# Wattage rating on a resistor

• HeyAwesomePeopl
Bryan.To make it a lot easier to work with, I could wire it sort of like this, but it uses a lot of mA. http://prntscr.com/5i4249This option would be easier to work with and less messy, but requires more mAmps to run. Does this affect how many mA will go to the Arduino?Second Option: For less of a mA usage, I could do it like this: http://prntscr.com/5i42jeThis would be more of a mess on the board, but only consumes 157 mA.f

#### HeyAwesomePeopl

Hello, I am new to any type of engineering and just have been working with Arduinos and other microcontroller lately for fun.

I am creating a prototype for a project that requires me to use a 9V power supply with 1Amp to power the Arduino(note, the arduino only needs around 300mA to run). I want to use some LED's(probably around 30) to be controlled by the arduino, BUT, I want the 9V, 1A power supply to power it. I have done some research and found that for controlling let's say, 30 LEDS, I would need the positive power to run through a 33Ohm resistor per 4 LEDs.

LED uses a max of 2.2V and a current draw of 20mA.

Okay. Easy, I just need a 33Ohm resistor now, but what wattage rating? I was told wattage can be calculated with:
P= I * R
or
Wattage = 9 * 1

Now doesn't this mean that the output of the power supply is going to be 9 Watts? The calculator I used said a 1/8th Watt resistor would work, but if the power supply is really supplying 9 Watts, wouldn't it burn up?
Thanks,
Bryan(HeyAwesomePeople)

The wattage rating of the resistor needs to be at least as large as the number of watts it will dissipate. Rule of thumb is approximately double. Your formula is wrong. P = I^2 * R. So, your resistor will dissipate about .132 watts. A quarter watt resistor is fine.

The wattage rating of the resistor needs to be at least as large as the number of watts it will dissipate. Rule of thumb is approximately double. Your formula is wrong. P = I^2 * R. So, your resistor will dissipate about .132 watts. A quarter watt resistor is fine.
Isn't I the voltage and R the amps? Wouldnt that turn out to be 81 watts? I know that is wrong, but how exactly did you calculate the dissipation?

I have done some research and found that for controlling let's say, 30 LEDS, I would need the positive power to run through a 33Ohm resistor per 4 LEDs.

LED uses a max of 2.2V and a current draw of 20mA.
Hi HAP. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What do you mean by "run through a 33Ohm resistor per 4 LEDs."? How did you calculate that you need 330Ω?

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Hi HAP. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What do you mean by "run through a 33Ohm resistor per 4 LEDs."? How did you calculate that you need 330Ω?
Yes good question. I had missed that.

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Hi HAP. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What do you mean by "run through a 33Ohm resistor per 4 LEDs."? How did you calculate that you need 330Ω?

33 Ohms, not 330.

I am trying to find a solution to connect 30 LEDS with a 9v 1.5Amp supply and an arduino. If I use one resistor to power all LEDs, it consumes like 600mA. The arduino has a recommended amp input of 300mA+. That's pushing it.

If I connect both the LEDs and the arduino to a 9v 1amp power supply, would the total consumed amps be the LED draw + arduino, or do they both get 1amp?

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If I connect both the LEDs and the arduino to a 9v 1amp power supply, would the total consumed amps be the LED draw + arduino?
that.

The 1A capability of the supply must not be exceeded.

33 Ohms, not 330.
Oops.

So my question becomes how did you calculate 33 ohms?

Oops.

So my question becomes how did you calculate 33 ohms?
Used a calculator. http://ledcalc.com/

First Option: To make it a lot easier to work with, I could wire it sort of like this, but it uses a lot of mA. http://prntscr.com/5i4249
This option would be easier to work with and less messy, but requires more mAmps to run. Does this affect how many mA will go to the Arduino?

Second Option: For less of a mA usage, I could do it like this: http://prntscr.com/5i42je
This would be more of a mess on the board, but only consumes 157 mA.

So here are my questions:

1. What is used to calculate the dissipation by resistor?
and
2. Say I split a 9volt 1amp power supply into two lines. One for an Arduino, which requires at least 7 volts and 250mA+, and one for the LEDs, let's say the first option. If the arduino used up say 400mA at one time, would that only leave 600mA for the LEDs? And vice versa?

