# Resistor Wattage

#### ronelle

Jan 5, 2011
4
Hi everyone....

i just want to know what wattage rating of a resistor should I use.

example, in a simple circuit....

5V(dc) supply
75ohms resistor
generic LED

what resistor wattage should I use and how to prove it in terms of computation....

Thank you

#### barathbushan

Sep 26, 2009
223
It would be more simple , if there was a circuit diagram

However consider you applied 3v to an LED , then it draws around 15 ma , so you can calculate the LED resistance as 3v/10ma = 300 ohms , however this calculation of mine is largely in-accurate and approximate , since LED's are designed to be driven by a current source , if not it would result in a thermal runaway condition (for more information visit
https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html )

But for now consider 375 ohms (300+75) of resistance connected to 5v , it would result in a draw of about approximately 13ma , and the maximum power dissipation at the resistor will be 13ma^2 * 75 = 3mw , hence you can use a standard 1/4 watt resistor

EDIT : calculationn error , the maximum power dissipated by the resistor is 63mw and not 3 mw

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#### ronelle

Jan 5, 2011
4
It would be more simple , if there was a circuit diagram

However consider you applied 3v to an LED , then it draws around 15 ma , so you can calculate the LED resistance as 3v/10ma = 300 ohms , however this calculation of mine is largely in-accurate and approximate , since LED's are designed to be driven by a current source , if not it would result in a thermal runaway condition (for more information visit
https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html )

But for now cosider 375 ohms (300+75) of resistance connected to 5v , it would result in a draw of about approximately 13ma , and the maximum power dissipation at the resistor will be 13ma^2 * 75 = 3mw , hence you can use a standard 1/4 watt resistor

Sir, please have patience in explaining things to... i have lot of questions....
how come that 13ma^2*75 is equals to 3mw and how did you rate that 3mw is under the 1/4watt resistor?

thank you :'(

#### barathbushan

Sep 26, 2009
223
sorry there was an error in calculation it is not 3mw

the right calculation is 13ma^2 * 375 = 64 mw (not 13 mw)
And coming to the calculation topic ,

The power dissipated by a resistor is given by

P = I² × R

or
P = V² / R

so put i=13ma and r=375 ohms in the first equation , then you get 64mw

1/4 watt resistor = 0.25 watt resistor = 250 mw resistor

since the maximum power dissipation is 63mw and well under
250 mw , you can use 1/4 watt resistor

#### ronelle

Jan 5, 2011
4
sorry there was an error in calculation it is not 3mw

the right calculation is 13ma^2 * 375 = 64 mw (not 13 mw)
And coming to the calculation topic ,

The power dissipated by a resistor is given by

P = I² × R

or
P = V² / R

so put i=13ma and r=375 ohms in the first equation , then you get 64mw

1/4 watt resistor = 0.25 watt resistor = 250 mw resistor

since the maximum power dissipation is 63mw and well under
250 mw , you can use 1/4 watt resistor

I see thank you vary much sir for explaining it to me...
I thought there are some other equations to get the wattage of a resistor... it is just a simple ranging...
I appreciate your help sir ^_^

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
There is no such thing as a "generic LED".

Let's assume it's a red LED with a forward voltage drop around 1.7V.

The voltage across the resistor is 5 - 1.7 = 3.3

The power is then given by V^2/R = (3.3*3.3)/75 = 1/7 Watt (approx) You would probably use a 1/4W resistor.

Let's assume it is a white LED with a voltage drop of 3.4 volts.

The voltage across the resistor is 5 - 3.4 = 1.6

The power is then given by V^2/R = (1.6*1.6)/75 = 1/30 Watt (approx) You would probably use a 1/4W resistor, or if it was surface mount you would probably use an 0603 resistor as they are commonly rated at around 1/16W (if it had a low duty cycle a 1/32W 0402 may be possible)..

#### ronelle

Jan 5, 2011
4
There is no such thing as a "generic LED".

Let's assume it's a red LED with a forward voltage drop around 1.7V.

The voltage across the resistor is 5 - 1.7 = 3.3

The power is then given by V^2/R = (3.3*3.3)/75 = 1/7 Watt (approx) You would probably use a 1/4W resistor.

Let's assume it is a white LED with a voltage drop of 3.4 volts.

The voltage across the resistor is 5 - 3.4 = 1.6

The power is then given by V^2/R = (1.6*1.6)/75 = 1/30 Watt (approx) You would probably use a 1/4W resistor, or if it was surface mount you would probably use an 0603 resistor as they are commonly rated at around 1/16W (if it had a low duty cycle a 1/32W 0402 may be possible)..

ok sir, i'll remember that...
here in our place generic LEDs are LEDs which are not branded, usually, they have different intensities of light and is not advisable to use in projects using many LEDs...
thank you sir for your correction... ^_^

#### barathbushan

Sep 26, 2009
223
ok sir, i'll remember that...
here in our place generic LEDs are LEDs which are not branded, usually, they have different intensities of light and is not advisable to use in projects using many LEDs...
thank you sir for your correction... ^_^

steve's calculation is accurate , you can follow steve's method next time you want to find the power dissipation of a resistor in series with an LED
Regarding the type of LED to use in your project , Just ask your LED vendor for a general purpose clear type 5 mm led , and if your project requires other types of led's refer this site
http://www.ledssuperbright.com/5mm-led-c-6

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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
A fairly universal test is to place a 1K resistor in series with the LED, and drive it from 5V. It will glow (perhaps dimly). Measure the voltage across the LED and use that in your calculations.

The voltage across the LED will be slightly higher at higher currents, but using this voltage will help you to be more conservative.

edit: and please don't call me Sir

Last edited:

#### neon

Oct 21, 2006
1,325
run of the mill LED is ~ 20ma @ 1.8v so 5v-1.8v /.02 160 ohms an 40 mw 160 ohms and .5 watt will do i.t

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