DC Characteristics of a CMOS InverterA 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 ResponseDC Response: Vout vs. Vin for a gateEx: InverterWhen Vin = 0 -> Vout = VDDWhen Vin = VDD -> Vout = 0In between, Vout depends ontransistor size and currentBy KCL, we thatIdsn = |Idsp|We could solve equationsBut graphical solution gives more insight

Transistor OperationCurrent depends on region of transistor behaviorFor what Vin and Vout are nMOS and pMOS inCutoff?Linear?Saturation?

nMOS Operation

CutoffLinearSaturatedVgsn

Vdsn

Vdsn >

nMOS Operation

CutoffLinearSaturatedVgsn < VtnVgsn > Vtn

Vdsn < Vgsn VtnVgsn > Vtn

Vdsn > Vgsn Vtn

nMOS OperationVgsn = VinVdsn = Vout

CutoffLinearSaturatedVgsn < VtnVgsn > Vtn

Vdsn < Vgsn VtnVgsn > Vtn

Vdsn > Vgsn Vtn

nMOS OperationVgsn = VinVdsn = Vout

CutoffLinearSaturatedVgsn < VtnVin < VtnVgsn > VtnVin > VtnVdsn < Vgsn VtnVout < Vin - VtnVgsn > VtnVin > VtnVdsn > Vgsn VtnVout > Vin - Vtn

pMOS Operation

CutoffLinearSaturatedVgsp >Vgsp Vgsp Vgsp VtpVgsp < Vtp

Vdsp < Vgsp Vtp

pMOS OperationVgsp = Vin - VDDVdsp = Vout - VDDVtp < 0

CutoffLinearSaturatedVgsp > VtpVgsp < Vtp

Vdsp > Vgsp VtpVgsp < Vtp

Vdsp < Vgsp Vtp

pMOS OperationVgsp = Vin - VDDVdsp = Vout - VDDVtp < 0

CutoffLinearSaturatedVgsp > VtpVin > VDD + VtpVgsp < VtpVin < VDD + VtpVdsp > Vgsp VtpVout > Vin - VtpVgsp < VtpVin < VDD + VtpVdsp < Vgsp VtpVout < Vin - Vtp

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

Current vs. Vout, Vin

Load Line AnalysisFor a given Vin:Plot Idsn, Idsp vs. VoutVout must be where |currents| are equal.

Load Line AnalysisVin = 0

Load Line AnalysisVin = 0.2VDD

Load Line AnalysisVin = 0.4VDD

Load Line AnalysisVin = 0.6VDD

Load Line AnalysisVin = 0.8VDD

Load Line AnalysisVin = VDD

Load Line Summary

DC Transfer CurveTranscribe points onto Vin vs. Vout plot

Operating RegionsRevisit transistor operating regions

RegionnMOSpMOSABCDE

Operating RegionsRevisit transistor operating regions

RegionnMOSpMOSACutoffLinearBSaturationLinearCSaturationSaturationDLinearSaturationELinearCutoff

Beta RatioIf bp / bn 1, switching point will move from VDD/2Called skewed gateOther gates: collapse into equivalent inverter

DC Characteristics of a CMOS InveterThe 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 = 0Vdsp = 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 saturationSaturation current Idsn is obtained by setting Vgs = Vin resulting in the equation:

CMOS Inverter DC Characteristics

CMOS Inverter Transfer CharacteristicsIn 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 CharateristicsIn Region E the input condition satisfies:

The p-type device is in cut-off: Idsp=0The n-type device is in linear modeVgsp = Vin VDD and this is a more positive value compared to Vtp.Vout = 0nMOS & pMOS Operating pointsACBDEVtpVtnVDD0VDD/2VDD+VtpVDDBoth in satnMOS in satpMOS in satOutput VoltageVout =Vin-VtpVout =Vin-Vtn