the full diagram and also edited the OP. \$\endgroup\$ – Dylan144GT Apr 12 '18 at 19:41
0
\$\begingroup\$

Norm's answer is correct.

Don't be fooled by the additional junctions. You can think that each time you have these junctions, the current that would otherwise go to the following LED is now equally split by 4 ( due to having 4 parallel strings).

So instead of 700mA you would have 700/4=175 mA for the rest of the LED's of the same string. BUT, this split happens at all junctions and therefore 175 mA from every other string will come and add up to the cuurent string to replenish the initial 700 mA.

\$\endgroup\$
0
\$\begingroup\$

You need to verify the 700 mA in your question.

A discrete LED is spec'd at a voltage 2 to 2.5 V at 20 mA. Since the LED is a diode so very little change in voltage change once your get pass the knee. It will take all the current your power source can give out.

In a typical design there will be resistor or a current regulator to prevent over driving. This can be built into your power supply or in your circuit.

A string of 5 LED implies a voltage requirement of 12.5 volts nominal.
A string of 14 LED implies a voltage requirement of 35 volts.

I have not seen any diode rated in the 3 digits but I could be wrong so please check your 700 mA. You can purchase a component call a current regulator which looks like a diode but keeps the current flow constant vs a resistor where the current voltage dependent. There should be one in each string. Check the minimum forward voltage requirements and adjust your power supply output spec's.

If your power source is AC, you need to add a 1N4004 or 1N4007 to protect against the reverse voltage. The 1N4007 (1000 DC volts blocking) is good enough to pass some lightning tests and it is very cheap (19 cents).

Note. The optimal operating point of your diode requires a simple test to find out what it needs. Just put a resistor in series and test for the brightness you require. This will determine your current requirements.

\$\endgroup\$
  • \$\begingroup\$ This is not true: "A discrete LED is spec'd at a voltage 2 to 2.5 V at 20 mA" Red LEDs are typically about 1.9-2.1V where blue, green and white LEDs are about 3V. 700mA is a very common current rating for LEDs used in lighting. \$\endgroup\$ – Misunderstood May 6 '18 at 0:32

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.