notes, M. Rodwell, copyrighted
ECE145A / 218A Notes : Basic Analysis of Analog Circuits Mark Rodwell University of California, Santa Barbara
[email protected] 805-893-3244, 805-893-3262 fax
Comment
notes, M. Rodwell, copyrighted
This (2009) is a transitional year: Next year 145abc will be reorganized, reorganized 145a: fundamentals (devices, analog & RF analysis, models) 145cb: RF systems at IC and system level This year: some students have taken 145c: already l d have h device d i models d l already know analog circuit analysis well some students have not must cover device models must review some circuit analysis methods These notes: shortened version (2009 only) of device models
Transistor Circuit Design
notes, M. Rodwell, copyrighted
This note set -reviews the basics -starts at the level of a first IC design course -moves very quickly
This will -establish establish a common terminology -accommodate capable students having minimal background in ICs.
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DC models d l DC bi bias analysis l i
Large-Signal Model For Bias Analysis Ib
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Ic
+ Vbe
-
Provided that Vce > 0, I c = I s exp(Vbe / VT ) and I b = I c / β , where VT = kT / q ...note that Vbe is specified internal to the emitter resistance Rex Th I e Rex drop The d is i significan i ifi t for f HBTs HBT operating ti att currentt densities d iti near that required for peak transistor bandwidth.
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DC Bias Example: Current Mirror
I ref
Ic2
Q2
Q1 I b 2 I b1
Rex 2
We have Vbe b 1 + I e1 ( Rex1 + Ree1 ) = Vbe b 2 + I e 2 ( Rex 2 + Ree 2 ) and Vbe1 = Vt ln( I c1 / I s1 ), Vbe 2 = Vt ln( I c 2 / I s 2 ) Assume that β >> 1, Ree 2 = 2 Ree1 & assume that AE1 = AE 2 ( AE is the emitter area). area) This implies Rex1 = Rex 2 / 2 , and I s1 = 2 I s 2 , from which we find I c 2 = I c1 / 2
Ree 2
Ie2
I e1
Rex1 Ree1
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Simpler DC Model for Bias Analysis Ib
Ib
Ic Vbe
Ic
Vbe
It is often sufficient in bias analysis to ignore the variation of Vbe b with I c and instead take Vbe b = Vbe b ,on = φ . Vbe,on depends d d upon currentt density d it and d technolog t h l y. Biased at current densities within ~ 10% of peak bandwidth bias, Vbe,on
⎧ 0.9 V Modern Si/SiGe HBTs ⎪ = φ ~ ⎨0.7 or 0.9 V InGaAs/InP HBTs ⎪ 1.4 V GaAs/GaInP HBTs ⎩
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Simple DC Bias Example Rc Ib2
Rb Q2
Ib11
Q1 Ree
-Vee
If we neglect the I b Rb drops, then Vb1 = Vb 2 = 0 Volts. Approximate Ve1 = Ve 2 = −φ ≅ −0.9 V (SiGe). I c1 + I c 2 = 2 I c1 = (−Vee − 0.9V ) / Ree I c1 = I c 2 = (−Vee − 0.9V ) / 2 Ree
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Efficiently Handling Base Currents In Bias Analysis If I b Rbdrop is singificant, one can solve simultaneous equations :
Rc Ib2
Rb
I c1 + I c 2 = 2 I c1 = ( −Vee − φ − I b1Rb ) / Ree where I b1 = I c1 / β ,
Q2
Q1 Ree
Q i k : find Quicker fi d by b iteration it ti : 1) solve I c1 = ( −Vee − φ ) / 2 Ree 2) solve I b1 ≅ I c1 / β
-Vee
3) use this value of I b to solve I c1 = ( −Vee − φ − I b1Rb ) / 2 Ree Works because any well - designed circuit has DC bias only weakly dependent upon β .
Ib1
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small-signal ll i l b baseband b d analysis l i
Hybrid-π Bipolar Transistor Model
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Ccbx B
Rbb
Ccbi
Rc
C
Rbe = β / g m
τ f = τb + τ c
Rbe Cbe,diff gmτ f Cje b diff =g
gm Vbee -jωτc
Vbe Rex E
Accurate model, but too detailed for quick hand analysis
Oversimplified Model for Quick Hand Analysis Cbe Rbe Ccb g mVbe B
Vbe
C
+
Rbe B
− E
g mVbe C
+
Vbe
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− E
In most high high-frequency frequency circuits circuits, the node impedance is low and Rce is therefore negligible. Neglecting Rbb in high-frequency high frequency analysis is a poor approximation but is nevertheless common in introductory treatments.
