# Lecture 23

```Lecture 23
ANNOUNCEMENTS
6
44
• Midterm #2 results (undergrad. scores only):
•
N=73; mean=63; median=63.5; std.dev.=8.42
• You may pick up your exam during any TA office hour.
•
exam.
• The list of misunderstood/forgotten points has been updated.
OUTLINE
• BJT Differential Amplifiers (cont’d)
– Cascode differential amplifiers
– Common‐mode rejection
– Differential pair with active load
EE105 Fall 2007
Lecture 23, Slide 1
Prof. Liu, UC Berkeley
21
Cascode Differential Pair
Rout = [1 + g m 3 (rO1 || rπ 3 )]rO 3 + rO1 || rπ 3
Rout ≅ g m 3 (rO1 || rπ 3 )rO 3 + rO1 || rπ 3
Half circuit for ac analysis
Av = −gm1Rout ≅ −gm1[gm3 (rO1 || rπ 3 )rO3 + rO1 || rπ 3 ]
EE105 Fall 2007
Lecture 23, Slide 2
Prof. Liu, UC Berkeley
Telescopic Cascode Differential Pair
Half circuit for ac analysis
Av ≈ − g m1 [g m3 rO 3 (rO1 || rπ 3 )] || [g m5 rO 5 (rO 7 || rπ 5 )]
EE105 Fall 2007
Lecture 23, Slide 3
Prof. Liu, UC Berkeley
Example
⎡
R1 ⎞ ⎤
R
⎛
Rop = ⎢1 + g m 5 ⎜ rO 7 || rπ 5 || ⎟ ⎥ rO 5 + rO 7 || rπ 5 || 1
2 ⎠⎦
2
⎝
⎣
Av = − g m1 ([g m 3 rO 3 ( rO1 || rπ 3 ) ] || Rop )
Half circuit for ac analysis
EE105 Fall 2007
Lecture 23, Slide 4
Prof. Liu, UC Berkeley
Effect of Finite Tail Impedance
• If the tail current source is not ideal, then when an input
common‐mode voltage is applied, the currents in Q1 and Q2
and hence the output common‐mode voltage will change.
∆Vout ,CM
∆Vin ,CM
EE105 Fall 2007
=−
(RC / 2)
1
+ REE
2gm
Lecture 23, Slide 5
=−
RC
1
+ 2 REE
gm
Prof. Liu, UC Berkeley
Effect of Input CM Noise
Ideal Tail Current
• There is no effect of the input CM noise at the output.
EE105 Fall 2007
Lecture 23, Slide 6
Prof. Liu, UC Berkeley
Effect of Input CM Noise
Non‐Ideal
Non
Ideal Tail Current
• The single‐ended outputs are corrupted by the input CM noise.
EE105 Fall 2007
Lecture 23, Slide 7
Prof. Liu, UC Berkeley
Comparison
Ideal Tail Current
Non-Ideal Tail Current
• The differential output
voltage
l
signall is the
h same
for both cases.
&AElig; For small input
p CM
noise, the differential
pair is not affected.
EE105 Fall 2007
Lecture 23, Slide 8
Prof. Liu, UC Berkeley
CM to DM Conversion; gain ACM‐DM
• If finite tail impedance and asymmetry (e.g. in load resistance)
are both present, then the differential output signal will
contain a portion of the input common‐mode signal.
∆VCM = ∆VBE + 2∆I C REE =
⇒ ∆I C =
∆I C
+ 2∆I C REE
gm
∆VCM
1
+ 2 REE
gm
∆Vout1 = −∆I C RC
∆Vout 2 = − ∆I C (RC + ∆RC )
∆Vout = ∆Vout1 − ∆Vout 2 = −∆I C ∆RC
∆Vout
∆RC
=
∆VCM (1 / g m ) + 2 REE
EE105 Fall 2007
Lecture 23, Slide 9
Prof. Liu, UC Berkeley
Example
ACM − DM =
EE105 Fall 2007
∆R C
1
+ 2{[1 + g m 3 ( R1 || rπ 3 )]rO 3 + R1 || rπ 3 }
g m1
Lecture 23, Slide 10
Prof. Liu, UC Berkeley
Common‐Mode Rejection Ratio
• CMRR is the ratio of the wanted amplified differential input
signal to the unwanted converted input common‐mode noise
that appears at the output.
CMRR ≡
EE105 Fall 2007
Lecture 23, Slide 11
ACM − DM
Prof. Liu, UC Berkeley
Differential to Single‐Ended Conversion
• Many circuits require a differential to single‐ended conversion.
• This topology is not very good; its most critical drawback is
pp y noise corruption,
p
, since no common‐mode cancellation
supply
mechanism exists. Also, we lose half of the voltage signal.
EE105 Fall 2007
Lecture 23, Slide 12
Prof. Liu, UC Berkeley
… A Better Alternative
• This circuit topology performs differential to single‐ended
conversion with no loss of gain.
g
(
vout
= g m ro , NPN ro , PNP
vin1 − vin 2
EE105 Fall 2007
Lecture 23, Slide 13
Prof. Liu, UC Berkeley
)
• With a current mirror as the load, the signal current produced
byy Q1 can be replicated
p
onto Q4.
• This type of load is different from the conventional “static
EE105 Fall 2007
Lecture 23, Slide 14
Prof. Liu, UC Berkeley
• The input differential pair decreases the current drawn from
RL by ∆I, and the active load pushes an extra ∆I into RL by
current mirror action; these effects enhance each other.
EE105 Fall 2007
Lecture 23, Slide 15
Prof. Liu, UC Berkeley