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Download DC Characteristics of a CMOS Inverter
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Transcript
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 the traces below. IV Curves for nMOS PMOS IV Curves 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 xaxis 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: I dsn n 2 Vun Vtn 2 CMOS Inverter DC Characteristics CMOS Inverter Transfer Characteristics • In region B Idsp is governed by voltages Vgs and Vds described by: V gs Vin VDD and Vds Vout VDD V VDD I dsp p Vin VDD Vtp Vout VDD out 2 2 Recall that : I dsn I dsp • n 2 Vin Vtn 2 p Vin VDD Vtp Vout VDD Vout VDD 2 2 • Region D is defined by the inequality VDD Vin VDD Vtp 2 • p-device is in saturation while ndevice is in its non-saturation region. p Vin VDD Vtp 2 ; Vin Vtp VDD I dsp 2 AND Region C has that both n- and p2 devices are in saturation. Vout I dsn n Vin Vtn Vout ; Vin Vtn • Saturation currents for the two 2 devices are: • Equating the drain currents allows us p to solve for Vout. (See supplemental Vin VDD Vtp 2 ; Vin Vtp VDD I dsp 2 notes for algebraic manipulations). AND 2 I dsn n Vin Vtn ; Vin Vtn 2 CMOS Inverter Static Charateristics Vin VDD Vtp • • • • • 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 A VD Vout =Vin-Vtp B D Output Voltage • Vout =Vin-Vtn Both in sat nMOS in sat pMOS in sat C D 0 Vtp Vtn E VDD/2 VDD+Vt VD p D