The "textbook" analyses which follow use this oversimplified model. These introductoryy treatments will later be refined.
notes, M. Rodwell, copyrighted
Common Emitter Stage: Basics 2) δVC = − RLeqq ⋅ δI C
RLeq
7) Rin ,T = δVb / δI b = β (re + Re ) 5) δI b = δI c / β 1) δI e ≅ δI c
+
4) δVbe = δI e / g m
−
3) δVe = RE ⋅ δI e 6) δVb = δI e ⋅ (re + Re ) = δI b ⋅ β (re + Re )
RE
8) δVout / δVin = δVout / δVB = − RLeq /(re + Re ) = − RLeq /( Re + 1 / g m ) Gain is - RLeq /( Re + 1 / g m ) ; Transistor Rin is β (re + RE )
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Emitter Follower Stage: Basics 6) Rin ,T = δVb / δI b = β (re + RLeq ) 4) δI b = δI c / β
1) δI e ≅ δI c
+
3) δVbe = δI e / g m
−
2) δVe = RLeq ⋅ δI e 5) δVb = δI e ⋅ (re + RLeqq ) = δI b ⋅ β (re + RLeqq )
RLeq
7) δVout / δVin = δVout / δVE = RLeq /(re + RLeq ) = RLeq /( RLeq + 1 / g m )
Gain is RLeq /( RLeq + 1 / g m ) ; Transistor Rin is β (re + RE )
notes, M. Rodwell, copyrighted
Common-Base Stage: Basics
2) δVout = RLeq ⋅ δI C
6) δVin = δI e ⋅ (re + Rb / β ) 7) Rin ,T = δVe / δI e = re + RB / β +
5) δVbe = δI e / g m − RB
1) δI e ≅ δI c RLeq
4) δVb = δI c ⋅ Rb / β 3) δI b = δI c / β 7) δVout / δVin = RLeq /(re + Rb / β ) Gain is RLeq /(re + Rb / β ) ; Transistor Rin is re + Rb / β
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Emitter Follower Output Impedance
ECE145C /218C notes, M. Rodwell, copyrighted
Common-Base Stage: Basics
2) δVout = RLeq ⋅ δI C
6) δVin = δI e ⋅ (re + Rb / β )
Rout ,emitter RB
Rout ,amp
7) Rin ,T = δVe / δI e = re + RB / β +
5) δVbe = δI e / g m − RB
4) δVb = δI c ⋅ Rb / β
1) δI e ≅ δI c RLeq
3) δI b = δI c / β
7) δVout / δVin = RLeq /(re + Rb / β )
REE
Gain is RLeq /(re + Rb / β ) ; Transistor Rin is re + Rb / β
E.F. output impedance is same problem as C.B. input impedance Rout ,emitter = re + RB / β = 1 / g m + RB / β
Rout ,amp = Rout ,emitter REE
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Including Bias Circuit Resistances
Rin , Amp = RB1 RB 2 RinT
Rgen
Vgen
RB1
RLeq = RC RL
RC
Rin ,T
Vin
RB1
Th are (trivially These (t i i ll ) added dd d in i parallel ll l with ith the th transisto t it r terminal impedances to determine the net circuit impedances. From which, Vin / Vgen = Rin ,amp /( Rin ,amp + Rgen ) , etc.
RL
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Baseband Analysis Of Multistage Circuits
For baseband analysis of multi-stage circuits, simply break into individual stages. Q1
Q2
Q3
Load impedance of the Nth stage includes the input impedance of the (N+1)th stage
Q4
Analysis is then trivial... trivial
Rin3
Rin2 Vin1
Q3
Q1 Vout1
Vout3 Q4
Q2 Rin3
Rin2
Rin4
Vout2
Vin2=V Vout1
Vout2=Vin3 Rin4
Vout3=Vin4
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small-signal ll i l b baseband b d analysis l i
High-Frequency Analysis: The General Problem Ccbx = 6.4 fF B
Rc = 6500 Ω Ccbi = 3.7 fF
Rbb = 49 Ω Vb'e
Rπ = 350 Ω
C
g mVb 'e
Cdiff = C je = 182 fF 38 fF
Rex = 4.3 Ω E B
Ccbx = 6.4 fF
Q2
Cdiff = C je = 182 fF 38 fF
Rc = 6500 Ω Ccbi = 3.7 fF
Rbb = 49 Ω Vb'e
Q1
notes, M. Rodwell, copyrighted
Rπ = 350 Ω
C
g mVb 'e Rex = 4.3 Ω
Vgen E
Analyzing frequency response is difficult: cannot separate stage-by-stage y accurate,, ggeneral,, tedious. Method #1: nodal analysis: Method #2: method of time constants: accurate, limited applicabilty, quick & intuitive
