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, Vo vs. Vin as shown in figure below.
Figure: Schematic Diagram of Simple MOS Inverter
Initially, with Vi = 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 VDD. Once the input voltage is raised to be equivalent to the threshold voltage, VT, the transistor starts to conduct and therefore the output voltage drops. As VDS > VGS – VT, the transistor operates initially in saturation region. Since the input voltage is further raised, the output voltage continues to drop until ultimately VDS < VGS – VT and the transistor comes out of the saturation region to operate in non-saturation region. Ultimately the input voltage reaches an utmost of VDD and the output reaches its minimum value of VOL as formerly computed.
Figure: Voltage Transfer Characteristic of Simple MOS Transistor Inverter
Critical Logic Voltages:The similar critical input and output logic voltages can be stated as for other logic families viz.:ViLMAX = maximum voltage acceptable as the logic LO input
ViHMIN = minimum voltage acceptable as the logic HI input.
VOLMAX = maximum voltage acceptable as the logic LO output.
VOHMIN = minimum voltage acceptable as the logic HI output.a) Critical Point ViL MAX, VOH 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:ID = Kn(VGS - VT)2However as VO = VDS and Vi = VGS and VO = VDD – iDRD then:Vo = VDD – KnRD (Vi - VT)2.......................... (a)On expanding it gives:Vo = VDD - KnRDVi2 + 2 KnRDViVT - KnRDVT2On differentiating:∂Vo/∂Vi = - 2 KnRDVi + 2 KnRDVTAt critical point ∂Vo/∂Vi = -1 with Vi = ViL MAX and VO = VOH MIN and hence:- 2 KnRDVi + 2 KnRDVT = - 12 KnRDViLMAX = 1 + 2 KnRDVTAnd hence,ViLMAX = VT + 1/(2 KnRD)This value is a slight higher than VT and for illustration given with VT = 1V, RD = 100kΩ and Kn = 100µAV-2, ViL MAX = 1.05V.Replacing back into equation (a) to find out the output voltage for this coordinate gives:VOHMIN = VDD – KnRD(ViL MAX - VT)2VOHMIN = VDD – KnRD[VT + (1/ KnRD) - VT]2And hence ultimately:VOHMIN = VDD – (1/4 KnRD)This value is a slight lower than VDD and for illustration given with VDD = 10V, VT = 1V, RD = 100kΩ and Kn = 100µAV-2, VOH MIN = 9.98V. The coordinate of critical point (a) is then:ViLMAX, VOHMIN = 1.05, 9.98 Vb) Critical Point ViH MIN, VOL MAXThis 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:ID = Kn[2(VGS - VT)VDS – V2DS]However again, as VO = VDS and Vi = VGS and VO = VDD – iDRD then:VO = VDD – 2 KnRD (Vi - VT) Vo + KnRDVo2On expanding:VO = VDD – 2 KnRDViVo + 2 KnRDVTVo + KnRDVo2On rearranging:VO [1 - KnRDVT] = VDD - 2 KnRDViVo + KnRDVo2There is a choice here to employ implicit differentiation to find ∂Vo/∂Vi or to re-arrange the expression as Vi in terms of VO and then determine ∂Vi/∂Vo. The latter is simpler as there is just one term in Vi. Then,2 KnRDViVo = VDD – [1 - 2 KnRDVT]Vo + KnRDVo2And hence,Vi = VDD/(2 KnRDVo) – [(1 - 2 KnRDVT)/ 2 KnRD) + (Vo/2)............... (b)Then,∂Vo/∂Vi = - VDD/(2 KnRDVo2) + (1/2)For ∂Vo/∂Vi = -1 we can employ ∂Vi/∂Vo= -1 and hence:- (VDD/2 KnRDVo2) + (1/2) = - 1(VDD/2 KnRDVo2) = 3/2Vo2 = VDD/3 KnRDBy taking the positive root as practical value gives:VOLMAX = √VDD/3KnRDThat for the illustration given with VDD = 10V, VT = 1V, RD = 100kΩ and Kn = 100µAV-2 , VOL MAX = 0.58V.This is significantly higher than the great value of VOL computed formerly. Then replacing this back into the expression for Vi in equation (b) above gives:
And therefore ultimately the critical input value is as shown:ViHMIN = VT + 2√VDD/3KnRD – (1/2 KnRD)That for illustration given with VDD = 10V, VT = 1V, RD = 100kΩ and Kn = 100µAV-2 provides ViH MIN = 2.1V. This gives the coordinates of critical point (b) as:ViHMIN, VOLMAX = 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:NMH = VOHMIN – ViHMIN = 9.98 – 2.1 = 7.88VNML = ViLMAX – VOLMAX = 1.05 – 0.58 = 0.47V
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