2
\$\begingroup\$

For clear any mistake, the question is: Why the LED started to blink after being replaced?

The goal was to replace the blue leds with red leds, after replacing the leds one of them started to blink, each blue led is connected to it own resistor of 100Ω

There are three tracks in this board with the following schematic. (Before Replacing there was just one track with resistor and led, on the other two there was no led (just resistor))

schematic

simulate this circuit – Schematic created using CircuitLab

And one track with this schematic

schematic

simulate this circuit

Based in the comments I did a calculation, but it was wrong, based in the answers I realized what was wrong and learned how to do it correctly.

First of all, I searched for the voltage of each LED, but I found a lot of tables with different information, so I found a video in portuguese in youtube with a small circuit to find the voltage of a LED.

schematic

simulate this circuit

Now I have:

enter image description here

Using the Kirchoff's Voltage Law, I've calculated the resistor voltage. enter image description here

After that I've applied the Ohm's Law to find the Current(I) in this circuit.

enter image description here

With the current(I) in mind I could do the inverse to find the resistence value for the red led.

enter image description here

"Just for fun", I've followed what "K H" said and calculated the current for the red LED using a 150Ω (what I will use) resistor:

enter image description here

Ok, with that I found the resistor I need to use, but "one track" have a place for a second led in parallel, so I calculated the resistor to place a second red led there.

enter image description here So I'm using a 68Ω to this circuit.

I know that this added 60mA (100mA Total) in the source, I could use something near to 400Ω to drop the current to 8mA per LED so I could drop the usage back to 40mA, but the brightness would be low.

\$\endgroup\$
  • \$\begingroup\$ Do you know Ohm's Law for Voltage drop across resistor? \$\endgroup\$ – Sunnyskyguy EE75 Jan 23 at 17:05
  • \$\begingroup\$ Remember that for each different colour (wavelength) there is a different voltage drop across the LED - so the resistor will also need to be different to accommodate for that. \$\endgroup\$ – QuickishFM Jan 23 at 17:09
  • \$\begingroup\$ @SunnyskyguyEE75, no I don't, I'm just trying to figure out how to calculate that. \$\endgroup\$ – André M. Faria Jan 23 at 17:12
  • \$\begingroup\$ @QuickishFM Ok, so I have to change the resistor, I was not sure of that because the other led is not blinking. \$\endgroup\$ – André M. Faria Jan 23 at 17:12
  • \$\begingroup\$ To be very honest, I don't know what I'm doing, I'm just trying to change the led color. \$\endgroup\$ – André M. Faria Jan 23 at 17:13
2
\$\begingroup\$

Diodes are non linear unlike resistors so when you're sizing resistors for LEDs, you normally do the calculations based on the forward voltage of the LED, total circuit voltage and resistance.

In your case, you have 160\$\Omega\$ ohm resistors in series with the original LEDs. This will let us estimate the original LED current(you don't want to overdraw whatever is powering the LED).

schematic

simulate this circuit – Schematic created using CircuitLab

A red LED takes about 1.8v and a blue LED takes about 3.3V. The forward voltage is determined by the materials required to make the color, so there are general voltage levels you can find to do equations like the following:

\$V_{F RED} = 1.8V, V_{F BLUE} = 3.3V\$

They were just pulled off random web pages, so do your best to identify the old and new LEDs you have and then re-do these equations with those values for better accuracy.

Kirchoff's Voltage Law tells us that the voltage across the series loads will add up to the source voltage:

\$V_{USB}=V_{R1}+V_{F BLUE}\$

\$V_{R1}=V_{USB}-V_{F BLUE}=5V-3.3V=1.7V\$

Resistors ARE linear, so knowing the voltage across the resistor, we can find current, using Ohm's Law, Which is that Current(I) is proportional to Voltage(V) divided by Resistance(R), or I=E/R

\$I_{BLUE}=V_{R1}/R1=1.7V/160\Omega=0.010625A\$

Now we know we shouldn't drive your new LED harder than 10mA off whatever the unknown switch is (The flashing LED could be caused by you browning out the power supply, overcurrent repeatedly tripping, a bunch of stuff that this will likely prevent.)

Now we just have to look at your intended circuit:

schematic

simulate this circuit

We can take the information we now have (intended current) and apply it to find the resistor you require.

