Led circuit Problem

[Shadow the ODST]

I'm having issues with my led circuit for my h3 radio pack. For some reason the marked row of leds keep blowing when I switch it on.

The blown leds have both been the top and bottom ones but the middle is always fine but all are now blown in that row meaning I have to once again replace them.

If it helps the white leds are marked as 3.0-3.4v and the yellow 2.0-2.2v running on a 9v battery.

Anyone got any ideas why it suddenly started killing leds in that row? As it was fine on initial testing.

 

[N8TEBB]

Hmm, It seems that there is a little too much current going through those yellow LEDs. This is because the yellow LEDs are rates for only 2V as you said, instead of the 3V the White LEDs are. To compensate for this, you need to add some resistance to each line of yellow LEDs. To figure out the resistance, you need to use Ohm's law V = I R Where V is the voltage supplied minus the voltage used, I is current needed for the LEDs, and R is resistance.

In this case the voltage supplied is 9V, and we use 6V (3*2V yellow LEDs)
9V - 6V = 3V
LEDs are typically rated for 20 mA or 0.02 A
Rearranging Ohm's law gives us R = V / I
substituting out values, we get
R = 3V / 0.02A
R = 150Ω

This means we need to add a 150 Ohm resistor to each line of yellow LEDs. Now, because you will have 2 resistors in parallel, we reduce this to 1 resistor. To find the equivalent resistance of 2 resistors, we can use the product over sum method:

(150 * 150) / (150 + 150) = 75Ω

This means you can simple add one 75Ω resistor before your yellow LEDs to reduce the current to a tolerable level for the yellow LEDs.

Hope this helps!

 

[Shadow the ODST]

Hmm, It seems that there is a little too much current going through those yellow LEDs. This is because the yellow LEDs are rates for only 2V as you said, instead of the 3V the White LEDs are. To compensate for this, you need to add some resistance to each line of yellow LEDs. To figure out the resistance, you need to use Ohm's law V = I R Where V is the voltage supplied minus the voltage used, I is current needed for the LEDs, and R is resistance.

In this case the voltage supplied is 9V, and we use 6V (3*2V yellow LEDs)
9V - 6V = 3V
LEDs are typically rated for 20 mA or 0.02 A
Rearranging Ohm's law gives us R = V / I
substituting out values, we get
R = 3V / 0.02A
R = 150Ω

This means we need to add a 150 Ohm resistor to each line of yellow LEDs. Now, because you will have 2 resistors in parallel, we reduce this to 1 resistor. To find the equivalent resistance of 2 resistors, we can use the product over sum method:
View attachment 334053
(150 * 150) / (150 + 150) = 75Ω

This means you can simple add one 75Ω resistor before your yellow LEDs to reduce the current to a tolerable level for the yellow LEDs.

View attachment 334055​

Hope this helps!

Wow I was expected a "well there's your Problem" answer and you went above and beyond I really appreciate your help.

I looked/researched into resistors before (made my thinker box hurt), never even knew the number of resistors could be reduced to just one like you suggested though. tbh I was seeing if I could get away with not buying them as its hard to just get one or two its all "buy in bulk" and not knowing which type to get.

would the resistor value have to be the same as the equation or could you get away with a slightly lower or higher one?

Lastly that bottom image, was it made using a program or just found off the Internet? I could do with some thing to make future circuit jobs easier.

Thanks again for this response, I really appreciate you doing the maths for me.

 

[N8TEBB]

If you don't want to buy resistors you can add a 4th LED to each yellow line. It wouldn't be as balanced visually, as you would have lines of 3 white LEDs and 4 yellow LEDs, but the extra yellow LEDs would act as resistors. You can probably get away with a resistance value as low as 70 if you can't find 75. Anything lower will reduce the lifespan of the LEDs. You can go higher too, but your LEDs will be dimmer.

Also, if you can't find 75, you can use multiple resistors in series. For example, you can use a 30 AND a 45. You can usually get a cheap set of resistors online with a variety of values. It's one of those things that you can buy once and use for the rest of your life if you don't do electronics very often.

The website I used to make that picture is called tinkerCAD. It's a free online program by autodesk that you can make basic circuits in! It's also a free CAD software. Tinkercad | From mind to design in minutes

 

[Shadow the ODST]

If you don't want to buy resistors you can add a 4th LED to each yellow line. It wouldn't be as balanced visually, as you would have lines of 3 white LEDs and 4 yellow LEDs, but the extra yellow LEDs would act as resistors. You can probably get away with a resistance value as low as 70 if you can't find 75. Anything lower will reduce the lifespan of the LEDs. You can go higher too, but your LEDs will be dimmer.

Also, if you can't find 75, you can use multiple resistors in series. For example, you can use a 30 AND a 45. You can usually get a cheap set of resistors online with a variety of values. It's one of those things that you can buy once and use for the rest of your life if you don't do electronics very often.

The website I used to make that picture is called tinkerCAD. It's a free online program by autodesk that you can make basic circuits in! It's also a free CAD software. Tinkercad | From mind to design in minutes

I'll probably get a pack of resistors with different values as you suggested, there's not much space to add anymore leds anyway I'm already having trouble fitting it all into the radio pack.

I'll definitely check out tinkerCAD, got to be better than Google sketch up which has been the only CAD I've learnt. Thanks again for the help