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N d l Analysis Nodal A l i
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Tutorial: Transfer Function Analysis: Nodal Analysis I Simple & very familiar example : common - emitter amplifier. Vout
Vin Rgen Vgen Rb
Vin
C L RL
B
Ccb
C
Vout
Rgen Vgen
E
Rb
Cbe Rbe
gmVbe
CL RL
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Tutorial: Transfer Function Analysis: Nodal Analysis II Reduced circuit :
Vin
Vout
( Ri = Rgen || Rbe || Rb ) Ii=Vgen/Rgen Cbe Ri
gmVbe CL RL
Step 1 : Write Nodal Equations from KCL
⎡Gi + sCbe + sCcb ⎢ g − sC m cb ⎣
− sCcb
⎤ ⎡ Vin ⎤ ⎡ I i ⎤ =⎢ ⎥ ⎥ ⎢ ⎥ GL + sCL + sCcb ⎦ ⎣Vout ⎦ ⎣ 0 ⎦
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Tutorial: Transfer Function Analysis: Nodal Analysis III Step 2 : Solve Nodal Equations : Vout / I in = N ( s ) / D ( s ) Gi + sCbe + sCcb N ( s) = g m − sCcb
1 = −( g m − sCcb ) 0
Gi + sCbe + sCcb D( s) = g m − sCcb
− sCcb GL + sCL + sCcb
D ( s ) = (Gi + sCbe + sCcb )(GL + sCL + sCcb ) − (g m − sCcb )(− sCcb ) Step 3 : Organize in powers of s D ( s ) = GiGL + s(GiCL + GiCcb + GLCbe + GLCcb + g mCcb + s 2 (CbeCL + CbeCcb + CcbCL + CcbCcb − CcbCcb )
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Tutorial: Transfer Function Analysis: Nodal Analysis IV Step 4 : Separate into dimensionless ratio - of - polynomials form, separating p g constants and g gains from the transfer function... Vout Vout N ( s) = = I in Vgen / Rgen D ( s ) − ( g m − sCcb ) = ⎛ GiGL + s(GiCL + GiCcbb + GLCbe b + GLCcb b + g mCcb b⎞ ⎜ ⎟ 2 ⎜ ⎟ + s ( C C + C C + C C ) be L be cb cb L ⎝ ⎠ − ( g m − sCcb ) Ri RL / Rggen Vout = Vgen ⎛1 + s( RLCL + RLCcb + RiCbe + RiCcb + g m Ri RLCcb ⎞ ⎟ ⎜ 2 ⎟ ⎜ + s ( C C + C C + C C ) R R be L be cb cb L i L ⎝ ⎠
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Tutorial: Transfer Function Analysis: Nodal Analysis V
Rin , Amp ( Rbe || Rb || Rgen ) Ri ( Rbe || Rb ) note that = = = Rggen Rggen ( Rbe || Rb ) + Rggen Rin , Ampp + Rggen so... ⎛ Rin , Amp ⎞ Vout ⎟(− g m RL ) =⎜ Vgen ⎜⎝ Rin , Amp + Rgen ⎟⎠ b1 − (1 − sC Ccb / g m ) × ⎛1 + s( RLCL + RLCcb + RiCbe + RiCcb + g m Ri RLCcb ) ⎞ ⎜ ⎟ 2 ⎜ ⎟ + s ( C C + C C + C C ) R R be L be cb cb L i L ⎝ ⎠
Vout ( s ) Vout ⇒ = Vgen ( s ) Vgen
mid − band
1 + b1s 1 + a1s + a2 s 2
a1
a2
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Tutorial: Transfer Function Analysis: Nodal Analysis VI Step 5 : Find the roots (poles & zeros) of the polynomial Vout ( s ) Vout = Vgen ( s ) Vgen
mid −band
1 + b1s Vout = 2 1 + a1s + a2 s Vgen
mid − band
1 + b1s (1 − s / s p1 )(1 − s / s p 2 )
what are efficient methods of finding the poles ?
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Finding Fi di Poles P l f from T Transfer f Functions F ti
Finding Poles and Zeros Ratio - of - Polynomial Form : Vout ( s ) Vout = Vgen ( s ) Vgen
at mid- band
2 1 + b s + b s + ... m 1 2 *s 1 + a1s + a2 s 2 + ...
Poles and Zeros : Vout ( s ) Vout = Vgen ( s ) Vgen
(1 − s / sz1 )(1 − s / sz1 )(1 − s / sz1 )... *s (1 − s / s p1 )(1 − s / s p1 )(1 − s / s p1 ))... m
at mid- band
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Finding Poles: Complex Poles Vout 1 + b1s + b2 s 2 + ... =k Vgen 1 + a1s + a2 s 2 + a3s 3 If a3 / a2