Kirchoff's Law to find the voltage on the unknown resistor:

\$V_{R?} = V_{USB} - V_{F RED} = 3.2V\$

Now we know the voltage across it and the current we want, so we can apply Ohm's Law to find resistance.

\$I_{R?}=V_{R?}/R?\$

\$R?=V_{R?}/I_{R?}=3.2V/0.010625A=301.176470\Omega\$

So you need a 301.176470\$\Omega\$ resistor, but live on the wild side and order a 300\$\Omega\$ one instead!

Just for fun, let's calculate the current for the red LED and the 160\$\Omega\$ resistor:

\$I_{R1}=V_{R?}/R1=3.2V/160\Omega=0.02A\$

So with the original resistor and the red LED, the LED is probably drawing about 20mA from whatever is driving it, just as much current as both blue LEDs combined were intended to.

\$\endgroup\$
  • \$\begingroup\$ Followed your instructions and, take a look in my question I've added all step by step. \$\endgroup\$ – André M. Faria Jan 28 at 16:01
2
\$\begingroup\$

UPDATE

The question has become a moving target.
NOW, what is your Question?

First of all, I searched for the voltage of each LED, but I found a lot of tables with different information, so I found a video in portuguese in youtube with a small circuit to find the voltage of a LED.

I previously told you; to calculate the Vf all you needed to do was measure the voltage on both sides of the resistor.


I'm sorry for my miss explanation, each LED has your own resistor.

If each LED has its own resistor then your explanation that you changed one of two blue LEDs to a red and the existing blue LED began blinking makes no sense.

You began with two separate circuits both blue.
In your original posting you said your replaced a blue with a red and the other blue began blinking.
How can changing the LED in one circuit affect the other causing it to blink?
Original you said it was mouse with the two blue LEDs. Now you seem to be using another PCB and the issue is not longer about blinking.

Now you state the supply is 5V and 5.14V and the resistors are 100 Ω and 220 Ω and the original 68R (68 Ω) no longer exists.

Do you want to know why the blue LED started blinking
Or do you want to know how to calculate a resistor value which in that case this question should be closed as a duplicate.

It should also be closed because you are no longer asking a question.



Regarding the original question I assume that was this circuit with two blue LEDs rather than an "empty" LED.

enter image description here

This circuit would work with two blue LEDs and would not likely work with a red and blue. Adding a red could cause the blue to blink as explained in my original answer to the original question.

END OF UPDATE

changed one to a red led and its fine, the other led started to blink after changed.

I am assuming you have two LEDs getting their current from the same resistor.
If each LED had its own resistor there would be no problem.

You do not have to change the resistor.
You need to use a blue LED.

If there is only one resistor, both anodes of the LEDs are wired together to the resistor.

The red LED's forward voltage is about 2V and the blue is about 3V.
Because they are connected together they both must be at the same voltage.
They are fighting with one another. The red one is winning the fight.

Use a blue LED and they will both be closer to the same forward voltage.
Ideally both blue LEDs should be identical but any blue LED will work MUCH better than a red.


two blue small leds connected to a resistor each one (160R)

This is a little unclear. Is it each LED to one or one resistor each LED?
I assumed one resistor for both LEDs.

If each has its own resistor then a piece of the puzzle is missing.
In this case there should be no issue with the blue LED.

\$\endgroup\$
  • \$\begingroup\$ I'm sorry for my miss explanation, each LED has your own resistor. \$\endgroup\$ – André M. Faria Jan 24 at 15:54
  • \$\begingroup\$ My poor english made the question become very incoherent. \$\endgroup\$ – André M. Faria Jan 29 at 17:57
  • \$\begingroup\$ First, I was reading your original answer and you don't say to meansure the two sides of the resistor. \$\endgroup\$ – André M. Faria Jan 29 at 18:03
  • \$\begingroup\$ Second, Not the other blue LED started to blink, was the red LED that started to blink. \$\endgroup\$ – André M. Faria Jan 29 at 18:04
  • \$\begingroup\$ Third, I'm still talking about a mouse, never changed that, the issue still was about why one of the red leds started to blink. \$\endgroup\$ – André M. Faria Jan 29 at 18:05

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.