bjt differential amplifier
TRANSCRIPT
Bipolar Differential Amplifiers: Qualitative Analysis
Different modes of operation of the BJT differential pair: (a) The differential pair with a common-mode input signal vCM. (b) The differential pair with a “large” differential input signal. (c) The differential pair with a large differential input signal of polarity opposite to that in (b). (d) The differential pair with a small differential input signal vi. Note that we have assumed the bias current source I to be ideal (i.e., it has an infinite output resistance) and thus I remains constant with the change in vCM.
Common Mode Differential Mode
Bipolar Differential Amplifiers: Large Signal Analysis
1 2
1 2
( ) / ( ) /1 2
( ) /1
2
The exponential relationship applied to each of the two transistors may be written as:
and
These two equations can be combined to obta
in
w
hic
B E T B E T
B B T
v v V v v VS SE E
v v VE
E
I Ii e i e
i ei
α α− −
−
= =
=
2 1 1 2
1
1 2( ) / ( ) /
1 2 1 2
1 2
1 (
2
1 11 1
1
h can be manipulated to yield
and
From the circuit we have
Which may be used to obtain the following expressions for and
B B T B B T
E Ev v V v v V
E E E E
E E
E v
E E
i ii i e i i e
i i I
ie
iI
i
− −= =+ + + +
+ =
=+ 2 1 1 22) / ( ) /
1 2
1 2
1and
and may be obtained by multiplying and by which is almost unity and
plotted as shown in the fi
gure
B B T B B TEv V v
C C
E E
v V
Iie
i ii i α
− −=+
Bipolar Differential Amplifiers: LinearizationFor Your Information
The transfer characteristics of the BJT differential pair (a) can be linearized (b) (i.e., the linear range of operation can be extended) by including resistances in the emitters.
What is the role of the degeneration resistance (Re)?
Bipolar Differential Amplifiers: DC Analysis (Example 1)
Problem: Find the Q-points of transistors in the shown differential amplifier.
Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100Analysis:
A3.95)3102(75
V7.015μ=
Ω×
−=
−=
⎟⎠⎞⎜
⎝⎛
EE2RBEVEEV
EI
100 94.4 A101
I I IE EC α μ= = =
94.4 A 0.944 A100
ICIBμ μ
β= = =
15 7.92V
7.92V-(-0.7V) 8.62V
V I RC C C
V V VECE C
= − =
= − = =
Due to symmetry, both transistors are biased at Q-point (94.4 μA, 8.62V)
Bipolar Common-mode Input Voltage Range
For symmetrical power supplies (VEE=VCC) , VEE >> VBE, and RC = REE,
0
2
1
1
V V I R VCB CC C C ICV V VIC BE EEIC REE
V VR EE BEC2R VEE CCV VIC CC RC
2REE
α
α
α
⎛ ⎞⎜ ⎟⎝ ⎠
= − − ≥
− +=
−−
∴ ≤+
53
VVCCV IC =≤
We want to find max. VIC while the C-B junction is reverse biased.
Problem: Find the max. VIC before saturation in the shown differential amplifier.
Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100Analysis:
Bipolar Differential Amplifiers: DC Analysis (Example 2)
Problem: Find vE ,vc1 , and vc2 in the shown differential amplifier.Given data: VCC=VEE= 5 V, REE=RC=1kΩ, α ≈, |VBE|=0.7 VAnalysis:
1
2
2 2
2 2
0.75 * 5 0 5
(5 0.7) /1 4.34.3
5 * 5 4.3 0.7
V V
mA mA
V
E
C1 C C
E E
C E
C C C
vv R II II Iv R I
α
+− + − + = −
= = − == =
= − + − + = −
== =
=
We can assume Q1 to be off and Q2 on
Differential-mode Gain and Input Resistance
( 1/ )(v v ) 1/ ve3 4v (1/ 2/ 2 ) 0 v 0e e
g r Rm EER r gEE mπ
π
+ + =
∴ + + = → =
2id
v4v −=
ev2id
v3v −= ev
2id
v4v −−=
Output signal voltages are:
2id
vc1v CRmg−=
2id
vc2v CRmg+=
idvodv CRmg−=∴
2id
v3v =∴
Emitter node in differential amplifier represents virtual ground for differential-mode input signals.
