Transcript Lecture 13

Lecture 13
ANNOUNCEMENTS
• Midterm #1 (Thursday 10/11, 3:30PM-5:00PM) location:
• 106 Stanley Hall: Students with last names starting with A-L
• 306 Soda Hall: Students with last names starting with M-Z
• EECS Dept. policy re: academic dishonesty will be strictly followed!
• HW#7 is posted online.
OUTLINE
• Cascode Stage: final comments
• Frequency Response
–
–
–
–
General considerations
High-frequency BJT model
Miller’s Theorem
Frequency response of CE stage
Reading: Chapter 11.1-11.3
EE105 Fall 2007
Lecture 13, Slide 1
Prof. Liu, UC Berkeley
Cascoding Cascode?
• Recall that the output impedance seen looking into the
collector of a BJT can be boosted by as much as a factor
of b, by using a BJT for emitter degeneration.
Rout  [1  g m1 (ro 2 || r 1 )]ro1  ro 2 || r 1
Rout  g m1ro1 ro 2 || r 1   bro1
• If an extra BJT is used in the cascode configuration, the
maximum output impedance remains bro1.
Rout  [1  g m1 ( bro 2 || r 1 )]ro1  bro 2 || r 1
Rout,max  g m1ro1r 1  bro1
EE105 Fall 2007
Lecture 13, Slide 2
Prof. Liu, UC Berkeley
Cascode Amplifier
• Recall that voltage gain of a cascode amplifier is high,
because Rout is high.
Av   g m1ro1 g m2 ro1 r 2 
• If the input is applied to the base of Q2 rather than the
base of Q1, however, the voltage gain is not as high.
– The resulting circuit is a CE amplifier with emitter degeneration,
which has lower Gm.
io
gm2
Gm 

vin 1  g m 2 ro1 ro 2 
EE105 Fall 2007
Lecture 13, Slide 3
Prof. Liu, UC Berkeley
Review: Sinusoidal Analysis
• Any voltage or current in a linear circuit with a sinusoidal
source is a sinusoid of the same frequency (w).
– We only need to keep track of the amplitude and phase, when
determining the response of a linear circuit to a sinusoidal source.
• Any time-varying signal can be expressed as a sum of
sinusoids of various frequencies (and phases).
 Applying the principle of superposition:
– The current or voltage response in a linear circuit due to a
time-varying input signal can be calculated as the sum of the
sinusoidal responses for each sinusoidal component of the
input signal.
EE105 Fall 2007
Lecture 13, Slide 4
Prof. Liu, UC Berkeley
High Frequency “Roll-Off” in Av
• Typically, an amplifier is designed to work over a
limited range of frequencies.
– At “high” frequencies, the gain of an amplifier decreases.
EE105 Fall 2007
Lecture 13, Slide 5
Prof. Liu, UC Berkeley
Av Roll-Off due to CL
• A capacitive load (CL) causes the gain to decrease at
high frequencies.
– The impedance of CL decreases at high frequencies, so that
it shunts some of the output current to ground.

1 

Av   g m  RC ||
jwCL 

EE105 Fall 2007
Lecture 13, Slide 6
Prof. Liu, UC Berkeley
Frequency Response of the CE Stage
• At low frequency, the capacitor is effectively an open
circuit, and Av vs. w is flat. At high frequencies, the
impedance of the capacitor decreases and hence the
gain decreases. The “breakpoint” frequency is 1/(RCCL).
Av 
EE105 Fall 2007
Lecture 13, Slide 7
g m RC
R C w 1
2
C
2
L
2
Prof. Liu, UC Berkeley
Amplifier Figure of Merit (FOM)
• The gain-bandwidth product is commonly used to
benchmark amplifiers.
– We wish to maximize both the gain and the bandwidth.
• Power consumption is also an important attribute.
– We wish to minimize the power consumption.
 1 

g m RC 
RC C L 
Gain  Bandwidth


Power Consumptio n
I CVCC
1

VT VCC C L
Operation at low T, low VCC, and with small CL  superior FOM
EE105 Fall 2007
Lecture 13, Slide 8
Prof. Liu, UC Berkeley
Bode Plot
• The transfer function of a circuit can be written in the
general form

jw 
jw 
1 
1 
  A is the low-frequency gain
0
w z1  w z 2 

wzj are “zero” frequencies
H ( jw )  A0



wpj are “pole” frequencies
j
w
j
w
1 
1 

 w  w 
p1 
p2 

• Rules for generating a Bode magnitude vs. frequency plot:
– As w passes each zero frequency, the slope of |H(jw)| increases
by 20dB/dec.
– As w passes each pole frequency, the slope of |H(jw)| decreases
by 20dB/dec.
EE105 Fall 2007
Lecture 13, Slide 9
Prof. Liu, UC Berkeley
Bode Plot Example
• This circuit has only one pole at ωp1=1/(RCCL); the slope
of |Av|decreases from 0 to -20dB/dec at ωp1.
1
w p1 
RC C L
• In general, if node j in the signal path has a smallsignal resistance of Rj to ground and a capacitance Cj to
ground, then it contributes a pole at frequency (RjCj)-1
EE105 Fall 2007
Lecture 13, Slide 10
Prof. Liu, UC Berkeley
Pole Identification Example
1
w p1 
RS Cin
EE105 Fall 2007
w p2
Lecture 13, Slide 11
1

