Stop LEDs from flickering

Thread Starter

Rokket2001

Joined Mar 19, 2023
9
My circuit board LEDs (scale model) flicker instead of staying steady. A few do not light at all (but have lit before). Here's my set up and some mini history:

18 white LEDs (3.3 Vf, 20mA ea, total 360mA)
5V mini power brick (2000mA, 5V, 1A)
18 x 100 ohm resistors (1w) in parallel. Ohm's Law and resistor calculator say 85ohm

Connections soldered. Bad connections unlikely.

I first tried a 12v camera battery and a tiny buck board to get to 5V: LEDs came on dim and flashed once, then nothing. Disconnected and reconnected, dim flash on reconnection each time.

Switched to a wall wart, 5V, 1A. The LEDs came on bright but flickered often, almost flashing. Original 0.25watt resistors very quickly became soooper hot.

Switched to a mini power brick as above. Lights came on bright and steady, but one LED flickers like it's on a flasher circuit, and a few don't come one at all (they have lit before). New resistors 1watt, warm but OK. Tried full sized bricks, 11,000 and 20,000mA, same.

I'm no super genius, but I think maybe something is wrong. The circuit wiring is 2 layers of copper tape (4mm or 3/16th"), about 960mm or 3ft. All the resistors are grouped on a mini circuit board.

I don't know what is happening. I thought the wall wart would solve everything.
 

Thread Starter

Rokket2001

Joined Mar 19, 2023
9
You have all the resistors in parallel, straight across your power supply, so it's no wonder the LED's are not lighting up! 1 resistor needs to be in series with 1 LED, and each pair then in parallel across the supply!
You've probably destroyed some of the LED's by now as well!
Thanks, I'll give it a try. LEDs are all still OK so far.
 

Jerry-Hat-Trick

Joined Aug 31, 2022
491
I'm counting 16 LEDs, not 18. Starting with a 12V suppply, no need to drop to 5.0V. If you connect three LEDs in series and expect 3.3V to drop across each that's a total of 9.9V drop. A series resistor with those three LEDs will therefore drop 2.1 Volts. For 20mA make that series resistor - near enough 100 ohms. Try it with one string first. Then put another 5 strings like the first one in parallel
 

Thread Starter

Rokket2001

Joined Mar 19, 2023
9
Old diagram (I added a couple LEDs and am using USB 5V power now). I'll mess around tomorrow and see what happens. I should have done a full breadboard test before solder and disappointment, but of course I didn't expect anything to go wrong. I've labeled my eyebrows for just such an emergency...
 

Ya’akov

Joined Jan 27, 2019
8,973
Welcome to AAC.

If you have actually wired this up per your drawing, as @sarahMCML has noted, you have ~6.7Ω directly across the buck converter. That means you are trying to draw ~750mA from that buck converter before even considering the load of the LEDs.

You are correct that it doesn’t matter whether the resistor is not connected directly to the LED—but a current limiting resistor has to be in series with the load. Resistors in series add, resistors in parallel reduce the resistance.

Do you know Ohm’s Law?
 

Thread Starter

Rokket2001

Joined Mar 19, 2023
9
Sorry, old diagram, I've switched to 5V USB. I got confused (small brain) about the parallel wiring - I put the LEDs in parallel, but also the resistors, trying to give common power. l'll do a breadboard test with two resistors wired correctly ("parallel" doesn't mean across resistors!) I think. Yes, used Ohm's Law to calc resistor:

5V(supply) - 3.3(LED) = 1.7V / .02mA(current) = 85 Ohms (100 for safety).

But putting all the resistors together was actually the problem. A new diagram and then breadboard!
 

Ya’akov

Joined Jan 27, 2019
8,973
Sorry, old diagram, I've switched to 5V USB. I got confused (small brain) about the parallel wiring - I put the LEDs in parallel, but also the resistors, trying to give common power. l'll do a breadboard test with two resistors wired correctly ("parallel" doesn't mean across resistors!) I think. Yes, used Ohm's Law to calc resistor:

5V(supply) - 3.3(LED) = 1.7V / .02mA(current) = 85 Ohms (100 for safety).

But putting all the resistors together was actually the problem. A new diagram and then breadboard!
When you put a resistor inline with an LED it decreases current in proportion to its value. When you put a resistor across the leads of an LED, it reduces the resistance in the combined circuit according to the formula:

\[ R_{eq}=\frac{1}{(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\ldots+\frac{1}{R_n})} \]
Often stated as the reciprocal of the sum of the reciprocals. This all stems from Ohm’s law. But you don’t really need to know th math, per se, if you just learn the rules for series and parallel circuits of resistors, capacitors, and inductors. Once that is internalized, you’ll find it intuitive and you can always look up the formulas if they are not easily recalled.

