# Resistor wattage

#### Propeller

Joined Nov 27, 2005
10
This might seem to be a stupid question.

I need a 100ohm 2W resistor.
Can add 2x 50ohm 1W resistors in series to acheive this?

#### Sebi

Joined Sep 24, 2005
59
Yes.

#### Azgard

Joined Dec 16, 2005
1
The series resistance will be 100 Ohms, but the resistors will still only be able to handle 1 watt, so no.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by Azgard@Dec 16 2005, 04:16 PM
The series resistance will be 100 Ohms, but the resistors will still only be able to handle 1 watt, so no.
[post=12500]Quoted post[/post]​
Umm... each resistor handles 1W, so both will be 2W.

P=I^2*R

I is the same because the total resistance is the same. With 50 ohm resistance the power is half the total power. So the total power dissipation is divided into two, half for each resistor.

#### alim

Joined Dec 27, 2005
113
Originally posted by n9352527@Dec 16 2005, 11:31 AM
Umm... each resistor handles 1W, so both will be 2W.

P=I^2*R

I is the same because the total resistance is the same. With 50 ohm resistance the power is half the total power. So the total power dissipation is divided into two, half for each resistor.
[post=12502]Quoted post[/post]​
The answer is YES. Resistors in series or in parallel the wattage adds.Some ohms LAW PRACTICE would resolve this.

#### chesart1

Joined Jan 23, 2006
269
Hi,

Lets use an actual example to illuistrate the point:

14.14 volts across a 100 ohm 2 watt resistor results in 2 watts dissipation.

Now lets substitute two 50 ohm resistors for the 100 ohm resistor and figure out the power dissipated by each resistor.

The curent is still .1414 amps.

So the power dissipated by each resistor is

P = .1414 amps * .1414 amps * 50 ohms
P = .02 * 50 = 1 watt

And this proves you can replace a 100 ohm 2 watt resistor with two 50 ohm 1 watt resistors in series.

John

#### n9xv

Joined Jan 18, 2005
329
Not a stupid question propeller. Thats what we're here for! Power is comulitive.

#### CoulombMagician

Joined Jan 10, 2006
37
Power is cumulative so long as they do not interfere with each others dissipation. Example 1, a surface mount resistor. You can solder one on top of another but together their power rating is not 2x that of one.
Example 2, a large wirewound adjustable power resistor with one of those metal slides on it for a wiper. If all the load current passes between the wiper and one terminal the power dissipation is concentrated in only a fraction of the length of the resistor and maximum power is that same fraction of the rated power of the whole resistor.

#### thingmaker3

Joined May 16, 2005
5,084
1/10 watt SM resistors I could stack, but one watters? You got a picture of that? I thought the pads were underneath...

De-rating is a consideration, though. There are plenty of situations where a one watt resistor stops becoming a full watt resistor.

The de-rated values still obey the math, of course.

#### chesart1

Joined Jan 23, 2006
269
Hi thingmaker3,

I understand your point about stacking resistors. You mentioned pads. Are you dealing with surface mount technology? I guess you'll have to search for the value, power rating and tolance you need.

With regard to derating, I would think that was considered when the circuit was designed. If you are designing the circuit, [as you well know] chose a power rating that allows a safety margin.

Derating is not the only issue. The other issue is tolerance especially when you have several resistors in series. If the value is critical than you may have to figure out the minimum and maximum possible resistance using the tolerance of each.

John

#### thingmaker3

Joined May 16, 2005
5,084
Im not following...

Why is tolerance more critical with resistors in series than with a single reisistor?

(100x10%)+(100x10%)=200x10% does it not?

#### chesart1

Joined Jan 23, 2006
269
thingmaker3,

Thank you for commenting about the resistors in series. My statement was not correct.

The only time that the tolerance is critical with resistors in series is when the designer is using the resistors as a voltage divider.

If each resistor has a 10% tolerance in a two resistor voltage divider, then one resistor could be 10% below it's rated value and the other resistor could be 10% above it's rated value yielding a 20% error [worst case] in the output voltage of the voltage divider.

One customer required a worst case analysis for a circuit I designed. As you know, performing a worst case analysis sometimes yields some interesting results.

John

#### alim

Joined Dec 27, 2005
113
Originally posted by CoulombMagician@Jan 28 2006, 01:54 AM
Power is cumulative so long as they do not interfere with each others dissipation. Example 1, a surface mount resistor. You can solder one on top of another but together their power rating is not 2x that of one.
Example 2, a large wirewound adjustable power resistor with one of those metal slides on it for a wiper. If all the load current passes between the wiper and one terminal the power dissipation is concentrated in only a fraction of the length of the resistor and maximum power is that same fraction of the rated power of the whole resistor.
[post=13585]Quoted post[/post]​
Hi power is additive ,whether in series or in parallel,in paralell circuits the resistases do not matter providing each can carry its own load. in series circuits -2-50ohms 1watt=100ohms 2watts, however in acase of unequal resistances it becomes practically incorrect. say youwant 100 ohms 2 watts, while 2- 50ohms 1waattwould work, an 80 ohm 1watt and 20ohm 1 watt would not work the 20ohm would carryonly 20%of the power and the 80ohm would have to carry 80%-more than 50% and would burn.