RC C L
Prof. Liu, UC Berkeley
High-Frequency BJT Model
• The BJT inherently has junction capacitances which
affect its performance at high frequencies.
Collector junction: depletion capacitance, Cm
Emitter junction: depletion capacitance, Cje, and also
diffusion capacitance, Cb.
C  Cb  C je
EE105 Fall 2007
Lecture 13, Slide 12
Prof. Liu, UC Berkeley
BJT High-Frequency Model (cont’d)
• In an integrated circuit, the BJTs are fabricated in the
surface region of a Si wafer substrate; another
junction exists between the collector and substrate,
resulting in substrate junction capacitance, CCS.
BJT cross-section
EE105 Fall 2007
BJT small-signal model
Lecture 13, Slide 13
Prof. Liu, UC Berkeley
Example: BJT Capacitances
• The various junction capacitances within each BJT are
explicitly shown in the circuit diagram on the right.
EE105 Fall 2007
Lecture 13, Slide 14
Prof. Liu, UC Berkeley
Transit Frequency, fT
• The “transit” or “cut-off” frequency, fT, is a measure
of the intrinsic speed of a transistor, and is defined as
the frequency where the current gain falls to 1.
Conceptual set-up to measure fT
I out  g mVin
I in 
Vin
Z in
 1 
I out
  1
 g m Z in  g m 
I in
 jwT Cin 
g
 wT  m
Cin
gm
2f T 
C
EE105 Fall 2007
Lecture 13, Slide 15
Prof. Liu, UC Berkeley
Dealing with a Floating Capacitance
• Recall that a pole is computed by finding the resistance
and capacitance between a node and GROUND.
• It is not straightforward to compute the pole due to Cm1
in the circuit below, because neither of its terminals is
grounded.
EE105 Fall 2007
Lecture 13, Slide 16
Prof. Liu, UC Berkeley
Miller’s Theorem
• If Av is the voltage gain from node 1 to 2, then a
floating impedance ZF can be converted to two
grounded impedances Z1 and Z2:
V1  V2 V1
V1
1

 Z1  Z F
 ZF
ZF
Z1
V1  V2
1  Av
V1  V2
V2
V2
1

 Z 2  Z F
 ZF
ZF
Z2
V1  V2
1 1
EE105 Fall 2007
Lecture 13, Slide 17
Av
Prof. Liu, UC Berkeley
Miller Multiplication
• Applying Miller’s theorem, we can convert a floating
capacitance between the input and output nodes of
an amplifier into two grounded capacitances.
• The capacitance at the input node is larger than the
original floating capacitance.
A0   Av
Z2 
ZF
1 1
1

Av
1
jwCF

1 1
jw 1  1 CF
A0
A0 

1
ZF
1
jw C F
Z1 


1  Av
1  A0
jw 1  A0 C F
EE105 Fall 2007
Lecture 13, Slide 18
Prof. Liu, UC Berkeley
Application of Miller’s Theorem
w p ,in
1

RS 1  g m RC C F
w p ,out 
EE105 Fall 2007
1

1
RC 1 
 g m RC

CF

Lecture 13, Slide 19
Prof. Liu, UC Berkeley
Small-Signal Model for CE Stage
EE105 Fall 2007
Lecture 13, Slide 20
Prof. Liu, UC Berkeley
… Applying Miller’s Theorem
w p ,in
1

RThev Cin  1  g m RC Cm 
w p ,out 
1


1
RC  Cout  1 
 g m RC

 
Cm 

 
Note that wp,out > wp,in
EE105 Fall 2007
Lecture 13, Slide 21
Prof. Liu, UC Berkeley
Direct Analysis of CE Stage
• Direct analysis yields slightly different pole locations
and an extra zero:
gm
wz 
C XY
1
w p1 
1  g mRC C XY RThev  RThevCin  RC C XY  Cout 

1  g m RC C XY RThev  RThev Cin  RC C XY  Cout 
w p2 
RThev RC CinC XY  CoutC XY  CinCout 
EE105 Fall 2007
Lecture 13, Slide 22
Prof. Liu, UC Berkeley
Input Impedance of CE Stage
1
Z in 
|| r
jwC  1  g m RC Cm 
EE105 Fall 2007
Lecture 13, Slide 23
Prof. Liu, UC Berkeley