AAC has some tools online that can make things easier:

Series Resistance Calculator
Parallel Resistance Calculator

[EDITED: cleaned up \( \LaTeX \) but no content changes.]
 
Last edited:

WBahn

Joined Mar 31, 2012
29,843
My circuit board LEDs (scale model) flicker instead of staying steady. A few do not light at all (but have lit before). Here's my set up and some mini history:

18 white LEDs (3.3 Vf, 20mA ea, total 360mA)
5V mini power brick (2000mA, 5V, 1A)
18 x 100 ohm resistors (1w) in parallel. Ohm's Law and resistor calculator say 85ohm

Connections soldered. Bad connections unlikely.

I first tried a 12v camera battery and a tiny buck board to get to 5V: LEDs came on dim and flashed once, then nothing. Disconnected and reconnected, dim flash on reconnection each time.

Switched to a wall wart, 5V, 1A. The LEDs came on bright but flickered often, almost flashing. Original 0.25watt resistors very quickly became soooper hot.

Switched to a mini power brick as above. Lights came on bright and steady, but one LED flickers like it's on a flasher circuit, and a few don't come one at all (they have lit before). New resistors 1watt, warm but OK. Tried full sized bricks, 11,000 and 20,000mA, same.

I'm no super genius, but I think maybe something is wrong. The circuit wiring is 2 layers of copper tape (4mm or 3/16th"), about 960mm or 3ft. All the resistors are grouped on a mini circuit board.

I don't know what is happening. I thought the wall wart would solve everything.
You are abusing Ohm's Law -- a pretty common thing for newbies.

In order for your calculations to mean anything, they must match the physical connections.

If you want to the following equation to calculate the value of your resistors:

R = (V_cc - V_led) / I_led = (5 V - 3.3 V) / 20 mA = 85 Ω

then your circuit must be constructed so that the voltage across the resistor is (5 V - 3.3 V) when the LED has 20 mA flowing through it.

That means that the resistor and the LED have to be in series.

If you are putting all of the resistors in parallel (and then all of the LEDs in parallel with that), the resistors have no effect on the LEDs (until they represent such a load on the power supply that it's output voltage starts dropping).

With all of the LEDs in parallel across the supply, it's hard to say what is happening because it all depends on the fine details of the individual LEDs and the power supply. In general, however, this is not good and will often lead to damaged/destroyed LEDs pretty quickly.
 

Tonyr1084

Joined Sep 24, 2015
7,766
Jumping right into the very first thing I saw was the parallel resistors. 18 of them as you state, though your diagram shows 16. Nevertheless, when you put two 100Ω resistors in parallel the resulting resistance is only 50Ω. when you put 18 of them in parallel the resulting resistance is 5.555•••Ω At 5 volts and 5 ohms, you're drawing 1 amp through them. At 1 amp and 5 volts you're dissipating 5 watts of power. Yikes! That's going to make the resistors hot to the touch. Fortunately when you parallel resistors you add their wattages together. So 18 quarter watt resistors is still 4.5 watts. Since you've replaced them with 1 watt resistors (overkill for the setup - though not done correctly) will survive 5 amps.

It's already been stated, the resistors need to be NOT in parallel BUT in series with each LED. That's the proper way to do it. With 5V and 20mA, you want a resistor (for each LED) to be:
(5V - 3.3V) ÷ 20mA (0.02A) = 85Ω. Using a 100Ω resistor will result in lower current to each LED, will result in 17mA. The difference in brightness will not be noticed by the human eye.

Your circuit is wired totally incorrectly.
 

Thread Starter

Rokket2001

Joined Mar 19, 2023
9
I now have a special certificate for being an idiot! I got totally confuzzled between parrallel LEDs and parallel resistors and stupidly made the resistors parallel too. The resistors DID get hot, egg frying style. I see now. I do wish I had a way to keep all the resistors together without a million wires though, part of my racing down the wrong path. Thank you for the clear diagram!
 

Tonyr1084

Joined Sep 24, 2015
7,766
Six resistors (105Ω) three LED's in series, six groups of series LED's in parallel calculated for 20mA. You will need a different power source, but if you don't want 18 resistors and 18 LED's you can series/parallel as shown and achieve the same thing. If you go with 100Ω resistors the amperage will be 21 mA. Well within safe operation for 5mm LED's.
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