Toggle LED's between blink and solid

I see a comment that the resistor can be attached to either the cathode or anode leg of the LED. Is there a preferable side?

The "Wizard" is stupid.
He doesn't know that the "3.3V" LEDs could actually be 3.8V which is listed on the datasheet.
Then three 3.8V LEDs in series with a 150 ohm resistor with a 12.6V supply will have a current of only 8mA and will be less if the resistor value is 5% high.

But if you calculate the resistor for two series 3.3V LEDs at 20mA with a 12.6V supply then its value is 300 ohms.
Then if the LEDs are actually 3.8V their current will be 16.7mA.

Your 6x6 array is a joke. If the "2V" LEDs are a little more like 2.3V each then they will not light with a 12.6V supply. They won't even light with a 13.6V supply.
But if you calculate the resistor for three series 2V LEDs at 20mA with a 12.6V supply then its value is 330 ohms.
If the LEDs are actually 2.3V then their current will be 17.3mA.
Maybe you can use four 2V LEDs in series. You calculate the currents.

Unfortunately, the wizard is a bit smarter than me at this point. I haven't got the variables organized to replicate your result of 8mA in 1st ppg. Do you use a wks template or software?

1 greenbar

Actually I got 7.6mA with (3.8 x 3) / 150

Or is it 12.6 / 150 = 8.4?

Just the part of computing the resistors has gotten significantly more confusing. It's my understanding to use the forward voltage of 2.0V. If the LED has a max fwd voltage of 2.4V should the max be used then? Also these superbrights are rated continuous forward current of 30mA. Why not use that instead of the standard 20mA. They have a peak forward current of 100mA.

Throwing around the different parameters is only making this part of my process more confusing. There is a best answer and that is all I'm looking for if anyone has it.

Frank88 said:

Unfortunately, the wizard is a bit smarter than me at this point. I haven't got the variables organized to replicate your result of 8mA in 1st ppg.
Actually I got 7.6mA with (3.8 x 3) / 150

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Ohm's law says that the current in a resistor is the voltage across it divided by its number of Ohms.
The voltage across the resistor IS NOT 3.8 x 3, it is THE DIFFERENCE between the 12.6V supply and 3.8 x 3.
Your answer is wrong, (3.8V x 3)/150= 76mA (not 7.6mA). My simple arithmatic is [12.6 - (3.8 x 3)]/150= 8mA.

Or is it 12.6 / 150 = 8.4?

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12.6V/150 would happen if the resistor is connected to the power supply WITHOUT ANY LEDS. Is that what you want?
Again your simple arithmatic is wrong because 12.6/150= 84mA, not 8.4.

It's my understanding to use the forward voltage of 2.0V. If the LED has a max fwd voltage of 2.4V should the max be used then?

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I didn't see the maximum voltage of 2.4V before.
Three 2.0V LEDs are 6.0V. The resistor has 12.6V - 6.0V= 6.6V across it and with a 330 ohm current-limiting resistor then the current is 6.6V/330= 20.0mA.
If the LEDs are actually 2.4V then three are 7.2V and the resistor has 12.6V - 7.2V= 5.4V across it. Then the 330 ohm resistor limits the current to 5.4V/330 ohms= 16.4mA.
There is a small difference between 20mA and 16.4mA and the LEDs will look the same. It is correct to calculate the value of the current-limiting resistor with the lowest forward voltage because then if the forward voltage is actually higher for some of your LEDs then they will have a little less current and will not burn out.

these superbrights are rated continuous forward current of 30mA.

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I think that is their absolute maximum allowed continuous current on a cool day when the LEDs are not enclosed so they are cooled by a breeze. I never operate any part at its absolute maximum rating.
I betcha its lifetime and brightness are rated at 20mA.

The peak current rating of 100mA is missing the details about the very short duration of the pulses and their frequency. The pauses between the pulses allows the LED to cool.

Here is link to superbright general specifications. Yellow is same as Red and Green is same as Blue. I appreciate the in-depth explanation of the computations.

https://www.superbrightleds.com/mor...egree-viewing-angle-flat-tipped-1200-mcd/279/

https://www.superbrightleds.com/mor...egree-viewing-angle-flat-tipped-1200-mcd/265/

If the LEDs are actually 2.4V then three are 7.2V and the resistor has 12.6V - 7.2V= 5.4V across it. Then the 330 ohm resistor limits the current to 5.4V/330 ohms= 16.4mA

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Why wouldn't I use a 270ohm resistor in this case to keep 20mA fwd current?

How is it handled then when 36 leds @ 2.4V max fwd voltage rating = 86.4? 12.6V - 86.4V = -73.8? I can set up a max 5 parallel strings with 7 in series (I'll just use 35 LED's in this example); then I would have .6V residual current to dissipate requiring a 30 ohm resistor on each parallel string?

