EE105 - Fall 2006 Microelectronic Devices and Circuitsee105/fa06/lectures/F06...1 EE105 - Fall 2006...
Transcript of EE105 - Fall 2006 Microelectronic Devices and Circuitsee105/fa06/lectures/F06...1 EE105 - Fall 2006...
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EE105 - Fall 2006Microelectronic Devices and Circuits
Prof. Jan M. Rabaey (jan@eecs)
Lecture 26: Bipolar Junction TransistorAmplifiers – Frequency Response
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Lecture Material
Last lecture– BJT amplifiers
This lecture– Emitter-degenerated CE Amplifier
– Frequency response of BJT Amplifiers
– Designing complex amplifiers
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Administrativia
Lab 8 this week
Lab 9 next week (no written report – just oral at end of lab session)
Last homework will be posted this weekNo lecture on Tu Dec 5 – Make-up lecture scheduled tentatively for December 11 at 2pm (203 McLaughlin)
Final: We December 13, 12:30-3:30pm, 105 North Gate
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Schedule for the rest of the semester
Tu 11/27: BJT Amplifiers (continued) – Complex amplifiers (start)
Th 11/29: Complex amplifiers; Advanced topics (start)
Th Dec 7: Advanced topics (cntd)
Mo Dec 11: Semester overview and review session
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One Last Exercise: Common-Collector Voltage Gain
KCL at the output node: note vπ = vt - vout
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Summary of Two-Port Parameters for CE/CS, CB/CG, CC/CD
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Frequency Response of Common-Emitter
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Frequency Response of Common-Emitter
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Assuming Rsig to be small
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BJT Biasing (Discrete)
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BJT Biasing (2)
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Common-Emitter with Emitter Degeneration
1. What is it?2. DC Bias3. Small-signal 2-port
model4. Output swing
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DC Bias for CEdeg
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CE(degen) Biasing
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Two-Port Model for CEdeg
Input looks like CC Rin =Output looks like CB (see p. 504 for details) Rout = Transconductance: Gm =
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Two-Port Model for CEdeg (cont.)
Find Gm
Voltage Gain:
Is it a good voltage amplifier (vs. CE)?
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Temperature Dependence
Gm of Bipolar strong function of temperature
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Analyzing Complex Amplifiers
Example: Discrete VHF Amplifier
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Multi-Stage Voltage Amplifier
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Cutting Through the Complexity
Two Approaches:
1. Eliminate “background” transistors to reduceclutter
2. Identify the “signal path” between the inputand output
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First Approach: Find I & V Sources
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What’s Left?
Voltage at base ofQ2 is set by totempole
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Second Approach: Find Signal Path
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Identifying the Stages
First stage (or two stages): CS/CB cascodeSecond stage (or two stages): CD/CC voltage buffer
Why does this make sense for a voltage amplifier?
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Find Key Two-Port Parameters
Output resistance of cascode:
( ){ }2222/, ||1(|| SmoocCBCSout RrgrrR π+=
( )666 1 Smoupoc RgrRr +==
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Two-Port Parameters (Cont.)
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Output Resistance and Voltage Gain
Source resistance of the CC stage is the output resistanceof the CD stage (small)
434
,
4,
1111
mommo
CCS
mCCoutout ggg
R
gRR ≈
β+=
β+==
Open-circuit voltage gain Av (last two stages have nearly unity gain):
( )( )76621 1|| omooomv rgrrgA +β−=
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Output Swing: VOUT,MIN
Minimum output voltage: M10, M3 , and Q2 are “suspects”
M10 goes into triode when VOUT = 0.5 V
M3 goes into triode when VSD3 = 0.5 V VOUT = 0.5 V – 0.7 V = -0.2 V
Q2 goes into saturation when VCE2 = 0.1 Vor VBC2 = 0.6 VVOUT = VB2 – VBC2 + VSG3 – VBE4
= 2 V – 0.6 V + 1.5 V – 0.7 VVOUT = 2.2 V
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Output Swing: VOUT,MAX
Maximum output voltage: Q4, M5, and M6 are “suspects”
Q4 goes into saturation when VCE4 = 0.1 V VOUT = 4. 9 V
M5 goes triode when VSD5 = 0.5 V VOUT = 3.8 V
M6 goes triode when VSD6 = 0.5 V VOUT = VS6 – 0.5 V + VSG3 – VBE4
= 3.5 – 0.5 + 1.5 – 0.7 V = 3.8 V
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Insight into the Frequency Response
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Qualitative Insight
Could always do “brute force” open-circuit time constants
CS*-CB is a wideband stage … so is the CD-CC buffer
Look for large RTxCx products: high-impedance nodesare likely candidates
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Node X
“High impedance node” is node X … look at RTxCx
Capacitance:
Cx = Cgd6 + Cμ2 + Cgd3 + CM3
Miller for CDstage (M3)
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Finding the Miller Capacitance CM3
Gain across Cgs3: 33
33 /1 Lm
LgsvC Rg
RA
+=
CD
X
Cgs3
RL3
RL3 = Rin4 =
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Dominant Pole of Voltage Amplifier
CDinCBoutinoutTx RRRRR ,,32 |||| ==
ooomoSmoocTx rrgrRrgrrR β+≅+= π 27662222 ||)1())||(1(||
Thévenin resistance for CX:
Dominant pole: xTxCR≈ω −11