Energy dissipated as heat in CMOS circuit

[EngIntoHW]

Hi,

Below is an example to a CMOS network.
I'd like to calculate the total energy dissipated as heat when A switches.

Is it:
E = 0.5 * Ctotal * VDD^2

or:

E = Ctotal * VDD^2

?

Thanks.


View attachment 60851

 

[alec_t]

Neither equation is necessarily correct for energy dissipated. The first equation gives the energy stored in capacitance Ctotal charged to a voltage Vdd.

 

[colin55]

The energy dissipated in a CMOS circuit will be miniscule.

 

[EngIntoHW]

I see, thank you.
Is there a formula for amount of energy dissipated as heat?

 

[EngIntoHW]

The energy dissipated in a CMOS circuit will be miniscule.

Hi Colin,
Just saw your message (didnt refresh page before posting earlier).

Isn't the dynamic Energy dissipated in CMOS quite much?

 

[crutschow]

Energy is dissipated in charging and discharging all the parasitic capacitances in a CMOS digital circuit.
The total energy dissipated for one charge and discharge cycle is CV^2.
The total power dissipated when the circuit is operating is equal to fCV^2.
Where:
f = switching frequency
C = total parasitic capacitance
V = switching voltage

All this energy is dissipated as heat in the circuit. That's why a modern CMOS PC microprocessor dissipates many tens of watts when operating at GHz switching speeds and requires a large dedicated heat sink with an integral cooling fan to keep it from burning up.

 

[colin55]

When you have millions of CMOS circuits or "gates" the energy may be significant, but a single network will be very small.
The energy dissipated by any circuit or chip is simply the voltage multiplied by the current.

 

[EngIntoHW]

Hi crutschow,
You helped me significantly.
You said: "The total energy dissipated for one charge and discharge cycle is CV^2.".
So it is true to say that when Input changes just once (for example, from 0V to VDD), the total energy dissipated as heat is CV^2, isn't it?

Energy is dissipated in charging and discharging all the parasitic capacitances in a CMOS digital circuit.
The total energy dissipated for one charge and discharge cycle is CV^2.
The total power dissipated when the circuit is operating is equal to fCV^2.
Where:
f = switching frequency
C = total parasitic capacitance
V = switching voltage

All this energy is dissipated as heat in the circuit. That's why a modern CMOS PC microprocessor dissipates many tens of watts when operating at GHz switching speeds and requires a large dedicated heat sink with an integral cooling fan to keep it from burning up.

 

[MrAl]

Hi,

It's not the capacitance itself that dissipates energy as heat, as capacitance is a storage container which can store and supply energy without loss. The problem is, it's hard to charge a capacitance without using some power because the various other components used to charge the cap have resistance, and that resistance is what eats up the energy when the capacitance is charged (or discharged).
Thus, as an exercise to figure out how much energy is dissipated you only need to figure out what gets lost in the resistance, and then generalize for any resistance, switching frequency, etc.
The resistance comes in the form of cap ESR and MOS transistor 'on' resistance, but also from the source impedance.

 

[crutschow]

Hi,

It's not the capacitance itself that dissipates energy as heat, as capacitance is a storage container which can store and supply energy without loss. The problem is, it's hard to charge a capacitance without using some power because the various other components used to charge the cap have resistance, and that resistance is what eats up the energy when the capacitance is charged (or discharged).
Thus, as an exercise to figure out how much energy is dissipated you only need to figure out what gets lost in the resistance, and then generalize for any resistance, switching frequency, etc.
The resistance comes in the form of cap ESR and MOS transistor 'on' resistance, but also from the source impedance.

That's true. But it is independent of the value of the resistance and depends only on the value of capacitance. The power loss can't be avoided unless you resonantly charge and discharge the capacitor using an inductor (as in a resonant circuit).

 

[MrAl]

That's true. But it is independent of the value of the resistance and depends only on the value of capacitance. The power loss can't be avoided unless you resonantly charge and discharge the capacitor using an inductor (as in a resonant circuit).

Hi,

Yes but part of finding that out was meant to be part of the exercise. It doesnt make as much sense until the individual goes through the exercise, that's why i suggested it.

 

[EngIntoHW]

Hi,
I really thank you and appreciate you helping me

Thanks a lot! it helped indeed!

 

[RCinFLA]

CMOS logic is charging and discharging capacitors. The CMOS device drivers' current to charge and discharge the capacitive load is what dissipates power. The drain to source resistance is where the heat is generated.

Power consumption will be C*V^2*F regardless of slew rate the Rds-ON of devices produce unless full voltage swing is not achieved for given drive frequency. Most digital logic have slew rates less then 10% of their maximum operating frequency. Power supply will have very high current spikes during these transitions and bypass caps are necessary to supply the high transition surge current. Since most all logic is clocked syncronously across the chip, many gates' transition occur at the same time, making the supply current surge worse.

This can become a significant power for low Rds-On, and higher voltage MOSFET's used for switching power supplies. Voltage being equal, low Rds-On means larger FET which has more gate drive capacitance. Current being equal, Rds-On being equal, higher voltage also mean long gate lengths which mean more gate drive capacitance.

Higher voltage may require an older larger geometry FET process to achieve the higher breakdown voltage.
Smaller device geometry can yield lower gate capacitance, but have lower Vds breakdown voltage.

At about 90 nm processors and below static leakage starts to become an issue. At about 25 nm to 35 nm the static leakage become just about as bad as dynamic current. Only effective mitigation is to literally shut the supply off to high speed logic sections when not actively in use. Dual and triple device gate threshold voltage devices are used. Low threshold is high speed but bad leakage. Higher threshold has better leakage performance but slow performance. The slow devices with low leakage are used to make registers that hold high speed logic registers during the power down process. Power up requires a few clock cycles to re-transfer the register states back into the high speed logic. The latest i3, i5, and i7 Intel processors have extensive power management process that adjust supply voltage and clocking speeds to accomplish these functional blocks' active/deactive process.