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741 Op-Amp Where we are going:
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Source Degeneration Why do this? Higher Linearity Possible Stability
GND Vout Vin Circuit Element Vout Vin Why not do this? gm Lower Bandwidth Higher Noise / Df GND
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Source Degeneration I I = Ieo e V1 /UT = Ieo e(Vin - V1 + Vout/Av )/UT
Neglect VA of Q1 and assume matched devices: Vout I Vin I = Ieo e V1 /UT = Ieo e(Vin - V1 + Vout/Av )/UT Vin V1 2 V1 = Vin + Vout / Av GND Q1 I = Ieo e(Vin + Vout/Av )/(2 UT) GND A similar result for MOSFETs
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Common Emitter Ibias Ibias = Ico eVin/UT eVout /VA Vout Vin
Common Emitter / Common Source Vdd Amplifies the input signal at the output Ibias 100mA Assuming an ideal current source: Vout Ibias = Ico eVin/UT eVout /VA Vin Vout = -VA ln(Ibias/Ico) + - (k VA / UT) Vin GND
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Common Drain Ibias Ibias = Ibias ekDVin/UT eDVout/VA
Vdd Amplifies the input signal at the output 100pA Ibias Ibias = Ibias ekDVin/UT eDVout/VA Vout Vin DVout = - (k VA / UT) DVin GND Input conductance = 0
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Common Drain Id = Ibias e-DVout/VAp = Ibias ekDVin/UT eDVout/VAn
We must account for the other current source: Vdd Vb Ibias Id = Ibias e-DVout/VAp = Ibias ekDVin/UT eDVout/VAn M6 Vout Vin M7 DVout = - (k (VAn // VAp) UT) DVin GND
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Common-Drain: Amplifier Measurements
Vdd V1 M6 Ibias Vout Mb M7 GND GND
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Common Drain Ibias Ibias = (K/2) ( Vin - VT )2 (1 + (Vout/VA) )
What about above-threshold operation: Vdd Operating region decreases (Vout > Vin - VT) Derive using quadratic functions: 100mA Ibias Vout Ibias = (K/2) ( Vin - VT )2 (1 + (Vout/VA) ) Vin GND Vout = VA( ) Amplifies the input signal at the output
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Common E / S: Resistive Load
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High-Gain Amplifier Experiments
Load-line Analysis
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Common Base Ibias Ibias = Ico e (Vb -Vin )/UT eVout /VA
Common Base / Common Gate Vdd Amplifies the input signal at the output (non-inverting gain) Ibias 100mA Assuming an ideal current source: Vout Ibias = Ico e (Vb -Vin )/UT eVout /VA Vb Vout = -VA ln(Ibias/Ico) + (VA / UT) Vin - (VA / UT) Vb Vin Gain = VA / UT = Av
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Common Gate Ibias = Io e (kVb -Vin )/UT eVout /VA Ibias
Vdd Using a subthreshold MOSFET : 100pA Ibias = Io e (kVb -Vin )/UT eVout /VA Ibias Vout = -VA ln(Ibias/Io) + (VA / UT) Vin - (k VA / UT) Vb Vout Vb Gain = VA / UT = Av Vin Problem: Large input current
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Common G: Resistive Load
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Cascode Circuits Use a common-gate/base transistor to:
1. Improve the output resistance of another transistor. 2. Reduce the Gate-to-Drain capacitance effect of another transistor. Vdrain Vin GND V1 Vgate Input resistance of common-gate is low Source is nearly fixed if connected to the drain of a transistor
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Cascode Circuits Vdrain Vdrain Vbias Vgate GND V1 Vgate GND Idrain = Io e kVgate/UT e kVbias /VA eVdrain / (Av VA ) Idrain = Io e (kVbias -V1 )/UT eVdrain /VA = Io e kVgate/UT eV1 /VA V1 ~ kVbias - kVgate + (UT/VA) Vdrain Drain is fixed Fixes the voltage at V1 or isolates V1 from the output
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Cascode Common-Drain Amp
Vdd One Pole V1 High Output Resistance / DC Gain biasp Vout Ibias biasn Vb Mb GND GND
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BJT Cascode Configuration
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MOS Cascode Circuit
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BJT - CMOS Cascode Circuits
Preserve High-gm/I
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Cascade Configurations
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Cascade Connection: Rout
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BJT-MOS Cascades A good way to get zero base current….
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Cascades: More stuff
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