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Difficulty finding the appropriate resistor with power calculation

t101

Jul 10, 2022
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So I have a circuit setup 15v dc and a 220 R. My goal is to calculate the correct power rating and replace the current 1/4 resistor if a new one is appropriate.

source:
15V/220R ohm = .068 amps

the resistor:
P = I^2R
P = (.068 amps )^2 x (220 ohms)
P = 1.017 watts

The P(watts) looks high to me. Is my P correct? If so is it supposed to be doubled for safety sakes? Also so far one has not been burned up.
 
Last edited:

bertus

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Nov 8, 2019
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Hello,

You can also calculate the power from the voltage and resistance:
P = v² / R

Ohm's_law_formula_wheel.JPG

Bertus
 

t101

Jul 10, 2022
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P = V^2/R
P = 15^2 volts / 220 ohms
P = 225 / 220
P = 1.02W

Same as the rest.
Equal to 1 watt.

I'm using 1/4 watt in this circuit and nothing has burned up yet
Do I "double" the power by 50% to put enough overhead = 2W?

Thank you
 

crutschow

May 7, 2021
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Is there actually the full 15V across the resistor or is one end connected to 15V and the other end connected to some other circuit components?
What does the resistor do?
 

CircutScoper

Mar 29, 2022
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Is there actually the full 15V across the resistor or is one end connected to 15V and the other end connected to some other circuit components?
What does the resistor do?

Yours is so important a point that it merits repeating (in boldface): The voltage that counts for the resistor power rating calculation
(P = V^2 / R) isn't the supply voltage, but is instead the voltage measured across the resistor.
 

t101

Jul 10, 2022
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Is there actually the full 15V across the resistor or is one end connected to 15V and the other end connected to some other circuit components?
What does the resistor do?
The intended voltage was across the resistor. The resistor will be connected to a multiplicity of IR LEDs and is to create current for the LEDs.
 

Bluejets

Oct 5, 2014
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The resistor in that case will be in series with the LED/LEDs so their voltage drop must be subtracted first.
i.e. if you have 2 red LEDs in series with your resistor, you must first subtract 2 x 1.7v (or there abouts) first.
Remainder is the voltage that will appear across the resistor.
If 12v supply, then 12 -(2 x 1.7v = 3.4v) = 8.6v.
You then need to calculate the resistor at 8.6v divided by whatever current through the LEDs (say 20mA)
So resistor is 8.6/0.20 = 430R.
As that is not a standard size, then go to 470R, you won't be able to see the difference.
Then the watage would be 8.6 (v) x 0.02 (A) = 0.172W .....so 1/4w resistor.
 

crutschow

May 7, 2021
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The resistor voltage that determines its current and power is the voltage across the resistor, not the voltage at one terminal.
So if there were 5V connected to both terminals, than obviously there would be no current and no power dissipated.
 

CircutScoper

Mar 29, 2022
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The intended voltage was across the resistor. The resistor will be connected to a multiplicity of IR LEDs and is to create current for the LEDs.

Really? Then I strongly suggest you be VERY careful while you work on it!

Because: If the designer of your "setup" was really willing to waste 15V across what's nothing but a heat-producing ballast resistor, then the total supply voltage across resistor plus series LED chain can be expected to total well above 100V, creating an obviously dangerous shock hazard.

Otherwise they're a very poor designer to create such an inefficient circuit, and the "setup" can be expected to be dangerous just from being a lousy design.
 
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Bluejets

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Otherwise they're a very poor designer to create such an inefficient circuit, and the "setup" can be expected to be dangerous just from being a lousy design.

I think it is meant to be a series resistor just explained incorrectly.
 

Audioguru

Sep 24, 2016
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What is missing is a schematic showing the power supply voltage and the forward voltages and wiring of the "multiplicity" of LEDs,.
 

CircutScoper

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What is missing is a schematic showing the power supply voltage and the forward voltages and wiring of the "multiplicity" of LEDs,.

Yeah. But what's also missing is a OP who understands the question: What's the voltage across the resistor?
 

Bluejets

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Yeah. But what's also missing is a OP who understands the question: What's the voltage across the resistor?

Seems to be many like this who simply don't seem to be able to understand basic English.
Something sadly rather common these days.
 

CircutScoper

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Seems to be many like this who simply don't seem to be able to understand basic English.
Something sadly rather common these days.

Well, be fair. Imagine trying to explain that question to an English Lit. major (which, for all we know, is actually the OP's background).

What's in play here is basic EEngineerish.
 

t101

Jul 10, 2022
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Really? Then I strongly suggest you be VERY careful while you work on it!

Because: If the designer of your "setup" was really willing to waste 15V across what's nothing but a heat-producing ballast resistor, then the total supply voltage across resistor plus series LED chain can be expected to total well above 100V, creating an obviously dangerous shock hazard.

Otherwise they're a very poor designer to create such an inefficient circuit, and the "setup" can be expected to be dangerous just from being a lousy design.
Really? Then I strongly suggest you be VERY careful while you work on it!

Because: If the designer of your "setup" was really willing to waste 15V across what's nothing but a heat-producing ballast resistor, then the total supply voltage across resistor plus series LED chain can be expected to total well above 100V, creating an obviously dangerous shock hazard.

Otherwise they're a very poor designer to create such an inefficient circuit, and the "setup" can be expected to be dangerous just from being a lousy design.
Well, be fair. Imagine trying to explain that question to an English Lit. major (which, for all we know, is actually the OP's background).

What's in play here is basic EEngineerish.
Well, be fair. Imagine trying to explain that question to an English Lit. major (which, for all we know, is actually the OP's background).

What's in play here is basic EEngineerish.
Well, be fair. Imagine trying to explain that question to an English Lit. major (which, for all we know, is actually the OP's background).

What's in play here is basic EEngineerish.
Ok ladies, I am an English Lit major and have a Masters of Health Science degree. So what. You would look better stuffed Mr. Owl. I am just trying to learn. Why would you deride that?
 

t101

Jul 10, 2022
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Really? Then I strongly suggest you be VERY careful while you work on it!

Because: If the designer of your "setup" was really willing to waste 15V across what's nothing but a heat-producing ballast resistor, then the total supply voltage across resistor plus series LED chain can be expected to total well above 100V, creating an obviously dangerous shock hazard.

Otherwise they're a very poor designer to create such an inefficient circuit, and the "setup" can be expected to be dangerous just from being a lousy design.

Why don't you explain what exactly you would change especially if you are going to posit how dangerous it is. That would be shocking.
I'd really like to know your thoughts.
 

t101

Jul 10, 2022
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The resistor in that case will be in series with the LED/LEDs so their voltage drop must be subtracted first.
i.e. if you have 2 red LED
Maybe a picture will help. What is the voltage across the resistor? Hint: You'll never know until you measure it with an actual voltmeter.

View attachment 55771
snap_design.png
s in series with your resistor, you must first subtract 2 x 1.7v (or there abouts) first.
Remainder is the voltage that will appear across the resistor.
If 12v supply, then 12 -(2 x 1.7v = 3.4v) = 8.6v.
You then need to calculate the resistor at 8.6v divided by whatever current through the LEDs (say 20mA)
So resistor is 8.6/0.20 = 430R.
As that is not a standard size, then go to 470R, you won't be able to see the difference.
Then the watage would be 8.6 (v) x 0.02 (A) = 0.172W .....so 1/4w resistor.

Here is my current scheme.
 
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