Resistor wattage


Joined Jan 23, 2006
Coulomb magician,

That is an excellent post [repeated below]! Heat transfer is a big part of the equation when considering appropriate wattage ratings. My calculations were based on ideal conditions.

Volumes of material exists on heat transfer in industrial electronic equipment, especially the equipment that does not employ fans.


Coulomb Magician wrote:
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.


Joined Jan 18, 2005
2 50-ohm, 1-watt resistors can be connected in series to replace a 100-ohm 2-watt resistor. Power is commulitive. Of course, the situation with the surface area of the SMD is another issue. That was covered quite ellequintly by the CoulombMagician and chesart1.

Power is additive. If a circuit with 10 resistors dissipate 1-watt each, the total power dissipation in the entire circuit will be 10-watts. Assuming of course that the voltages and currents meet the requirement for maximum dissipation of 1-watt per resistor. Series or parallel has no effect on power dissipation. You can rearrange the voltage and/or current configuration but the power dissipation remains unchanged.


Configuration 1:

2 100-watt (120-volt) bulbs are parallel connected across a 120-volt line. A current of 833-mA flows through each bulb. Each bulb has a resistance of 144-ohms.

Configuration 2:

2 100-watt (120-volt) bulbs are series connected across a 240-volt line. A current of 833-mA flows through the series string. Each bulb "drops" 120-volts across it. Each bulb has a resistance of 144-ohms.

Ohm's Law still applies;

I * I * R = P

.833 * .833 * 144 = 100

The light bulb is a good example because its nothing more than a resistor that lights up when you pass current through it.


Joined Jan 18, 2005
Originally posted by thingmaker3@Feb 4 2006, 12:56 AM
How does one stack SMD resistors that are in series with each other? :huh:
[post=13795]Quoted post[/post]​
Thats classified! We could tell you, but then we'd have to . . .

With SMD, other factors come into play. A good opportunity to "brush up" on the physics of heat transfer and the science heat sinking.

Thread Starter


Joined Nov 27, 2005
I gather the easy answer to the original question is Yes...
Chesart1 post explains it well...

Cheers to all.
(Hope this ends the loop...)