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## Theory of Static Characteristics II of MOS Transistor Inverter

MOS Transistor Inverter: Static Characteristics IIMOS Inverter Voltage Transfer Characteristic:The schematic figure of simple MOS transistor inverter with a resistive load is repeated in figure shown below. Since with the simple bipolar transistor inverter, the transfer characteristic can be plotted as output voltage against input voltage, V

_{o}vs. V_{in}as shown in figure below.Figure: Schematic Diagram of Simple MOS Inverter

Initially, with V

_{i}= 0 the input voltage to transistor is beneath threshold voltage and the transistor is OFF or non-conducting and therefore the output voltage is pulled up to the supply voltage V_{DD}. Once the input voltage is raised to be equivalent to the threshold voltage, V_{T}, the transistor starts to conduct and therefore the output voltage drops. As V_{DS}> V_{GS}– V_{T}, the transistor operates initially in saturation region. Since the input voltage is further raised, the output voltage continues to drop until ultimately V_{DS}< V_{GS}– V_{T}and the transistor comes out of the saturation region to operate in non-saturation region. Ultimately the input voltage reaches an utmost of V_{DD}and the output reaches its minimum value of V_{OL }as formerly computed.Figure: Voltage Transfer Characteristic of Simple MOS Transistor InverterCritical Logic Voltages:The similar critical input and output logic voltages can be stated as for other logic families viz.:

V

_{iLMAX}= maximum voltage acceptable as the logic LO inputV

_{iHMIN }= minimum voltage acceptable as the logic HI input.V

_{OLMAX}= maximum voltage acceptable as the logic LO output.V

_{OHMIN}= minimum voltage acceptable as the logic HI output.a)

:Critical Point V_{iL MAX}, V_{OH MIN}This is the point on upper left-hand portion of transfer characteristic where the slope is -1. At this point, the transistor can be taken to operate in the saturation region where, neglecting the consequences of channel length modulation for simplicity, the drain current is explained as:

I

_{D}= K_{n}(V_{GS }- V_{T})^{2}However as V

_{O}= V_{DS}and V_{i}= V_{GS}and V_{O }= V_{DD}– i_{D}R_{D}then:V

_{o}= V_{DD}– K_{n}R_{D}(V_{i }- V_{T})^{2}.......................... (a)On expanding it gives:

V

_{o}= V_{DD}- K_{n}R_{D}V_{i}^{2}+ 2 K_{n}R_{D}V_{i}V_{T}- K_{n}R_{D}V_{T}^{2}On differentiating:

∂V

_{o}/∂V_{i }= - 2 K_{n}R_{D}V_{i }+ 2 K_{n}R_{D}V_{T}At critical point ∂V

_{o}/∂V_{i}= -1 with V_{i }= V_{iL MAX}and V_{O}= V_{OH MIN}and hence:- 2 K

_{n}R_{D}V_{i }+ 2 K_{n}R_{D}V_{T}= - 12 K

_{n}R_{D}V_{iLMAX}= 1 + 2 K_{n}R_{D}V_{T}And hence,

V

_{iLMAX}= V_{T}+ 1/(2 K_{n}R_{D})This value is a slight higher than V

_{T}and for illustration given with V_{T}= 1V, R_{D}= 100kΩ and K_{n}= 100µAV^{-2}, V_{iL MAX }= 1.05V.Replacing back into equation (a) to find out the output voltage for this coordinate gives:

V

_{OHMIN}= V_{DD}– K_{n}R_{D}(V_{iL MAX}- V_{T})^{2}V

_{OHMIN}= V_{DD}– K_{n}R_{D}[V_{T}+ (1/ K_{n}R_{D}) - V_{T}]^{2}And hence ultimately:

V

_{OHMIN}= V_{DD}– (1/4 K_{n}R_{D})This value is a slight lower than VDD and for illustration given with V

