DIFFERENTIAL AMPLIFIER, SINGLE-ENDED OUTPUT
Analysis of a symmetrical differential single-ended output amplifier
THE CIRCUIT
VCC
RC
SINGLE-ENDED OUTPUT AMPLIFIER
RC
vout
vi1
Q1
Q2
R
IBIAS
vi2
Differential amplifiers are used in integrated circuit
design. In IC design, large capacitors and large (Ω)
resistors are impractical.
-VEE
DC ANALYSIS
VCC
Since the circuit is symmetrical, we are looking at only ½ the
circuit, through which ½ the bias current flows. The bias
resistance R is ignored. We find the value of re for our AC circuit
model using these formulas:
RC
C
IB
B
IC
IC =
β
β
I E = 0.5
I BIAS
β +1
β +1
g m = 40 I C
IE
rπ =
β
gm
re =
E
rπ
β +1
0.5 IBIAS
-VEE
AC ANALYSIS
For the AC analysis, we divide the input signals vi1 and vi2 into differential and common-mode
components. The inputs are written as
vi1 = vi cm +
vid
2
vi 2 = v i cm −
v id
2
We solve for the differential output vd and the common mode output vcm separately using different
models and then combine the results.
Tom Penick
tomzap@eden.com
www.teicontrols.com/notes 12/06/98
DIFFERENTIAL MODE ANALYSIS - solving for the differential mode voltage gain Ad
DM MODEL
Since the circuit is symmetrical and the differential mode inputs are
180° out of phase, a ground potential results at the emitters. The two
halves of the circuit are isolated. Since there is an output only on the
right-hand half, this is the only part of the circuit we include in this
analysis.
RC
ic
β ib
vout
ie = −
vid
2
vi d
2
÷ re = −
v d out = −ic RC =
ie
re
vi d
ic =
2re
β  vi d

β + 1  2re

 RC

β  vi d
−
β + 1  2re
Ad =



v d out
vi d
Ad is the differential mode gain [V/V]
COMMON MODE ANALYSIS - solving for the common mode voltage gain Acm
AC ANALYSIS
RC
RC
Q1
2R
ie =
vi cm
re + 2 R
Q2
CM MODEL
RC
ic
βib
vi cm
Even though we have looked only
at the right-hand side in both
phases of the analysis, the
presence of the left-hand side
does significantly affect the
outcome.
2R
ic =
Because of the symmetry of the
circuit, we can write R as two
parallel 2R resistors and observe
that no current flows between
them. Therefore this connection
can be eliminated and we
separate the circuit into halves.
Once again, we need only look at
the right-hand half since that
contains the output.
β  vi cm 


β + 1  re + 2 R 
vcm out = −ic RC =
β  vid

β + 1  2re
vout
ie
re
2R

 RC

Acm =
vcm out
vi cm
COMMON MODE REJECTION RATIO (CMRR)
CMRR =
Ad
Acm
CMRR dB = 20 log 10
Tom Penick
tomzap@eden.com
Ad
Acm
www.teicontrols.com/notes 12/06/98