[SOLVED] Does increasing the IPC on a chip increase the heat output?

[pedrolopez]

Increasing the number of cores, and increasing the clock speed both increase performance, but also increase the thermal output. IPC increases the performance, but does it also increase the heat output?

 

[siaan312]

Increasing clock speed does not increase thermal output. neither does ipc.

What increases the thermal output is voltage, and wattage.
More volt=more wattage (usually)=more thermal.

To achieve higher clock speeds, the voltage usually needs to be adjusted, making it output more heat, but the increase of the clockspeed itself does not release more heat.
In some cases, you can up the clockspeed, but put the voltage lower, making thermal output lower even with increased performance.

Increasing the number of cores also doesn't reaaaally matter to thermal output directly, but a chip that has more cores will usually take more wattage.

But i guess clockspeed and cores affect thermal output INdirectly.
So following that logic, IPC increase does not increase thermal output, neither directly or indirectly.

 

[hotaru.hino]

IPC isn't a direct factor in heat output as it's reflective of the design of the part. Manufacturers will take what IPC they've obtained, then adjust the clock speed to meet some requirement like offering the same performance as their competitor but with less power dissipation or more performance for the same power dissipation.

The three main things that affect heat output are voltage, clock frequency, and the insulating material used to separate electrically conductive components in the integrated circuit. The equation is generally simplified to P = C * V^2 * f.

 

[drea.drechsler]

Increasing the number of cores, and increasing the clock speed both increase performance, but also increase the thermal output. IPC increases the performance, but does it also increase the heat output?

With all else equal increasing IPC means the processor is doing more work. Doing more work inherently means increasing current draw and, therefore, greater wattage (watts = current * voltage) which increases heat.

But I can't believe it's ever 'all else equal' when improving something so architecture dependent as IPC. There could also be other changes that improve efficiency much more than IPC, for instance, so specific heat output might actually decrease.

 

[Math Geek]

heat output is simple to understand. energy can not be created nor destroyed. therefore every bit of energy put into the chip will also leave the ship.

100w in, means 100w heat output. 50w in means 50w heat output and so on. volts x amps = wattage. increase the volts or amps in and you increase the wattage is and thus the wattage out. does not matter why the voltage/amps are increased, but if they are, then so is output.