_{DD}= 10V, V_{T}= 1V, R_{D}= 100kΩ and K_{n}= 100µAV^{-2}, V_{OH MIN}= 9.98V. The coordinate of critical point (a) is then:V

_{iLMAX}, V_{OHMIN}= 1.05, 9.98 Vb)

Critical Point V_{iH MIN}, V_{OL MAX}This is the point on lower right-hand portion of the characteristic where slope is -1. At this point, the transistor can be taken to operate in non-saturation region where the drain current is explained as:

I

_{D}= K_{n}[2(V_{GS}- V_{T})V_{DS}– V^{2}_{DS}]However again, as V

_{O}= V_{DS}and V_{i }= V_{GS}and V_{O}= V_{DD}– i_{D}R_{D }then:V

_{O}= V_{DD}– 2 K_{n}R_{D}(V_{i}- V_{T}) V_{o}+ K_{n}R_{D}V_{o}^{2}On expanding:

V

_{O}= V_{DD}– 2 K_{n}R_{D}V_{i}V_{o}+ 2 K_{n}R_{D}V_{T}V_{o}+ K_{n}R_{D}V_{o}^{2}On rearranging:

V

_{O}[1 - K_{n}R_{D}V_{T}] = V_{DD }- 2 K_{n}R_{D}V_{i}V_{o}+ K_{n}R_{D}V_{o}^{2}There is a choice here to employ implicit differentiation to find ∂V

_{o}/∂V_{i}or to re-arrange the expression as V_{i}in terms of V_{O}and then determine ∂V_{i}/∂V_{o}. The latter is simpler as there is just one term in V_{i}. Then,2 K

_{n}R_{D}V_{i}V_{o}= V_{DD}– [1 - 2 K_{n}R_{D}V_{T}]V_{o}+ K_{n}R_{D}V_{o}^{2}And hence,

V

_{i }= V_{DD}/(2 K_{n}R_{D}V_{o}) – [(1 - 2 K_{n}R_{D}V_{T})/ 2 K_{n}R_{D}) + (V_{o}/2)............... (b)Then,

∂V

_{o}/∂V_{i}= - V_{DD}/(2 K_{n}R_{D}V_{o}^{2}) + (1/2)For ∂V

_{o}/∂V_{i}= -1 we can employ ∂V_{i}/∂V_{o}= -1 and hence:- (V

_{DD}/2 K_{n}R_{D}V_{o}^{2}) + (1/2) = - 1(V

_{DD}/2 K_{n}R_{D}V_{o}^{2}) = 3/2V

_{o}^{2}= V_{DD}/3 K_{n}R_{D}By taking the positive root as practical value gives:

V

_{OLMAX}= √V_{DD}/3K_{n}R_{D}That for the illustration given with V

_{DD}= 10V, V_{T}= 1V, RD = 100kΩ and K_{n}= 100µAV^{-2}, V_{OL MAX}= 0.58V.This is significantly higher than the great value of VOL computed formerly. Then replacing this back into the expression for V

_{i}in equation (b) above gives:And therefore ultimately the critical input value is as shown:

V

_{iHMIN}= V_{T }+ 2√V_{DD}/3K_{n}R_{D}– (1/2 K_{n}R_{D})That for illustration given with V

_{DD}= 10V, V_{T }= 1V, R_{D }= 100kΩ and K_{n}= 100µAV^{-2}provides V_{iH MIN }= 2.1V. This gives the coordinates of critical point (b) as:V

_{iHMIN}, V_{OLMAX}= 2.1, 0.58 VNoise Margins:Ultimately, the noise margins for simple MOS inverter can be computed approximately from the critical points evaluated from the transfer characteristic as shown:

NM

_{H}= V_{OHMIN}– V_{iHMIN }= 9.98 – 2.1 = 7.88VNM

_{L }= V_{iLMAX }– V_{OLMAX}= 1.05 – 0.58 = 0.47VLatest technology based Electrical Engineering Online Tutoring AssistanceTutors, at the

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