Why does a faster clock require more power?

Question

If you overclock a microcontroller, it gets hot.

If you overclock a microcontroller, it needs more voltage.

In some abstract way it makes sense: it is doing more computation, so it needs more energy (and being less than perfect, some of that energy dissipates as heat).

However, from a just plain old Ohm's law level electricity and magnetism, what is going on?

Why does the clock frequency have anything to do with power dissipation or voltage?

As far as I know, the frequency of AC has nothing to do with its voltage or power, and a clock is just a super-position of a DC and a (square) AC. Frequency doesn't affect the DC.

Is there some equation relating clock frequency and voltage or clock frequency and power?

I mean does a high speed oscillator need more voltage or power than a low speed one?

Answer 1

Voltage required is affect by significantly more than clock speed, but you are correct, for higher speeds you will need higher voltages in general.

Why does power consumption increase?

This is a lot messier than a simple circuit, but you can think about it being similar to an RC circuit.

RC circuit equivilent

At DC an RC circuit consumes no power. At a frequency of infinity, which is not attainable, but you can always solve this theoretically, the capacitor acts as a short and you are left with a resistor. This means you have a simple load. As frequency decreases the capacitor stores and discharges power causing a smaller amount of power dissipated overall.

What is a microcontroller?

Inside it is made up of many many MOSFETs in a configuration we call CMOS.

If you try to change the value of the gate of a MOSFET you are just charging or discharging a capacitor. This is a concept I have a hard time explaining to students. The transistor does a lot, but to us it just looks like a capacitor from the gate. This means in a model the CMOS will always have a load of a capacitance.

Wikipedia has an image of a CMOS inverter I will reference.

The CMOS inverter has an output labeled Q. Inside a microcontroller your output will be driving other CMOS logic gates. When your input A changes from high to low the capacitance on Q must be discharged through the transistor on bottom. Every time you charge a capacitor you see power use. You can see this on wikipedia under power switching and leakage.

Why does voltage have to go up?

As you voltage increases it makes it easier to drive the capacitance to the threshold of your logic. I know this seems like a simplistic answer, but it is that simple.

When I say it is easier to drive the the capacitance I mean that it will be driven between the thresholds faster, as mazurnification put it:

With increased supply drive capability of the MOS transistor also increases (bigger Vgs). That means that actual R from RC decreases and that is why gate is faster.

In relation to power consumption, due to how small transistors are there is a large leakage through the gate capacitance, Mark had a bit to add about this:

higher voltage results in higher leakage current. In high transistor count devices like a modern desktop CPU leakage current can account for the majority of power dissipation. as process size gets smaller and transistor counts rise, leakage current becomes more and more the critical power usage statistic.

Answer 2

In general, CMOS gates only use current when they switch states. So the faster the clock speed is, the more often gates are switching, thus more current is switched, and more power is consumed.

Answer 3

Well, it's all about logic level transitions.

When any single bit of an output changes... the electrical value must slew from high to low, or low to high. This pulls power from the power supply, or dumps some power back onto the ground plane. It also generates a little waste heat due to inefficiencies.

If you increase the clock rate, you increase the number of these transitions per unit time, therefore you use more power to feed these logic level transitions.

Increased voltage requirements are a little different. The time it takes a signal to transition from low to high is called the rise time. To safely operate at any given frequency the logic must be able to consistently make this transition before the next clock samples the new value. At a certain point, the logic will not be able to meet the rise time requirements of a particular frequency. This is where upping the voltage will help, as it decreases rise time.

Heat is fairly simple. The chip is designed to handle a certain amount of heat generated by a certain clock rate. Increase the number of transitions by increasing the clock rate, and you're going to get more waste heat. When overclocking, you can easily outpace the cooling system's ability to remove that heat.

Answer 4

Think of a basic RC circuit where the R and C are in parallel. Our goal is to have a clock at the output of this circuit - a 0-5V 1KHz square wave. So when we want the clock to be high we turn on our voltage source and it charges up the capacitor until the output is at 5V, and when we want 0V we turn it off and let it discharge. The charge/discharge time is determined by the RC constant of the circuit. There's a problem - the circuit doesn't charge up quick enough for a 1KHz clock. What do I do?

We can't change the RC constant of the circuit - it's fixed. So we have to charge the capacitor up quicker somehow, but still have the same charged voltage. To do this we need an active circuit that monitors the output voltage of the RC circuit and varies the current going into the capacitor to charge it up quicker. More current means more power.

When you want a faster clock, you need to charge up the capacitor faster. You charge up a capacitor by pushing current into it. Current * voltage = power. You need more power!

Everything in a digital system is tied to the clock and everything has capacitance. If you have 100 TTL chips on one clock it has to drive a lot of current to charge all of them, then draw a lot of current to pull them down. The fundamental reason ohms law isn't holding is because these are active devices, not passive. They do electrical work to force the clock to be as close to a perfect square wave as possible.

If you overclock a microcontroller it gets hot

Yes - quicker change means more current flowing and power is voltage * current. Even if voltage stays the same, current used increases, so more power dissipation, more heat.

If you overclock a microcontroller it needs more voltage

Partially true - it needs more power, not necessarily more voltage. The microcontroller is in some way converting the extra voltage to more current to achieve its needs.

As far as I know, the frequency of AC has nothing to do with its voltage or power, and a clock is just a super-position of a DC and a (square) AC. Frequency doesn't affect the DC.

Only for a purely resistive load. There's a lot of trickery happening with AC power.

Is there some equation relating clock frequency and voltage or clock frequency and power?

Probably not a consistent one, but it's related to the simple equations Q=CV, V=I*R, P=I*V

Just remember: Higher frequency => faster rise time => must fill up capacitors quicker => more charge => more current => more power.

Answer 5

Power = switching factor * Capacitance * (VDD^2) * frequency.

As the fast clock has higher switching factor, and also higher frequency, thus higher dynamic power consumption.