differential amp with one input - carleton university › ~rmason › teaching › 283-f.pdf ·...
TRANSCRIPT
Differential Amp with One Input
Differential Amp with One Input
• Common to implement circuit on an IC with Q1 and Q2 being matcheddevices
• Transistors are biased in the constant current region. Bias isestablished by a constant DC current source Io
• vg has no DC component such that the DC bias values vBE1 and vBE2 areexactly the same. Io is split evenly between Q1 and Q2 causing them tohave the same parameters gm and rπ
• Amplification is performed by Q1. Its emitter is incr5ementallybypassed to ground by the incremental Thevenin resistance lookinginto the emitter of Q2.
ivr
ivr
rvi
rg
bTEST
TESTTEST
THTEST
TEST m
22 2
22
2
1
11
= − = +
= =+
≅
πο
π
π
ο
β
β
( )
• If gm2 is large the value rTH2 will be small andeffectively short node E to ground
iv
r rv
r r
ivr
r r
v i Rvr
Rg R
v
Avv
g R
bIN
TH
IN
bIN
OUT b CIN
Cm C
IN
VOUT
IN
m C
11 2
12
11
2 1
11
1 1 1
2
2 2
2
=+ +
=+ + +
= =
= − = − =−
= =−
π ο π οπ
ο
ππ π
ο οπ
β β β
β β
( ) ( ) ( )
where
• The gain is half that of a single BJT inverter(−gmRC). The gain does not degrade at lowfrequencies as in the case of a shunt capacitor.
Differential Amp with Two Inputs
Differential Amp with Two Inputs
• With a simple relocation of input and output terminals Q2
can function as the inverter and Q1 as the bypass device
• In general input signals may be simultaneously applied toboth devices. Since the small signal models are linear theresponse can be analyzed using superposition
• The diff. amp amplifies the difference component of theinput signals by a large gain factor and multiplies theaverage component by a small gain factor
v v v vv v
v vv v v v v
v
v vv
v v v vv v
v vv
v vv
idm icm
icmidm
icmidm
odm o o ocmo o
o ocmodm
o ocmodm
= − =+
= + =+
+−
=
= −
= − =+
= + = −
1 21 2
11 2 1 2
1
2
1 21 2
1 2
2
2 2 2
2
2
2 2
Similarly
Differential Mode Response
v v t vv v
v v t v v v v
a icma a
a idm a a a
1
2
20
2
= =+ −
=
= − = − − =
( )( )
( ) ( )
• Solving for vo1 and vo2
v v v v i r i r
r r r
i i
ivr
vr
ivr
a a b b
b b
ba a
ba
− − = − = −
= = = == −
= = = −
( )
;
π π π π
ο ο ο π π π
π π π
β β β
1 2 1 1 2 2
1 2 1 2
2 1
1 222
For matched transistors
v i R i Rr
v R
v i R i Rr
v R
o c C b C a C
o c C b C a C
1 1 1
2 2 2
= − = − = −
= − = − =
β β
β β
οο
π
οο
π
• The differentially driven circuit is symmetric aboutits center line. Its base currents and dependentsource currents are always equal and opposite. Noincremental current flows through ro. As a result,the node E remains at incremental ground orvirtual ground.
