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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.

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IV Curves for nMOS

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PMOS IV Curves

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**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

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**Transistor Operation Current depends on region of transistor behavior**

For what Vin and Vout are nMOS and pMOS in Cutoff? Linear? Saturation?

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**nMOS Operation Cutoff Linear Saturated Vgsn < Vgsn > Vdsn <**

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**nMOS Operation Cutoff Linear Saturated Vgsn < Vtn Vgsn > Vtn**

Vdsn < Vgsn – Vtn Vdsn > Vgsn – Vtn

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**nMOS Operation Cutoff Linear Saturated Vgsn < Vtn Vgsn > Vtn**

Vdsn < Vgsn – Vtn Vdsn > Vgsn – Vtn Vgsn = Vin Vdsn = Vout

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**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

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**pMOS Operation Cutoff Linear Saturated Vgsp > Vgsp < Vdsp >**

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**pMOS Operation Cutoff Linear Saturated Vgsp > Vtp Vgsp < Vtp**

Vdsp > Vgsp – Vtp Vdsp < Vgsp – Vtp

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**pMOS Operation Cutoff Linear Saturated Vgsp > Vtp Vgsp < Vtp**

Vdsp > Vgsp – Vtp Vdsp < Vgsp – Vtp Vgsp = Vin - VDD Vdsp = Vout - VDD Vtp < 0

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**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

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

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Current vs. Vout, Vin

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**Load Line Analysis For a given Vin: Plot Idsn, Idsp vs. Vout**

Vout must be where |currents| are equal.

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Load Line Analysis Vin = 0

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Load Line Analysis Vin = 0.2VDD

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Load Line Analysis Vin = 0.4VDD

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Load Line Analysis Vin = 0.6VDD

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Load Line Analysis Vin = 0.8VDD

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Load Line Analysis Vin = VDD

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Load Line Summary

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DC Transfer Curve Transcribe points onto Vin vs. Vout plot

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**Operating Regions Revisit transistor operating regions Region nMOS**

pMOS A B C D E

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**Operating Regions Revisit transistor operating regions Region nMOS**

pMOS A Cutoff Linear B Saturation C D E

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**Beta Ratio If bp / bn 1, switching point will move from VDD/2**

Called skewed gate Other gates: collapse into equivalent inverter

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**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:

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**CMOS Inverter DC Characteristics**

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**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).

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**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

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