Transcript CH6

BJT Differential Pa
Transistors Q1, Q2 are matched
Responds to differential mode signals
(a) The differential pair with a common-mode input signal
vCM. (b) The differential pair with a “large” differential input
signal.
6-1
BJT Differential Pair
(c)
(d
)
(c) The differential pair with a large input signal of polarity opposite
to that in (b). (d) The differential pair with a small differential input
signal vi.
6-2
BJT Differential Pair
6-3
BJT Differential Pair
Small-Signal Operat
6-4
Differential Voltage Gain
Vd less
DC
DC
Differential output
Single-ended output
Differential output
with emitter
Resistance RE
6-5
Differential Half-Circuit
vC2 = -vC1
voltage gain = gain of ½ cct
(output taken differentially)
Fig. 6.7 Equivalence of the differential amplifier (a) to the two
common-emitter amplifiers in (b). This equivalence applies only for
differential input signals (unless R >> re). Either of the two
common-emitter amplifiers in (b) can be used to evaluate the
differential gain, input differential resistance, frequency response,
and so on, of the differential amplifier (e.g. using hybrid pi model).
6-6
Common-Mode Gain
Analyzing the equiv
Circuit (b)
For output taken differentially
common-mode gain = 0
For output taken single-ended
= 0 and
6-7
Common-Mode Gain (Cont’d)
Consider the case of a collector resistor mismatch
Q1 has load
Q2 has load
(differential component)
(common-mode component)
6-8
(P)
Input Common-Mode Resistance
Input CM resistance is large so we must take into account
CM gain is small, so assume 0V at collector then r0 appears in
parallel with 2R in the emitter and ru appears between input
and ground.
“emitter
reflectance
rule”
6-9
(P)
Example6.1 pg. 502
15V
10k
10k
- vo +
vs
150
200k
150
1mA
6-10
(P)
Example6.1 pg. 502 (Cont’d)
(c) The worst case common-mode gain if the two collector
resistances are accurate to within 1%
where DRC = 0.02 RC (worst case)
(d) The CMRR in dB
Using Ad = 40 from (b) and ACM = 5E-4 from ( c)
(e) The input common mode resistance
(assuming VA = 100 and rm = 10 b r0 )
Eqn. 6.44 p 502