#### stuzman

Joined Jan 29, 2006
1
Originally posted by Propeller@Dec 16 2005, 01:59 AM
This might seem to be a stupid question.

I need a 100ohm 2W resistor.
Can add 2x 50ohm 1W resistors in series to acheive this?
[post=12491]Quoted post[/post]​
No...If you want an equivalent 100 ohm 2W resistor, put two 200ohm resistors in parallel. If you put them in series, as you've said, it would equate to a 1W resistor. For an example, if you put a 1W and a 10W in series, the wattage would equate to the smaller wattage of 1W.

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by stuzman@Jan 29 2006, 05:35 PM
No...If you want an equivalent 100 ohm 2W resistor, put two 200ohm resistors in parallel. If you put them in series, as you've said, it would equate to a 1W resistor. For an example, if you put a 1W and a 10W in series, the wattage would equate to the smaller wattage of 1W.
[post=13629]Quoted post[/post]​
I think i'm stuck in a loop. This topic would never ever end!!!

#### chesart1

Joined Jan 23, 2006
269
Stuzman and Propeller,

My calculations shown in one my previous messages and repeated here for your conveneience prove that two 50 ohm 1 watt resistors could replace one 100 ohm 2 watt resistor:

Lets use an actual example to illustrate the point:
14.14 volts across a 100 ohm 2 watt resistor results in 2 watts dissipation.
Now lets substitute two 50 ohm resistors for the 100 ohm resistor and figure out the power dissipated by each resistor.
The curent is still .1414 amps.
So the power dissipated by each resistor is
P = .1414 amps * .1414 amps * 50 ohms
P = .02 * 50 = 1 watt
And this proves you can replace a 100 ohm 2 watt resistor with two 50 ohm 1 watt resistors in series.

The incorrect statement I referenced in my previous message was about resistance tolerance, not power.

Coulomb Magician,

I agree with you only with regard to the true power per the proof I provided in my calculations.

However, wire wound resistors also have a finite amount of inductance that affects AC operation. If you apply a sinusoidal waveform to the two wire wound resistors, you may be correct. However, if you apply any non-sinusoidal waveform with sharp rise times to the two wire wound resistors, then I am uncertain about your statement.

John

#### alim

Joined Dec 27, 2005
113
Originally posted by chesart1@Jan 29 2006, 02:16 PM
Stuzman and Propeller,

My calculations shown in one my previous messages and repeated here for your conveneience prove that two 50 ohm 1 watt resistors could replace one 100 ohm 2 watt resistor:

Lets use an actual example to illustrate the point:
14.14 volts across a 100 ohm 2 watt resistor results in 2 watts dissipation.
Now lets substitute two 50 ohm resistors for the 100 ohm resistor and figure out the power dissipated by each resistor.
The curent is still .1414 amps.
So the power dissipated by each resistor is
P = .1414 amps * .1414 amps * 50 ohms
P = .02 * 50 = 1 watt
And this proves you can replace a 100 ohm 2 watt resistor with two 50 ohm 1 watt resistors in series.

The incorrect statement I referenced in my previous message was about resistance tolerance, not power.
Coulomb Magician,

I agree with you only with regard to the true power per the proof I provided in my calculations.

However, wire wound resistors also have a finite amount of inductance that affects AC operation. If you apply a sinusoidal waveform to the two wire wound resistors, you may be correct. However, if you apply any non-sinusoidal waveform with sharp rise times to the two wire wound resistors, then I am uncertain about your statement.

John
[post=13633]Quoted post[/post]​
I fully agree with the calculations of chesart I personally avoid long posts (keyboard challenged), and would think anyone can follow the presentation.

#### n9xv

Joined Jan 18, 2005
329
Excellent solve gentleman!

#### PointGiven

Joined Sep 10, 2005
7
The simple answer is yes you can do it..

#### CoulombMagician

Joined Jan 10, 2006
37
Originally posted by alim@Jan 29 2006, 02:10 PM
I fully agree with the calculations of chesart I personally avoid long posts (keyboard challenged), and would think anyone can follow the presentation.
[post=13639]Quoted post[/post]​
My point is really quite simple. The power rating of a resistor is based on how much heat can be removed under normal operating conditions such as the maximum operating temperature and given maximum ambient temperature, the transfer mechanism of heat out of the resistor such as still air convection, and the size of the physical package. That's why the rating of an 0805 resistor is the same whether it's 1k or 100k. If you solder one SMD on top of another the effective area for dissipating heat has not really increased very much so your trying to dissipate twice as much heat through about the same area. If the resistors are physically separated you have about twice the area and twice the power dissipation so about the same temperature rise and you're still within rating.
The wirewound example has to do with the fact that with a pot where all of the leg terminal current is flowing through the wiper and none is flowing through the other leg terminal all the power is dissipated in the small area between the first leg and the wiper and so you have the same problem of smaller area through which to transfer the power. Take a 100W 5 Ohm and set the wiper for 1 Ohm wiper-leg and try to dissipate 100Watt in that wiper-leg connection, you'll get the picture and go from theoretical to practical very quickly!
The discussion up to here has involved lumped circuit parameters and has ignored the distributed nature of current density and heat dissipation, that's my point.