The single ended gain seen from output 1 to groundis equal to:
Avv
v v v vdm seo
idmidm a− = = − =1
11 2 2
Av R
rv
Rr
g Rdm se
aC
a
C m C− = − = − =
−1 2 2 2
β βοπ ο
π
The differential gain is equal to:
Av v
v
v Rr v R
rg Rdm diff
o o
idm
aC
aC
m C− =−
=− −
= −1 2
2
β βοπ
οπ
Common Mode Response
v v vv v
v
v v v v v
b icm b
b idm
11 2
2 1 2
20
= =+
=
= = − =
i i i i i i
v i r i r i i i
iv
r rr r r
ro b b b b b
b b b o b b b
bb
o
= + + + = +
= + + = == =
=+ +
= =
( ) ( ) ( )
( )
( )
1 1 1 2 2 2 1
1 2
1 2
1 2
2 1
2 1
2 1
β β β
ββ β β
β
ο ο ο
π ο
ο ο ο
π οπ π π
( )
( )
v i RR
r rv v
Avv
vv
Rr r
Av v
vR R
r r
o b CC
ob o
cm seo
icm
o
b
C
o
cm diffo o
icm
C C
o
1 1 2
11 1
1 2
2 1
2 1
2 10
= − =−
+ +=
= = =−
+ +
=−
=− − −
+ +=
−
−
β ββο
ββ
β ββ
οο
π
ο
π ο
ο ο
π ο
( )
( )
• Differential amp amplifies differential signals andrejects (does not amplify) common signals. Thedegree to which common signals are rejected iscalled the Common Mode Rejection Ratio(CMRR)
CMRR =AA
dm
cm
• The CMRR is often measured in decibels
CMRR = 20 10logAA
dm
cm
• For a perfectly balanced diff amp
CMRRA
A
Rrdm diff
cm diff
C−
−
= = ∞βο
π
0
• Real diff amps have typical CMRR of 60 - 100 dB
Input and Output Resistance
• Incremental input resistance is different fordifferential and common mode signals
( )
rv v
iv v
i
ivr
iv
r
rv v
vr
r
rv v
i
v v
i
IN dm
a a
IN dma a
a
IN cm
−
−
−
=−
=−
−
= =−
=− −
=
=+
=+
1 2
1
1 2
2
1 2
1 2
1
1 2
2
2
2 2
where i and i are the signal currents into Q and Q1 2 1 2
π π
π
π
( ) ( )
i iv
r rv v v
r r r
r r R
b
ob
IN cm o
OUT se OUT se C
1 2 1 2
1 2
2 1
2 1
= =+ +
= =
= + +
= =
−
− −
π ο
π ο
β
β
( )
( )
• The differential output resistance can be computedusing the test source method
iv
R Rv
R
rvi
R
TESTTEST
C C
TEST
C
OUT diffTEST
TESTC
=+
=
= =−
2
2
Differential Amp Swing Range
• If an input signal causes any one device to reach alimit to its linear operation, the output voltage (orcurrent) will reach a swing limit
• Consider the following with differential modeexcitation. Note that RC1 and RC2 may havedifferent values. If one transistor could operateindependently VCC ≥ Vo ≥ VE + VSAT, however, Q1
and Q2 operate in tandem such that one transistormay reach cutoff before the other reachessaturation or vice versa
• For the previous circuit for the saturation of Q1
iV V V
R
V
V V
i i
i i I
iV V V
R
i I i
C SATCC E SAT
C
SAT
E BE
C E
C C o
C SATCC E SAT
C
C o C SAT
11
1 2
22
1 2
10 0 7 0133
32
01
0 7
2
10 0 7 015 6
19
01
−
−
−
=− +
=− − +
=
=≅ − = −
≅
+ = =
∴
=− +
=− − +
≅
= − =
( ) [( . ) . ].
.
.
.
( ) [( . ) . ].
.
.
kmA
Assume V
V
mA
Q will reach cutoff before Q reaches saturation
kmA
When Q reaches saturation mA
2 1
2
Ω
Ω
• When Q2 goes into cutoff iC2 = 0 and Q1 willremain in active region with iC1 = Io = 2 mA
v V i R
v V i R
OUT CC C C
OUT CC C C
2 2 2
1 1 1
10 0 10
10 2 34
= − = − =
= − = − =
V V
V mA)(3.3k V( ) .Ω
• When Q2 goes into saturation with iC2 = 1.9 mA,iC2 = 0.1 mA
v
v
OUT
OUT
2
1
10 0 6
10 01 33 9 7
= − = −
= − =
V (1.9 mA)(5.6 k V
V mA) k V
Ω
Ω
) .
( . ( . ) .
For the basic BJT diff amp of the previous page
A
CMRR dB
Diff Swing V
r
r
Power Supply V
dm SE
IN diff
OUT SE
−
−
−
=
>= ±
>
<= ±
100
60
3
1
1
10
.
k
k
Ω
Ω
Large Signal Performance
dvdv
g RAOUT
idm
m Cdm se
1 112
=−
= −
Power Amplification Stages• In many designs an amplifier is required to deliver large amounts of
power to a passive load. The power may be a large current to a smallresistance or a large voltage to a moderate resistance (impedance).
• Using a linear amplifier the power wasted in the active device iscomparable to the power delivered to the load. Devices in the outputstage must be capable of dissipating this excess power.
• Alternative configurations offer increased efficiency at the expense oftrue linear operation.
Complementary Pair (Class B) Output
• When an amplifier is required to deliver large load currents it isdesirable to bias the voltage of its output terminal near ground. Thisminimizes the bias power dissipated in both the load element and theactive devices of the output stage.
Linearly Biased (Class A) Output
Minimally Biased (Class AB) Output