Thanks

Hi HAP. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

What do you mean by "run through a 33Ohm resistor per 4 LEDs."? How did you calculate that you need 330Ω?
I love your immediate abbreviation of my name. I've only seen a few people automatically do that, most I must tell. But that is what I go by. :)

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I still don't see the value 33 Ω appearing anywhere. When operating LEDs in series, you need to allow plenty of volts across their current-limiting resistor. 3 is the max you can use here.

When operating LEDs in parallel, you can't operate all 20 off one resistor. Individual differences mean that some will hog the current, some will get little. 3 or 4 in parallel should be okay if you don't run them too hard, but if any individual LED seems noticeably dull then swap it out.

Say I split a 9volt 1amp power supply into two lines. One for an Arduino, which requires at least 7 volts and 250mA+, and one for the LEDs, let's say the first option. If the arduino used up say 400mA at one time, would that only leave 600mA for the LEDs? And vice versa?
Yes.
Good luck! http://thumbnails111.imagebam.com/37333/a62855373324849.jpg [Broken]

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I still don't see the value 33 Ω appearing anywhere. When operating LEDs in series, you need to allow plenty of volts across their current-limiting resistor. 3 is the max you can use here.

When operating LEDs in parallel, you can't operate all 20 off one resistor. Individual differences mean that some will hog the current, some will get little. 3 or 4 in parallel should be okay if you don't run them too hard, but if any individual LED seems noticeably dull then swap it out.

Yes.
Good luck! http://thumbnails111.imagebam.com/37333/a62855373324849.jpg [Broken]
Okay well let me tell you what I am planning to do. It is actually quite a simple idea that I came up with. My family just moved into a new house and my mom still gets out of the car before turning it off to check if she is far enough inside the garage so the door will not hit the car. I said I could create a stoplight type system to help her park. Using an ultrasonic distance sensor I figured it would be a fairly easy project.
Red = Pull up
Yellow = Go slowly
Green = Inside garage fully

Seems pretty simple. But I want to run this all off of a 9v power supply. I said 30 LEDs because I want 10 per red/yellow/green. So what would you recommend I do?
Use a 120Ohm resistor for every 3 LEDs?

Thanks

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Grouping series LEDs in threes would work. How much current will 120 Ω give?

Grouping series LEDs in threes would work. How much current will 120 Ω give?

Used the calculator. http://prntscr.com/5i4b3g
How would I calculate the current with the Voltage and Resistance?
V / R = I

for 3 LEDs, current = ( 9 - 6.6 ) / 120

⅛ watt resistors should be okay

for 3 LEDs, current = ( 9 - 6.6 ) / 120

⅛ watt resistors should be okay
Now let's say I used a 12Volt power supply, with 1A. How would I calculate the resistor ohms I would need between the power and leds?

R = V / I

Would "I" be the current recommend for an LED and "V" be input voltage? This would mean that to power from a 12 Volt supply, I could have 5 LEDs(11Volts) with a 60 Ohm resistor(or 68 Ohm because that is the closest rounding up)?

EDIT: Wait, wouldn't I only be able to have a max of 4 LEDs, because after the resistor of 68Ω I would have only roughly 10.2 Volts to work with? Is this math correct?
Final Voltage = ( 12 / .2 )*( 12 / 68) = 10.2 Volts

I am using
Final Voltage = ( supplyvoltage / loadamperage )*( supplyvoltage / resistorvalue)
to find the voltage after a resistor.
Supply Voltage = 12v
Resistor Value = 68Ω
Load Amperage = Total current of combined LED's? (In this case, for 4, 0.08A?)

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5 LEDs would use about 11V, leaving roughly 1V across R and that is not quite enough for comfort, unless you make an effort to match the LED strings for their voltage drops.

Looking at 4 LEDs, using 8.8V leaves 3.2V across R.
For 20mA, choose R = 3.2/0.02
= 160 ohms
Then choose the nearest preferred value, 150 ohms

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5 LEDs would use about 11V, leaving roughly 1V across R and that is not quite enough for comfort, unless you make an effort to match the LED strings for their voltage drops.

Looking at 4 LEDs, using 8.8V leaves 3.2V across R.
For 20mA, choose R = 3.2/0.02
= 160 ohms
Then choose the nearest, 150 ohms

So to find the resistance to use, I would use:
Resistance = (supplyvoltage / totalcurrent)

So if I got this right, then to power 3 LEDS on a 9 volt supply, I would use a resistor of 150 Ohms(or the closest one)?

