# Help calculating the resistors

#### Carlos Eduardo Costa

Mar 16, 2015
10
Hello everyone, this is my first time posting in this forum.

I have basically zero experience calculating circuits on my own, usualy, I just do the manual labor (soldering, etc)
My current client needs the following circuit in his product, and I'm having trouble calculating the resistance of the resistors.

The circuit is pretty simple, it's three groups of LEDs (two white, and one red), with a power switch, and a momentary button for the white LEDs.
Space is limited, so I can only fit some 3V coin cells, or two AAA batteries inside. The red LEDs need 2.2V and 20mA, and the white ones need 3.5V and 20mA.

So, if I power it using two AAA batteries will it work properly? The white LEDs will be a little bit weak, since I'll get 3V maximum. Or should I go for a couple of 3V cells? It will get me 6V and require less space than the batteries.

And what's the math I need to do to know the resistance of the resistors? I found some online calculators, but they only calculate on the ideia that all the LEDs are equal. Since I'm using two different types of LEDs, I'm can't find the answer.

Can someone give me the formula?

PS: Is it an option to use the AAA bateries and connect the white LEDs whitout resistors? The voltage will do, but I fear the current will damage the LEDs.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
And check your circuit. You'll find the battery is drawn the wrong way around

#### Carlos Eduardo Costa

Mar 16, 2015
10
I've read it, but I think I'm too stupid, because I still can't figure out how to calculate if I'm using different colors of LEDs.

I mean, if I got 6V to power 3 parallel arrays of red and white LEDs, is this the right math?

obs: I'm using 0.06A because it's 3 LEDs with 20mA each

1st white array:
(6 - 3.5) / 0.06 = 41ohms for each white led (closest commercial value is 43)

2nd white array:
same as above

Red array:
(6 - 2.2)/ 0.06 = 63ohms for each red led (closest is 68)

Resulting in:

Can anyone check for me if this is correct?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
The calculation is correct except that you need to use 0.02 because that's the actual current through the led. In your case each of the six white less and resistors are in parallel, there is no grouping into threes electrically.

And your battery is still shown back to front.

#### davenn

Moderator
Sep 5, 2009
14,304
And check your circuit. You'll find the battery is drawn the wrong way around

ohhh I missed that, Steve ... nice spotting

#### Carlos Eduardo Costa

Mar 16, 2015
10
Another thing, will the circuit work if I power it using two AAA batteries?

I know the white LEDs will be a little dim, but the space is very limited.

#### Carlos Eduardo Costa

Mar 16, 2015
10
BTW, I've inverted the LEDs, is it right now?

#### Arouse1973

Dec 18, 2013
5,178
The calculation is correct except that you need to use 0.02 because that's the actual current through the led. In your case each of the six white less and resistors are in parallel, there is no grouping into threes electrically.

And your battery is still shown back to front.

Or the LEDs are reversed

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
As previously noted, your resistors need to be three times the value you calculated, but yes, reversing the LEDs is the same as reversing the battery (just a little harder to change on the diagram).

If your switch is connected to the battery holder in such a way that you can't easily swap the battery around then this would be the correct solution. Otherwise both are equally correct.

#### Geonovast

Aug 26, 2012
16
Is there any reason you need a separate resistor for each LED? If space is a concern, you can easily get away with using two 1/2 W resistors in the entire circuit. If the Reds are always on together, and the Whites are always on together, you can use a single resistor for each group (which is what you were actually calculating for when you used total current)

Multisim didn't have white LEDs, so I used Blue, which had a 3.45 Voltage drop.

#### Attachments

• LEDs.png
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#### Carlos Eduardo Costa

Mar 16, 2015
10
Is there any reason you need a separate resistor for each LED? If space is a concern, you can easily get away with using two 1/2 W resistors in the entire circuit. If the Reds are always on together, and the Whites are always on together, you can use a single resistor for each group (which is what you were actually calculating for when you used total current)

Multisim didn't have white LEDs, so I used Blue, which had a 3.45 Voltage drop.

Well, I've read you need a resistor for each led when you're using them in parallel. And the space is only limited in the place where the battery is gonna be, in the other places won't make much of a difference.

Does multisim calculate the resistor values?
Could you please check for me the resistance if I'm using a 9v battery?

Edit: I've installed Multisim but it's harder than I thought. If it's not much trouble, could you send me the file you made of my circuit? Then I can experiment on it. Thanks

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Carlos, you are correct about the resistors.

#### Carlos Eduardo Costa

Mar 16, 2015
10
This is what I got, can anyone check for me?
I've changed the battery to 9V.
The resistance is correct now?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
For the white leds the resistance should be (9 - 3.5) / 0.02

For the red leds it should be (9 - 2.2) / 0.02

However, now that you're using 9v you can achieve better efficiency by placing three red leds in series with a single resistor and pairs of white leds with a single resistor

The calculation for the required resistor for the white leds becomes (9 - (3.5 + 3.5)) / 0.02

And the resistor for the red leds (9 - (2.2 + 2.2 + 2.2)) / 0.02

#### Carlos Eduardo Costa

Mar 16, 2015
10
For the white leds the resistance should be (9 - 3.5) / 0.02

For the red leds it should be (9 - 2.2) / 0.02

However, now that you're using 9v you can achieve better efficiency by placing three red leds in series with a single resistor and pairs of white leds with a single resistor

The calculation for the required resistor for the white leds becomes (9 - (3.5 + 3.5)) / 0.02

And the resistor for the red leds (9 - (2.2 + 2.2 + 2.2)) / 0.02

That's the math I did, except I divided by the number of LEDs. For example, with the reds ( (9 - 2.2) / 0.02) / 3 and for the whites and divided by 6.

But for the series option, Is there another way besides putting the white ones in pair? That's because I need 3 in one spot, and 3 on another spot.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
The additional divide by 2 or 3 is incorrect.

With a 9v supply the most white less you can place in series is 2 because three timed 3.5 is greater than 9.

For each group of physically grouped white less you can place two leds in series and have another single led or have each one on its own like you have now.

#### Carlos Eduardo Costa

Mar 16, 2015
10
Take a look now, did I finally get it done?

#### BobK

Jan 5, 2010
7,682
Yes, but you ignored the advice to put as many as possible of the LEDs in series, which would make your battery last almost twice as long (9/5 times as long to be exact.)

Bob

#### Carlos Eduardo Costa

Mar 16, 2015
10
I'll see if I can put a least the red ones in series. Thanks.

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