This arrangement would put 7 LED's in series. If one light went out would the remaining 6 go out? I don't want that. My preference will be to have only one LED out.

AG, with the LED specs I shared what parameters would you use:

The forward voltage or the max fwd voltage?
20mA or the continuous fwd current indicated of 30mA?
And for the 12V power supply would you add the 5% tolerance factor to use 12.6V?

How is resistor wattage determined? The wizard explains as:

The wizard uses your drive current to calculate the power dissipated in the array resistors. Standard resistors are available in 1/8W, 1/4W, 1/2W and 1W, and some values are available above that. The wizard picks a resistor wattage for you based on the guideline of operating at less than 60% of the rated power.

Rated power of what?

If each resistor dissipates 132 mW, than a 1/4 watt resistor is ok?

Frank88 said:

Why wouldn't I use a 270ohm resistor in this case to keep 20mA fwd current?

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You don't know the actual forward voltage of your LEDs unless you measure and label all of them. The forward voltage is somewhere from 2.0V to 2.4V but most will be 2.0V.
If you calculate a current-limiting resistor for LEDs that have a forward voltage of 2.4V then the current will be higher when they are 2.0V. You do the simple calculation to see if they will burn out.

How is it handled then when 36 leds @ 2.4V max fwd voltage rating = 86.4? 12.6V - 86.4V = -73.8?

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Of course not.
If you have more than 5 LEDs in series then they will not light from a 12.6V supply.
I showed that 3 LEDs in series allow the current to be reasonable when the LED voltage is 2.4V or 2.0V.

I can set up a max 5 parallel strings with 7 in series (I'll just use 35 LED's in this example); then I would have .6V residual current to dissipate requiring a 30 ohm resistor on each parallel string?

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The same problem as one hundred replies ago:
1) Calculating with 2.4V LEDs.
7 in series total 16.8V. 20mA in 30 ohms is 0.6V so the power supply is 16.8V + 0.6V= 17.4V.
But if all 7 LEDs are actually 2.0V then their current will be 113mA and they will instantly burn out.

2) Calculating with 2.0V LEDs.
7 in series total 14V. A 30 ohm current-limiting resistor has 0.6V at 20mA so the power supply is 14.6V.
But if all 7 LEDs are actually 2.4V then they will not light.

This arrangement would put 7 LED's in series. If one light went out would the remaining 6 go out?

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When LEDs fail, they usually go open, they do not short.
So all the LEDs in a series string all stop lighting when one burns out because they are in series.
If you buy quality LEDs from a Name-Brand manufacturer then they will be reliable.
But if you buy Cheap Chinese LEDs on E-Bay then some will probably not work in the beginning and the remaining will probably fail soon.

I misunderstood. I thought the number of parallel runs is used to compute the total diode voltage to subtract from the power supply voltage, but it is the number of diodes in the series. Thank you.

Ok if I'm progressing at all with understanding this I should use 330 ohm resistors for the Red and Yellow LED's. That would be 36 LED's for each color, each having a forward voltage of 2, setting up 12 parallel arrays with 3 LED's in series in each array, using a 12.0V power supply with 5% tolerance, and a diode fwd current of 20mA.

Since each resistor will dissipate 132mW a 1/4 watt resistor is ok?

And for the Blue and Green LED's using their fwd voltage of 3.3V, I'd set up the same array which leaves a residual voltage for dissipation of 2.7V, to get 20mA diode fwd current I'd need 135Ohm resistors (or the next size above reducing the diode fwd current somewhat).

Since each resistor dissipates 54 mW a 1/8 watt resistor would suffice?

Just out of curiosity for the Reds and Yellows why wouldn't you set up 9 parallel arrays with 4 LED's in each and use a 230 resistor?

Frank88 said:

Ok if I'm progressing at all with understanding this I should use 330 ohm resistors for the Red and Yellow LED's. That would be 36 LED's for each color, each having a forward voltage of 2, setting up 12 parallel arrays with 3 LED's in series in each array, using a 12.0V power supply with 5% tolerance, and a diode fwd current of 20mA.

Since each resistor will dissipate 132mW a 1/4 watt resistor is ok?

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Yes and yes.

And for the Blue and Green LED's using their fwd voltage of 3.3V, I'd set up the same array which leaves a residual voltage for dissipation of 2.7V, to get 20mA diode fwd current I'd need 135Ohm resistors (or the next size above reducing the diode fwd current somewhat).

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I said before that if each LED is actually 3.8V but the resistor was calculated for 3.3V at 20mA then three LEDs would have a current of only 8mA which is not enough. I calculated before that two 3.3V LEDs in series will need a 300 ohm resistor for 20mA. If the LEDs are actually 3.8V each then their current will be 16.7mA which is fine. The power dissipated in the resistor will be 120mW so a 1/4W resistor is fine.

Thank you for all your help! It's been very informative.