Differential-mode gain for balanced output, is:
If either vc1 or vc2 is used alone as output, output is said to be single-ended.
Differential-mode Gain and Input Resistance (contd.)
CRmgddA −=
=
=
0ic
vidvod
vc2vc1vodv −=
220
icvid
vc1
v1
ddA
CRmg
ddA =−=
=
=22
0ic
vidvc2
v2
ddA
CRmg
ddA −==
=
=
Differential-mode input resistance is small-signal resistance presented to differential-mode input voltage between the two transistor bases.
If vid =0, . For single-ended outputs,
πridR 2b1i/idv ==∴
CRorCRodR 2)(2 ≅= CRodR ≅πr
)2/id
v(b1i =
Common-mode Gain and Input Resistance
Both arms of differential amplifier are symmetrical. So terminal currents and collector voltages are equal. Characteristics of differential pair with common-mode input are similar to those of a C-E amplifier with large emitter resistor.
Output voltages are:
vicib 2( 1)r REEβπ=
+ +
v v i vc1 c2 b ic2( 1)RCRC r REE
ββ
βπ
−= =− =
+ +
v 2( 1)ie b2( 1)
v vic ic2( 1)
REEREE
r REE
βββπ
= +
+= ≅
+ +
Common-mode gain is given by:
For RC=REE, common-mode gain =0.5. Thus, common-mode output voltage and Acc is 0 if REE is infinite. This result is obtained since output resistances of transistors are neglected. A more accurate expression is:
Therefore, common-mode conversion gain is found to be 0.
Common-mode Gain and Input Resistance (contd.)
vocv 2( 1) 2ic v 0id
R RC CAcc r R REE EE
β
βπ= =− ≅−
+ +=
0c2vc1vodv =−=
1 12
A Rcc C r Ro EEβ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
≅ −
v 2( 1)ic ( 1)2i 2 2b
r R rEER REEicβπ π β
+ += = = + +
Common-Mode Rejection ratio (CMRR)
Represents ability of amplifier to amplify desired differential-mode input signal and reject undesired common-mode input signal.For differential output, common-mode gain of balanced amplifier is zero, CMRR is infinite. For single-ended output,
/ 2 1C M R R1 12 2r g
A Adm dd g Rm EEA Acm ccg Ro m m EEβ
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
≅= = =−
Analysis of Differential Amplifiers Using Half-Circuits
Half-circuits are constructed by first drawing the differential amplifier in a fully symmetrical form- power supplies are split into two equal halves in parallel, emitter resistor is separated into two equal resistors in parallel.None of the currents or voltages in the circuit are changed.For differential mode signals, points on the line of symmetry are virtual grounds connected to ground for ac analysisFor common-mode signals, points on line of symmetry are replaced by open circuits.
Bipolar Differential-mode Half-circuits
Applying rules for drawing half-circuits, the two power supply lines and emitter become ac grounds. The half-circuit represents a C-E amplifier stage.
2id
vc1v CRmg−=
2id
vc2v CRmg+=
v v v vc1 c2od idg Rm C= − =−
Direct analysis of the half-circuits yield:
πridR 2b1i/idv == )(2 orCRodR =
CRmgddA −=
=
=
0ic
vidvod
v
220
icvid
vc1
v1
ddA
CRmg
ddA =−=
=
=
Bipolar Common-mode Half-circuits
Applying rules for drawing half-circuits, the points at the line of symmetry are open circuited. The half-circuit represents a C-E amplifier stage with an emitter resistance.
v v v 0c1 c2od = − =
Direct analysis of the half-circuits yield:
v v i vc1 c2 b ic2( 1)RCRC r REE
ββ
βπ
−= =− =
+ +
v 2( 1)ic ( 1)2i 2 2b
r R rEER REEicβπ π β
+ += = = + +
/2CMRR
A Adm dd g Rm EEA Acm cc≅= =
vocv 2( 1) 2ic v 0id
R RC CAcc r R REE EE
β
βπ= =− ≅−
+ +=
Biasing with Electronic Current Sources
Differential amplifiers are biased using electronic current sources to stabilize the operating point and increase effective value of REE to improve CMRRElectronic current source has a Q-point current of ISSand an output resistance of RSS as shown.DC model of the electronic current source is a dc current source, ISS while ac model is a resistance RSS.
SSR
0V
SSIDCI −=