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Re: An interesting prototype has turned up
12/22/12 02:54 AM
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OK, now you're way out. In a CMOS chip, I^2*R losses in the channels are a relatively small contributor to heat. The channel and gate of each transistor separated by the oxide layer form a capacitor. Turning the transistor on requires charging this capacitor, and turning the transistor off requires discharging it. The majority of heat from running the chip comes from losses in charging/discharging gate capacitance. If you clock the chip faster, you're be charging/discharging these capacitors more often, and pump out more heat. Power dissipation depends largely on clock speed.
The temperature of the chip depends on ambient temperature, power dissipation, and how effectively you're getting heat out of the package into the environment. Ambient temperature isn't necessarily room temperature - it's often higher as your chip is in the vicinity of other heat-producing components. Then you're got the inefficiency of getting heat from the die to the package, from the package to the heatskink, and from the heatsink to the air. Even if you're well within voltage and clock speed limits, if your ambient temperature is too high or you aren't removing heat effectively you'll end up with very high temperatures.
Yes, you can damage semiconductors by applying excessive voltage. This can result in "punch-through" collector/emitter shorts in BJTs, and gate/channel shorts from sparking across the oxide layer in MOSFETs. This is something that happens suddenly, it's not what I'm talking about.
Chemical reactions towards lower energy states happen naturally. You don't need to supply large amounts of energy to get this to happen. Most chemical reactions can be accelerated by higher temperatures. Chips just degrade, and they degrade faster if you use them - it's a fact of life.
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