 # DC Characteristics of a CMOS Inverter

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DC Characteristics of a CMOS Inverter
A complementary CMOS inverter consists of a p-type and an n-type device connected in series. The DC transfer characteristics of the inverter are a function of the output voltage (Vout) with respect to the input voltage (Vin). The MOS device first order Shockley equations describing the transistors in cut-off, linear and saturation modes can be used to generate the transfer characteristics of a CMOS inverter. Plotting these equations for both the n- and p-type devices produces voltage-current characteristics shown below.

IV Curves for nMOS

PMOS IV Curves

DC Response DC Response: Vout vs. Vin for a gate Ex: Inverter
When Vin = 0 -> Vout = VDD When Vin = VDD -> Vout = 0 In between, Vout depends on transistor size and current By KCL, we that Idsn = |Idsp| We could solve equations But graphical solution gives more insight

Transistor Operation Current depends on region of transistor behavior
For what Vin and Vout are nMOS and pMOS in Cutoff? Linear? Saturation?

nMOS Operation Cutoff Linear Saturated Vgsn < Vgsn > Vdsn <

nMOS Operation Cutoff Linear Saturated Vgsn < Vtn Vgsn > Vtn
Vdsn < Vgsn – Vtn Vdsn > Vgsn – Vtn

nMOS Operation Cutoff Linear Saturated Vgsn < Vtn Vgsn > Vtn
Vdsn < Vgsn – Vtn Vdsn > Vgsn – Vtn Vgsn = Vin Vdsn = Vout

nMOS Operation Cutoff Linear Saturated Vgsn < Vtn Vin < Vtn
Vdsn < Vgsn – Vtn Vout < Vin - Vtn Vdsn > Vgsn – Vtn Vout > Vin - Vtn Vgsn = Vin Vdsn = Vout

pMOS Operation Cutoff Linear Saturated Vgsp > Vgsp < Vdsp >

pMOS Operation Cutoff Linear Saturated Vgsp > Vtp Vgsp < Vtp
Vdsp > Vgsp – Vtp Vdsp < Vgsp – Vtp

pMOS Operation Cutoff Linear Saturated Vgsp > Vtp Vgsp < Vtp
Vdsp > Vgsp – Vtp Vdsp < Vgsp – Vtp Vgsp = Vin - VDD Vdsp = Vout - VDD Vtp < 0

pMOS Operation Cutoff Linear Saturated Vgsp > Vtp
Vin > VDD + Vtp Vgsp < Vtp Vin < VDD + Vtp Vdsp > Vgsp – Vtp Vout > Vin - Vtp Vdsp < Vgsp – Vtp Vout < Vin - Vtp Vgsp = Vin - VDD Vdsp = Vout - VDD Vtp < 0

I-V Characteristics Make pMOS wider than nMOS such that bn = bp

Current vs. Vout, Vin

Load Line Analysis For a given Vin: Plot Idsn, Idsp vs. Vout
Vout must be where |currents| are equal.

Load Line Analysis Vin = 0

Load Line Analysis Vin = 0.2VDD

Load Line Analysis Vin = 0.4VDD

Load Line Analysis Vin = 0.6VDD

Load Line Analysis Vin = 0.8VDD

Load Line Analysis Vin = VDD

DC Transfer Curve Transcribe points onto Vin vs. Vout plot

Operating Regions Revisit transistor operating regions Region nMOS
pMOS A B C D E

Operating Regions Revisit transistor operating regions Region nMOS
pMOS A Cutoff Linear B Saturation C D E

Beta Ratio If bp / bn  1, switching point will move from VDD/2
Called skewed gate Other gates: collapse into equivalent inverter

DC Characteristics of a CMOS Inveter
The DC transfer characteristic curve is determined by plotting the common points of Vgs intersection after taking the absolute value of the p-device IV curves, reflecting them about the x-axis and superimposing them on the n-device IV curves. We basically solve for Vin(n-type) = Vin(p-type) and Ids(n-type)=Ids(p-type) The desired switching point must be designed to be 50 % of magnitude of the supply voltage i.e. VDD/2. Analysis of the superimposed n-type and p-type IV curves results in five regions in which the inverter operates. Region A occurs when 0 leqVin leq Vt(n-type). The n-device is in cut-off (Idsn =0). p-device is in linear region, Idsn = 0 therefore -Idsp = 0 Vdsp = Vout – VDD, but Vdsp =0 leading to an output of Vout = VDD. Region B occurs when the condition Vtn leq Vin le VDD/2 is met. Here p-device is in its non-saturated region Vds neq 0. n-device is in saturation Saturation current Idsn is obtained by setting Vgs = Vin resulting in the equation:

CMOS Inverter DC Characteristics

CMOS Inverter Transfer Characteristics
In region B Idsp is governed by voltages Vgs and Vds described by: Region C has that both n- and p-devices are in saturation. Saturation currents for the two devices are: Region D is defined by the inequality p-device is in saturation while n-device is in its non-saturation region. Equating the drain currents allows us to solve for Vout. (See supplemental notes for algebraic manipulations).

CMOS Inverter Static Charateristics
In Region E the input condition satisfies: The p-type device is in cut-off: Idsp=0 The n-type device is in linear mode Vgsp = Vin –VDD and this is a more positive value compared to Vtp. Vout = 0 nMOS & pMOS Operating points Vout =Vin-Vtp A VDD B Vout =Vin-Vtn Both in sat C nMOS in sat Output Voltage pMOS in sat D E VDD/2 VDD+Vtp VDD Vtp Vtn