Digital systems and design

Total power dissipation in CMOS inverter

Total power dissipation in CMOS inverter

In this post we calculate the total power dissipation in CMOS inverter. The total power of an inverter is combined of static power and dynamic power.

Let’s consider the inverter representation depicted on the figure below, and let’s imagine that there is a square alternating wave on the input of the inverter.

When the voltage of the square wave is low, the MOSFET is OFF. The load capacitor CL is charged up to the voltage VS via the load resistor RL.

Total power dissipation in CMOS inverter

When the MOSFET is ON, the load capacitor discharges through the MOSFET resistance, and finally the capacitor voltage will reach the voltage level VSRON(RON+RL). (figure below)

The total power dissipated on the inverter can be found as p=ω1+ω2T1+T2.

  1. Let’s calculate what energy will dissipate during interval of time T1 when the signal is high. In this case we can get the equivalent circuit depicted below, and in addition we can also transform this circuit in accordance to Thenevin theorem.

Total power dissipation in CMOS inverter

Total power dissipation in CMOS inverter

Similarly to calculations made before, we can find the nodal voltage vC as the solution of the differential equation, and the the result vC=VTH+(VSVTH)etRTHCLVTH=VSRONRON+RL, RTH=RLRONRON+RL.  Knowing that at the moment t=0 capacitor voltage was VS, when t= the capacitor charges till voltage VTH=VSRONRON+RL. So we can get the expression for the energy ω1=v2SaT1+v2SRL2CL2a2, where a=RON+RL.

2. Now let’s calculate the energy dissipated during the interval T2 when the inverter signal is  low.  In this case the equivalent circuit looks as below:

And the vC nodal voltage  can be found as vC=VSRONRON+RL+(VS+VSRONRON+RL)(1etRLCL).  Here when the t=0 the vCVTH, and when t= the vC=VS. Then dissipating energy for the period of time T2 is ω2=VS2RL2CL2a.

Then the total dissipated energy is ω=ω1+ω2=VS2T1a+VS2RL2CLa, then the total power dissipation of the CMOS inverter is p=VS2T1a(T1+T2)+VS2RL2CLa(T1+T2). Referring to the beginning of the discussion that the dissipated power consist of static and dynamic power, we can conclude that pstatic=VS2T1a(T1+T2) and dynamic power pdynamic=VS2RL2CLa2(T1+T2), where a=RON+RL.

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