Transcript No Slide Title

MALVINO
Electronic
PRINCIPLES
SIXTH EDITION
Differential
Amplifiers
Chapter 17
A differential amplifier
+VCC
RC
vc1
> vv221
vv112 =
vout = 0
vout
v1
RC
vc2 vout = vc2 - vc1
v2
RE
-VEE
A differential amplifier with single-ended output
+VCC
Block symbol
v1
A
vout
RC
v2
vout
vout = A(v1 - v2)
v1
(A is one-half the
differential output value)
v2
RE
-VEE
Single-ended output and single-ended input.
+VCC
v1
A
vout
RC
vout = Av1
vout
v1
RE
-VEE
Inverting configuration
+VCC
vout
A
v2
RC
vout
vout = -Av2
v2
RE
-VEE
IT, the tail current,
flows in RE
IT =
VEE - VBE
IE =
dc analysis
+VCC
RC
RC
RE
IT
2
VC = VCC - IERC
RE
-VEE
RC
A=
2re’
ac analysis
+VCC
RC
Differential
A=
re’
output case:
RC
vout
zin = 2bre’
v1
RE
-VEE
The input transistors are not exactly the same.
IB1 = IB2
Iin(bias) =
IB1 + IB2
2
Iin(off) = IB1 - IB2
IB1 = Iin(bias) +
IB1
IB2
Iin(off)
2
IB2 = Iin(bias) -
Iin(off)
2
Base current can produce a false signal.
+VCC
RC
RC
Verror
Vin = IB1RB
RB
RE
-VEE
Vin and Verror are minimized with equal base resistors.
+VCC
Iin(off) < Iin(bias)
Vin = Iin(off) RB
RC
RC
Verror
RB
RB
RE
-VEE
Input offset voltage
• The RC values may be different.
• VBE for each transistor may be different.
• Other parameters may differ slightly on
each side of the diff amp.
• The input offset voltage would produce
the same error in an ideal amplifier.
• Vin(off) = Verror/A
• Both bases are grounded when Verror is
measured.
Combined dc error
•
•
•
•
•
•
•
V1err = (RB1 - RB2)Iin(bias)
V2err = (RB1 + RB2)Iin(off)/2
V3err = Vin(off)
Verror = A(V1err + V2err +V3err)
V1err eliminated with equal base resistors
V2err reduced with smaller base resistors
V3err reduced with premium op amps
Common-mode gain
+VCC
RC
ACM =
2RE
RC
vout
Vin(CM)
Vin(CM)
RE
-VEE
Common-mode rejection ratio
•
•
•
•
•
CMRR = A/ACM
CMRRdB = 20log CMRR
The higher the CMRR, the better
A typical op amp has CMRRdb = 90 dB
Much interference is common-mode and
a high CMRR means an amplifier will be
effective in rejecting interference
The current mirror
+VCC
VCC - VBE
IR =
R
R
IC = I R
Compensating diode
In ICs, a current mirror sources the tail current
+VCC
RC
vout
R
V2
V1
VCC + VEE - VBE
IC =
R
-VEE
The impedance of a current source is very high.
+VCC
RC
vout
R
V2
V1
Zcurrent source
CMRR =
re’
-VEE
Active loading produces a very high voltage gain.
+VCC
R
vout
V2
V1
-VEE
Loaded diff amps
2RC
vout
Thevenin equivalent circuit
for differential output
RL
vL
RL
vL Thevenin equivalent circuit
for single-ended output
RC
vout