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EDIT No, not now that you edited your post and made it wrong. Yes

Different colour LEDs have different voltage drops. You'll probably find that 4 LEDs of each colour will be ample. Poking out of a scrap of black plastic there will be such good contrast that they can't be missed.

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Yes.

Different colour LEDs have different voltage drops. You'll probably find that 4 LEDs of each colour will be ample. Poking out of a scrap of black plastic there will be such good contrast that they can't be missed.
Great! Thanks!

One last question:
I know that splitting the supply(1 amp) into 2 parts, and using up 700mA leaves 300mA for the other split, but does that apply for volts as well? Say for example, the Arduino requires at least 7.5 volts, and no more than 12V. If I use a 12Volt power brick, would 12V be dedicated to just the arduino, or does a split keep voltage same across both splits?

Also, I seem to calculate that powering 5 LEDs that require 25mA and 2 volts to run with a Resistance of 82Ω has a dissipation wattage of ~1.2W. This would mean I need a resistor rated for more than 1.2 watts, correct? Like a 5 Watt resistor would work, right?

HeyAwesomePeopl said:
I know that splitting the supply(1 amp) into 2 parts, and using up 700mA leaves 300mA for the other split, but does that apply for volts as well? Say for example, the Arduino requires at least 7.5 volts, and no more than 12V. If I use a 12Volt power brick, would 12V be dedicated to just the arduino, or does a split keep voltage same across both splits?
The voltage does not split up. Each gets the full 12V from a 12V supply.

Also, I seem to calculate that powering 5 LEDs that require 25mA and 2 volts to run with a Resistance of 82Ω has a dissipation wattage of ~1.2W. This would mean I need a resistor rated for more than 1.2 watts, correct? Like a 5 Watt resistor would work, right?
Power is I2.R so 25mA does not come near 1W.

I did point out that 5 LEDs in series is too many to operate with a current-limiting resistor from 12V.

You will need to re-do the calculations for each colour, these differ for each colour LED with its different voltage.

So to find the resistance to use, I would use:
Resistance = (supplyvoltage / totalcurrent)
No. You had it right, then as I was replying you edited your post and now it's wrong. Can you fix it to how it was? See post #16 where I showed how to calculate the value of the series resistor.

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No. You had it right, then as I was replying you edited your post and now it's wrong. Can you fix it to how it was? See post #16 where I showed how to calculate the value of the series resistor.
Okay I see now

Power is I2.R so 25mA does not come near 1W.

I did point out that 5 LEDs in series is too many to operate with a current-limiting resistor from 12V.

You will need to re-do the calculations for each colour, these differ for each colour LED with its different voltage.

Using
P = I^2 * R
for 5 LEDs with a current use of 30mA, with a 39Ω resistor, I calculated 0.0297W... would this be the correct value? P = 0.03^2 * 39

I did recalculations and all that, but I am confused on why 1 volt left over has you so worried. Could you explain that please?

The voltage across each LED won't be precisely 2.20V. Devices are never identical. This means if you cannect 5 in series, you can't say exactly how much of a volt this will leave across the resistor. Not knowing the resistor voltage means you can't know what value to use to set the current to even approximately 30mA.

But if you connect fewer LEDs in series so that more than 2V is "left over" then even if the LEDs are not all exactly 2.2V you can still rely on there being a potential difference across the resistor close to what you calculated, so your resistor calculation will still be close to the mark.

So you should am to limit the number of series connected diodes so as to leave at least 2V across their series resistor.

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The voltage across each LED won't be precisely 2.20V. Devices are never identical. This means if you cannect 5 in series, you can't say exactly how much of a volt this will leave across the resistor. Not knowing the resistor voltage means you can't know what value to use to set the current to even approximately 30mA.

But if you connect fewer LEDs in series so that more than 2V is "left over" then even if the LEDs are not all exactly 2.2V you can still rely on there being a potential difference across the resistor close to what you calculated, so your resistor calculation will still be close to the mark.

So you should am to limit the number of series connected diodes so as to leave at least 2V across their series resistor.
Okay I see. What happens if the total voltage needed goes over 12 volts?

" Using
P = I^2 * R
for 5 LEDs with a current use of 30mA, with a 39Ω resistor, I calculated 0.0297W... would this be the correct value? P = 0.03^2 * 39 "
Thanks for everything so far

Okay I see. What happens if the total voltage needed goes over 12 volts?

