ece 255 l37 review for final exam - nanohub... · review for final exam mark lundstrom school of...
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ECE 255: L37
Review for Final Exam
Mark Lundstrom School of ECE
Purdue University West Lafayette, IN USA
Spring 2019 Purdue University
Lundstrom: 2019
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Announcements
2 Lundstrom: 2019
HW11 Due 5:00 PM Friday, April 26 in EE-209 dropbox (note submission sheet)
Final Exam: Thursday, May 2, 7:00 – 9:00 PM, CL50
Please complete the online course evaluation
No class on Friday, April 26!
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About the final exam
The final exam is comprehensive, but will emphasize topics since Exam 3
1) Multi-stage amplifiers
2) Differential amplifiers: -BJT diff amp with RC -MOS diff amp with RD -MOS diff amp with active load
3) Op Amps
-General definitions / characteristics -2-stage MOS op amp
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About the final exam
4) Low-frequency response -Short-circuit time constant method -Finding Rth for each C -Dominant pole -Overall fL: combining poles (adding frequencies)
5) High-frequency response -Hybrid-pi transistor models -Open-circuit time constant method -Finding Rth for each C -Dominant pole -Overall fH: combining poles (adding time
constants) -Miller effect
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Review problem 1a)
RSS = 100 k
+VDD
RD = 5 k
+υid 2M1 M2
υo2
RD = 5 k
RS RS
−υid 2
RS = 0.5 k
gm = 2 mS
Find the gain: Adm =
υo2
υid
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Review problem 1a)
RS = 0.5 k
gm = 2 mS
υo2 = −
gmRD
1+ gmRS
−υid 2( )
υo2
υid
= 12
gmRD
1+ gmRS
Adm = 1
22×5
1+ 2× 0.5⎛⎝⎜
⎞⎠⎟= 10
4
From the half-circuit:
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Review problem 1b)
RSS = 100 k
+VDD
RD = 5 k
+υic
M1 M2
υo2
RD = 5 k
RS RS
+υic
RS = 0.5 k
gm = 2 mS
Find the common mode gain:
Acm =
υo2
υic
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Review problem 1b)
RS = 0.5 k
gm = 2 mS
υo2 = −
gmRD
1+ gm RS + 2RSS( ) υic( )
From the half-circuit:
RSS = 100 k
Acm =
υo2
υic
= −gmRD
1+ gm RS + 2RSS( )
Acm = − 2×5
1+ 2 0.5+ 200( ) =10402
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Review problem 1c)
RSS = 100 k
+VDD
RD = 5 k
+υid 2M1 M2
υo2
RD = 5 k
RS RS
−υid 2
RS = 0.5 k
gm = 2 mS
Find the common mode rejection ratio:
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Review problem 1c)
CMRR =
Adm
Acm
= 10 410 402
= 4024
≈100
CMRR = 40 dB
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Review problem 2)
IO
−VSS
+VDD
M1 M2
M3 M4
−υid 2 +υid 2 υo
ro = 100 k
gm = 2 mS
Find the gain: Adm =
υo
υid
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Review problem 2)
ro = 100 k
gm = 2 mS Adm =
υo
υid
= +gm ro4 || r02( )
Adm = +2 50( ) = +100
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Review problem 3)
VDD
2 k 200 k
1k
Rsig = 1k
υs
CC1 = 0.4 µF
CC2 = 0.4 µF
CE = 4 µF
RL = 2 k
120 k
rπ = 1k
gm = 77 mS
Find the dominant pole and estimate fL
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Review problem 3)
rπ = 1k
gm = 77 mS
ω L =
1RthC
ω L =1
RthCE
= 1
1k ||rπ + 200 ||120 ||1
β +1⎡
⎣⎢
⎤
⎦⎥
⎛
⎝⎜⎞
⎠⎟CE
ω L ≈1
1k || 1+178
⎡⎣⎢
⎤⎦⎥
⎛⎝⎜
⎞⎠⎟
CE ω L ≈
11k || 0.026 k( )CE
ω L ≈
126( )× 4×10−6 ≈104
fL ≈
104
2π= 1600 Hz
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Review problem 3)
rπ = 1k
gm = 77 mS ωCE
≈104
Let’s check the other two poles:
ωC1
= 11k + 200 k ||120 k ||1 k( )⎡⎣ ⎤⎦ ×10−6
ωC1
≈ 12 k⎡⎣ ⎤⎦ ×10−6 = 500
ωC2
= 12 k + 2 k⎡⎣ ⎤⎦ ×10−6 = 250
So CE does produce the dominant pole
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Review problem 4)
VDD
50 k 2 MΩ
RS
1kυs
CC1
CC2
CS
50 k
Cgd = 0.1 pF
Cgs = 0.5 pF
CL
1 MΩ
gm = 2 mS CL = 0.05 pF
Estimate fH
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Review problem 4)
VDD
50 k 2 MΩ
1kυs
CC1
CC2
50 k
Cgd = 0.1 pF
Cgs = 0.5 pF
CL
1 MΩ
gm = 2 mS CL = 0.05 pF
Estimate fH
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Review problem 4)
VDD
50 k 2 MΩ
1kυs
50 k
Cgd = 0.1 pF
Cgs = 0.5 pF
CL
1 MΩ
gm = 2 mS CL = 0.05 pF
Estimate fH
Cgd
Cgs
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Review problem 4)
VDD
50 k 2 MΩ
1kυs
50 k
Cgd = 0.1 pF
Cgs = 0.5 pF
1 MΩ
gm = 2 mS CL = 0.05 pF
Cgd
ω H = 1
RthC
Expected dominant pole
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Review problem 4)
Cgd = 0.1 pF
Cgs = 0.5 pF
gm = 2 mS CL = 0.1 pF
Miller capacitance will probably dominate
τ gd = 1k || 2M ||1M( )⎡⎣ ⎤⎦CM + 50k ||50 k⎡⎣ ⎤⎦Cgd
CM = 1+ Aυ
⎡⎣ ⎤⎦Cgd = 1+ gm 50k ||50 k( )⎡⎣ ⎤⎦Cgd
τ gd = 1000⎡⎣ ⎤⎦5.1×10−12 + 25000× 0.1×10−12 = 7.6×10−9
CM = 51⎡⎣ ⎤⎦0.1×10−12 = 5.1×10−12
ω H = 1
τ gd
= 1.3×108
fH = 1
τ gd
= 21×106 Hz