Lecture 4_ Single Device Amplifiers-1.ppt

Lecture 4_ Single Device Amplifiers-1.ppt - V1 Inp Out 1...

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James Morizio 1 Single Stage Amplifiers Inp Out V 1 1) Common Source Amplifiers 2) Inverting Amplifiers 3) Source Follower Amplifiers

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James Morizio 2 Amplifier Analysis Concepts 1) Large Signal Analysis for Large Signal Voltage Transfer and Vout Swing Limits Set Vin to Vss and Vdd limits Assume Saturation and Linear Regions pending limits Equate Currents and solve for Vout 2) Small Signal AC Analysis for Voltage Gain, Input/Output Resistances, BW Frequency Response Use Small Signal Model of Mosfets Assume all dc bias voltages, power supplies are gnd Solve Voltage Gain for Vout / Vin Solve BW using Network analysis
James Morizio 3 Common Source Amplifiers (NMOS) V out V in NW, NL, M gnd R load V dd V in NW, NL, M gnd V dd PW, PL, M V out V in NW, NL, M gnd V dd PW, PL, M V out R Ref PW, PL, M Common Source  Resistor Load  Common Source  Active Load  Common Source  Current Source Load  Assume all devices are in saturation!

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James Morizio 4 Common Source Amplifiers (PMOS) V out V in gnd V dd V in NW, NL, M gnd V dd PW, PL, M V out V in NW, NL, M gnd V dd PW, PL, M V out R Ref Common Source  Resistor Load  Common Source  Active Load  Common Source  Current Source Load  R load PW, PL, M NW, NL, M Assume all devices are in saturation!
James Morizio 5 Common Source with Resistor Load V out V in NW, NL, M R L V out = V dd – (R load * I ds ) Where I ds = u n C ox W(V in -V th ) 2 V out > V in -V t 2L or I ds = u n C ox W[(V in -V th )V ds - V ds 2 ] V out < V in -V t L 2 Large Signal Characteristics I ds Small Signal Model when M1 is in saturation v in + - g m V in r ds + - V in v out Derive voltage gain A v = v out / v in v out = - g m v gs * r ds || R L v out = - g m v in * r ds || R L A v = - g m * (r ds || R L ) A v = - g m * r out R L V dd = 5v v in + - g m V gs1 r ds + - V gs1 R L v out V out (max) when Vin = 0v, I ds = 0 V out (max) = 5v V out (min) when Vin = 5v, V out (min) = 5 I ds max * R L = 5 - u n C ox W[ (5-V th )V ds - V ds 2 ]R L L 2

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James Morizio Common Source with Active Load  Swing Limit Vout Maximum (Vin = 0v)  V in NW = 15um NL = 3um 5v PW = 15um PL = 3um V out M2 M1 I2 I1 6
James Morizio Common Source with Active Load  Swing Limit Vout Minimum (Vin = Vdd) V in NW = 15um NL = 3um 5v PW = 15um PL = 3um V out M2 M1 I2 I1 M2 Saturation, M1 linear Solving for I1 = I2 7

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James Morizio 8 Common Source with Active Load  V in NW = 15um NL = 3um 5v PW = 15um PL = 3um V out M2 M1
James Morizio Common Source with Active Load   Output Interpretation

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