0
\$\begingroup\$

I made a led circuit in parallel with a power supply of 12v 2a.

But there is the problem, I’ve changed the power supply. Still the same value of 12v 2a but my circuit doesn’t work. I’ve tried a lot of power supply with the same value but still the same.

The leds blink on and off quickly. I don’t understand why? enter image description here

R1 to R20 : 27ohm resistors

Leds : https://www.digikey.fr/products/fr?keywords=1214-1616-1-ND

\$\endgroup\$
  • \$\begingroup\$ I suggest you start with showing us you schematic. \$\endgroup\$ – Oldfart Apr 27 '19 at 12:22
  • \$\begingroup\$ Do you have a datasheet for the regulator? \$\endgroup\$ – RoyC Apr 27 '19 at 12:37
  • \$\begingroup\$ digikey.fr/product-detail/fr/texas-instruments/… \$\endgroup\$ – CharlieWhite Apr 27 '19 at 12:47
  • 1
    \$\begingroup\$ The 10 volts across 27 ohms, is 0.5 amps into each LED. You want this? \$\endgroup\$ – analogsystemsrf Apr 27 '19 at 13:10
  • \$\begingroup\$ The datasheet explain i = 500 mA at Vf min = 3,0 V. Vf = 3,3 V provides a 395 nm. That’s enough for me. Do you mean leds don’t have enouh current and this is why they are blinking? Resistors should be around 18 ohm in this case. \$\endgroup\$ – CharlieWhite Apr 27 '19 at 13:22
1
\$\begingroup\$

The datasheet you have linked in the comments (it should be in the question) is for "LMS1585A 5A/LMS1587 5A and 3A Low Dropout Fast Response Regulators". At best this can give out 5 A.

enter image description here

Figure 1. The VI curves for the LEDs. You're using the 385 nm (red curve).

Your LEDs are going to drop 3.4 V at 0.5 A. That means your resistors will drop about 8.6 V. (We'll home in on a closer value later.) The current through each resistor is \$ I = \frac {V}{R} = \frac {8.6}{27} = 0.32 \ \text A \$. With 20 LEDs you will need a supply capable of 6.4 A. The voltage regulator is dropping out on over-current or over-temperature.

Returning to the graph we can see that at 0.32 A the forward voltage is a little lower so we could recalculate if we wanted but it's not going to make much difference.

Your circuit is not efficient since 1/3 of the power is used by the LEDs and 2/3 wasted as heat in the resistors. Instead you can put two or three LEDs in series on each branch. Recalculate the resistor values for your chosen layout. With three in each branch you will require seven branches and a total of 7 x 0.32 A = 2.24 A. Close!

I have written about using a Loadline resistance graphic tool and it may be of interest.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ But why they are blinking? Normally with low current the blinkness will be low or normally turn off \$\endgroup\$ – CharlieWhite Apr 27 '19 at 14:27
  • \$\begingroup\$ As I said, the power-supply is shutting down on current limit and, presumably, restarting again. This cycling will cause the blinking. \$\endgroup\$ – Transistor Apr 27 '19 at 14:55
  • \$\begingroup\$ and another scenario. The cycle can be produced by the linear regulator? As a ripple - noise. Is that possible? Thank you for your help \$\endgroup\$ – CharlieWhite Apr 27 '19 at 15:17
  • \$\begingroup\$ Sorry, Charlie, but you're going to have to write proper sentences. I don't understand what you are asking. Why are you asking about a "linear regulator"? You haven't got one. \$\endgroup\$ – Transistor Apr 27 '19 at 16:13
  • \$\begingroup\$ @CharlieWhite - "As a ripple - noise" No. Ripple and noise don't occur at visible rates. Transistor is correct. You are grossly abusing your power supply, and what you see is the regulator doing its best to function when it can't. Look - if you're going to mess around with this stuff, get a cheap DMM. Start with a single LED, and measure your 12 volts. It will be perfectly stable. Then increase the number of LEDs and watch what happens. Stop flailing about in the dark and start measuring things. \$\endgroup\$ – WhatRoughBeast Apr 27 '19 at 16:41

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.