" Using
P = I^2 * R
for 5 LEDs with a current use of 30mA, with a 39Ω resistor, I calculated 0.0297W... would this be the correct value? P = 0.03^2 * 39 "
Thanks for everything so far
Red LEDs will still give out a small amount of light if each has a bit less than 2.2V, but it's not dependable.

Yes, that is the right power in a 39 ohm resistor carrying 30mA. Buy a resistor of at least 3 times that power rating so that its surface won't get unduly hot in operation.

As has been pointed out - each different color LED most likley has a different characteristic - if you are buying them new refer to the datasheets. My guess each of the Red/Yellow/Green will probably need a different solution to get the max brightness.

For the Arduino - you will want to have an output transistor, and the transistor will be the switch (one for each color) REF LINK - this will also help keep you from connecting 12V directly to your arduino.

The Transistor will add a V drop - but probably less then the arduino would have.

As for the variation from unit to unit - the difference in V drop is less of a concern than the difference in current for a given voltage - so LEDs in parallel without resistors are more risky than LEDs in series.

Red LEDs will still give out a small amount of light if each has a bit less than 2.2V, but it's not dependable.

Yes, that is the right power in a 39 ohm resistor carrying 30mA.
Ahh I see. Okay. Thanks a bunch

As has been pointed out - each different color LED most likley has a different characteristic - if you are buying them new refer to the datasheets. My guess each of the Red/Yellow/Green will probably need a different solution to get the max brightness.

For the Arduino - you will want to have an output transistor, and the transistor will be the switch (one for each color) REF LINK - this will also help keep you from connecting 12V directly to your arduino.

The Transistor will add a V drop - but probably less then the arduino would have.

As for the variation from unit to unit - the difference in V drop is less of a concern than the difference in current for a given voltage - so LEDs in parallel without resistors are more risky than LEDs in series.

Okay led me try to understand the transistor part. Let's say I use a NPN transistor. The base would be an Arduino pin. The collector would be the 12 volt supply. The emitter would be going to the LED's(with a resistor of course). If I understand this right, triggering the pin with the transistor on the Arduino will allow the 12volts to pass through the transistor with no power/amperage loss? Would this resistor be okay: https://www.futurlec.com/Transistors/2N6043pr.shtml ?

In the link, with an NPN, you keep the Emitter and Arduino close to 0V (GND) .

12+
Current Limiting Resistor
LEDs
Transistor
GND

In the link, with an NPN, you keep the Emitter and Arduino close to 0V (GND) .

12+
Current Limiting Resistor
LEDs
Transistor
GND
So I want my wire that is at the end of the LEDs to go into the Collector part of the Transistor. Base to the Arduino, and the Emitter to Negative?
Virtual Example(Drawings of circuits really help me): Is this how it is supposed to work? The arduino just completes the circuit, correct?

HAP...Thank you for taking a few minutes to review ( so many people that come here for answers -- well just want answers!) Yes your sketch is what we need as a concept. -- I have not done the math...
As a good practice the 12V and the arduino should share the same GND or Negative Bus - in this schema these details are not shown...

OK back - from my sketch pad... 4 x 2.2 V + 0.2V ( transistor) = 9V... 12 V supply... the difference is ? ... 3 V ... series ckt is 30 mA ..

3 V and 30 mA --- what is the necessary resitance to drop that V?...

Then use the I^2 * R or V*I to get W, and that is the resistance and size (wattage) of the resistor you are looking for... now repeat for the other colors...

OK back - from my sketch pad... 4 x 2.2 V + 0.2V ( transistor) = 9V... 12 V supply... the difference is ? ... 3 V ... series ckt is 30 mA ..

3 V and 30 mA --- what is the necessary resitance to drop that V?...

Then use the I^2 * R or V*I to get W, and that is the resistance and size (wattage) of the resistor you are looking for... now repeat for the other colors...
Thanks for this. And yes, I am looking to learn how to do this on my own, instead of coming here every time I need an answer to something.

So when you were calculating the transistor, you used: (Vf * ledcount) + 0.2V = 9V
What exactly is this meaning? Why do you add the 0.2V?

So then when there is 3V and 30 mA left over, you add a resistor just to drop the voltage to completely 0? (3*30 = 90) This would mean I would put a 100Ω(Closest to 90 rounding up) to the negative then to ground or negative?
Note: Power source in diagram is meant to be DC, not AC Last edited: