# BJT amps

Lecture 11
ANNOUNCEMENTS
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HW#6 has been posted.
Review session: 3-5PM Friday (10/5) in 306 Soda (HP Auditorium)
Midterm #1 (Thursday 10/11):
• Material of Lectures 1-10 (HW# 2-6; Chapters 2, 4, 5)
• 2 pgs of notes (double-sided, 8.5”&times;11”), calculator allowed
OUTLINE
• Review of BJT Amplifiers
• Cascode Stage
EE105 Fall 2007
Lecture 11, Slide 1
Prof. Liu, UC Berkeley
Review: BJT Amplifier Design
• A BJT amplifier circuit should be designed to
1. ensure that the BJT operates in the active mode,
2. allow the desired level of DC current to flow, and
3. couple to a small-signal input source and to an output “load”.
 Proper “DC biasing” is required!
(DC analysis using large-signal BJT model)
• Key amplifier parameters:
(AC analysis using small-signal BJT model)
– Voltage gain Av  vout/vin
– Input resistance Rin  resistance seen between the input
node and ground (with output terminal floating)
– Output resistance Rout  resistance seen between the output
node and ground (with input terminal grounded)
EE105 Fall 2007
Lecture 11, Slide 2
Prof. Liu, UC Berkeley
Large-Signal vs. Small-Signal Models
• The large-signal model is used to determine the DC
operating point (VBE, VCE, IB, IC) of the BJT.
• The small-signal model is used to determine how the
output responds to an input signal.
EE105 Fall 2007
Lecture 11, Slide 3
Prof. Liu, UC Berkeley
Small-Signal Models for Independent Sources
• The voltage across an independent voltage source does
not vary with time. (Its small-signal voltage is always zero.)
It is regarded as a short circuit for small-signal analysis.
Large-Signal
Model
Small-Signal
Model
• The current through an independent current source does
not vary with time. (Its small-signal current is always zero.)
It is regarded as an open circuit for small-signal analysis.
EE105 Fall 2007
Lecture 11, Slide 4
Prof. Liu, UC Berkeley
Comparison of Amplifier Topologies
Common Emitter
• Large Av &lt; 0
• Moderate Rin
- Increased by RB
- Increased by RE(b+1)
Common Base
• Large Av &gt; 0
• Small Rin
- Increased by RB/(b+1)
- Decreased by RE
• Rout  RC
• Rout  RC
but impedance seen
looking into the collector
can be “boosted” by
emitter degeneration
EE105 Fall 2007
Emitter Follower
• 0 &lt; Av ≤ 1
• Large Rin
(due to RE(b+1))
• Small Rout
- Effect of source
impedance is
reduced by b+1
- Decreased by RE
but impedance seen
looking into the collector • ro decreases Av, Rin,
can be “boosted” by
and Rout
emitter degeneration
Lecture 11, Slide 5
Prof. Liu, UC Berkeley
Common Emitter Stage
VA  
VA  
vout
 RC

1
RB
vin
 RE 
gm
b 1
Rin  RB  r  ( b  1) RE
Rout  RC
EE105 Fall 2007
Rout  RC || rO 1  g m ( RE || r )
Lecture 11, Slide 6
Prof. Liu, UC Berkeley
Common Base Stage
VA  
vout
RC
RE


1
RB RS  RE
vin
 RS RE 
gm
b 1
 1
RB 
 RE
Rin   
 gm b 1 
Rout  RC
EE105 Fall 2007
VA  
Rout  RC || rO 1  g m ( RE || r )
Lecture 11, Slide 7
Prof. Liu, UC Berkeley
Emitter Follower
VA  
VA  
vout
RE

vin R  1  RS
E
gm b 1
Rin  r  (1  b ) RE
 1
Rs 
 || RE
Rout   
 gm b  1 
EE105 Fall 2007
vout
RE || rO

vin R || r  1  RS
E
O
gm b 1
Rin  r  b  1RE || rO 
Rout
Lecture 11, Slide 8
 Rs
1 
 || RE || rO
 

 b 1 gm 
Prof. Liu, UC Berkeley
Ideal Current Source
Circuit Symbol
I-V Characteristic
Equivalent Circuit
• An ideal current source has infinite output impedance.
How can we increase the output impedance of a BJT that
is used as a current source?
EE105 Fall 2007
Lecture 11, Slide 9
Prof. Liu, UC Berkeley
Boosting the Output Impedance
• Recall that emitter degeneration boosts the impedance
seen looking into the collector.
– This improves the gain of the CE or CB amplifier. However,
Rout  1  g m RE || r rO  RE || r
EE105 Fall 2007
Lecture 11, Slide 10
Prof. Liu, UC Berkeley
Cascode Stage
• In order to relax the trade-off between output
impedance and voltage headroom, we can use a
transistor instead of a degeneration resistor:
Rout  [1  g m (rO 2 || r 1 )]rO1  rO 2 || r 1
Rout  g m1rO1 rO 2 || r 1 
I C 2  I E1  I C 2 if b1  1
• VCE for Q2 can be as low as ~0.4V (“soft saturation”)
EE105 Fall 2007
Lecture 11, Slide 11
Prof. Liu, UC Berkeley
Maximum Cascode Output Impedance
• The maximum output impedance of a cascode is
limited by r1.
If rO 2  r 1 :
Rout,max  g m1rO1r 1  b1rO1
EE105 Fall 2007
Lecture 11, Slide 12
Prof. Liu, UC Berkeley
PNP Cascode Stage
Rout  [1  g m1 (rO 2 || r 1 )]rO1  rO 2 || r 1
Rout  g m1rO1 rO 2 || r 1 
EE105 Fall 2007
Lecture 11, Slide 13
Prof. Liu, UC Berkeley
False Cascodes
• When the emitter of Q1 is connected to the emitter of
Q2, it’s not a cascode since Q2 is a diode-connected
device instead of a current source.
Rout

 1

1
 1  g m1 
|| rO 2 || r 1  rO1 
|| rO 2 || r 1
g m2
 g m2


Rout

g m1 
1
rO1 
 1 
 2rO1
g m2 
g m2

EE105 Fall 2007
Lecture 11, Slide 14
Prof. Liu, UC Berkeley
Short-Circuit Transconductance
• The short-circuit transconductance of a circuit is a
measure of its strength in converting an input
voltage signal into an output current signal.
iout
Gm 
vin
EE105 Fall 2007
Lecture 11, Slide 15
vou t  0
Prof. Liu, UC Berkeley
Voltage Gain of a Linear Circuit
• By representing a linear circuit with its Norton
equivalent, the relationship between Vout and Vin can
be expressed by the product of Gm and Rout.
Norton Equivalent Circuit
Computation of
short-circuit
output current:
EE105 Fall 2007
vout  iout Rout  Gm vin Rout
vout vin  Gm Rout
Lecture 11, Slide 16
Prof. Liu, UC Berkeley
Example: Determination of Voltage Gain
Determination of Gm
Gm 
iout
vin
 g m1
vo ut  0
Determination of Rout
Rout 
vx
 ro1
ix
Av   g m1rO1
EE105 Fall 2007
Lecture 11, Slide 17
Prof. Liu, UC Berkeley
Comparison of CE and Cascode Stages
• Since the output impedance of the cascode is higher
than that of a CE stage, its voltage gain is also higher.
vout   g m1vinrO1
VA
Av   g m1rO1  
VT
EE105 Fall 2007
Av   gm1rO 2 gm2 rO1 r 2 
Lecture 11, Slide 18
Prof. Liu, UC Berkeley
Voltage Gain of Cascode Amplifier
• Since rO is much larger than 1/gm, most of IC,Q1 flows
into diode-connected Q2. Using Rout as before, AV is
easily calculated.
iout  g m1vin  Gm  g m1
Av  Gm Rout
  g m1 1  g m 2 rO1 || r 2 rO 2  rO1 || r 2 
  g m1 g m 2 rO1 || r 2 rO 2 
EE105 Fall 2007
Lecture 11, Slide 19
Prof. Liu, UC Berkeley
Practical Cascode Stage
• No current source is ideal; the output impedance is finite.
Rout  rO3 || g m2 rO 2 (rO1 || r 2 )
EE105 Fall 2007
Lecture 11, Slide 20
Prof. Liu, UC Berkeley
Improved Cascode Stage
• In order to preserve the high output impedance, a
cascode PNP current source is used.
Rout  g m3rO 3 (rO 4 || r 3 ) || g m 2 rO 2 (rO1 || r 2 )
Av   g m1Rout
EE105 Fall 2007
Lecture 11, Slide 21
Prof. Liu, UC Berkeley