Question about Dynamic Power in CMOS

[videogames101]

Prompted by a question in a recent interview I had, I was thinking about dynamic power consumption in CMOS. I know the formula well, as it relates to the charging and discharging cycle of the load capacitance. P=afCV^2

However, my question has become where is the power actually dissipated? Rabaey et al. mentions it is dissipated in the NMOS and PMOS which are turned on to charge/discharge the capacitance. There are short derivations which demonstrates that the power consumed is independent of the resistance of that NMOS or PMOS, which involves energy calculations when you charge an RC circuit with a step function. Something seems strange to me though. Lets say I replace the PMOS and NMOS with a similar devices except they have superconducting behavior with a R_on of truly 0 Ohms. Where is the power being dissipated now? The RC constant is now 0, which breaks most of the math involved in the most cited derivations of power consumption and capacitor charging because basic circuit theory says dv/dt must be finite across the capacitor. I must be missing something very fundamental here.

I'm also assuming an ideal load capacitance with 0 real impedance, which in reality would be non-zero. In which case I propose the same question with a hypothetical circuit where all material is superconducting so there is 0 resistance at all points.

Although I can follow the equations which yield this result with a finite R, it seems fundamentally strange that a capacitor only stores half the energy that was used to charge it. It seems I had forgotten this idea since basic circuit theory.

The best explanation I've found isn't even in any of my circuit textbooks, it's on hyperphysics:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c4

"Though it will not be shown here, if you proceed further with this problem by making the charging resistance so small that the initial charging current is extremely high, a sizable fraction of the charging energy is actually radiated away as electromagnetic energy. This crosses the threshold into antenna theory because not all the loss in charging was thermodynamic - but still the loss in the process was half the energy supplied by the battery in charging the capacitor."

Given I haven't studied more than basic EM theory, maybe someone could direct me to some further explanation of this phenomena? Any organic explanations would also be appreciated.

 

[uclabachelor]

What you're going into is the physics of devices... but from an application level, the current-voltage relationship of a capacitor is independent of resistance, ie, I = C * dv/dt.

In a perfect world where R = 0 for everything, dv/dt would be instantaneous (since dt will be 0) and I = infinity... however the charge in the capacitor is still finite, along with dV, which is the essentially the voltage of its terminal in this case.

As for power loss in a capacitor, you still have to discharge all that energy (C) into ground, so there is still power loss - it's just not lost through resistance (converted into heat) in the device.

 

[TuxDave]

However, my question has become where is the power actually dissipated?

Mostly heat. If you have a simple battery connected to a resistor and nothing else, the battery is clearly using power (voltage * current) and you can ask yourself the same question, where did that energy go? The typical answer is to point at the resistor and basically when you have any current going over a resistor, that will consume energy (and due to conservation of mass/energy), that energy has to go somewhere and it's either heat or, as you mentioned later, electromagnetic waves.

Back to your RC circuit, half goes to the capacitor and half gets wasted on the resistor. For extra homework, instead of using a step function as your input voltage, if the slope of the voltage source slows down towards infinity, how much energy is wasted in the resistor?

Lets say I replace the PMOS and NMOS with a similar devices except they have superconducting behavior with a R_on of truly 0 Ohms. Where is the power being dissipated now?

If resistance is zero, your RC time constant is zero and your battery used infinite current in zero time (sort of like a dirac delta function except I'm guessing the integral of this delta function = energy stored on the capacitor). How much energy do you think the battery used?

 

[Mark R]

Although I can follow the equations which yield this result with a finite R, it seems fundamentally strange that a capacitor only stores half the energy that was used to charge it. It seems I had forgotten this idea since basic circuit theory.

It's not correct that the capacitor only stores half the energy used to charge it, generally. This applies if you use a fixed voltage source.

This is because the voltage source will deliver CV^2 energy, but the capacitor can only store 1/2 CV^2, because its voltage is related to charge.

The remainder of the energy will be dissipated in the impedance of other circuit elements. In the case of a resistor, it will be lost as heat. In the case of an inductor it will be stored (and form a tuned oscillator with the capacitor - so that the capacitor and inductor end up alternately storing CV^2 energy, but on average each will store 1/2 CV^2).