Download Analysis of a simple BJT OpAmp INEL 4202

Analysis of a simple BJT OpAmp
INEL 4202 - Electronics II
+15V
Q3
v1
QD
Q4
Q1
Q2
QE
QF
Q5
v2
Q6
RA
vOUT
QC
QA
RC3
QB
RE4
QC
-15V
Differential-mode analysis
Assume β = 100. Apply v1 = +vd /2, v2 = −vd /2. Invoke virtual ground at the emitter
of Q1,2 :
ic1
ic2
vb5
vb5
vd
= gm1 (vd /2) = ic3 = ic4
= gm2 (−vd /2) ⇒ io1 = ib5 = ic4 − ic2 = gm1 (vd /2) − gm1 (−vd /2) = gm1 vd
= ib5 (ro2 ∥ro4 ∥Rin,b5 ) = gm1 vd (ro2 ∥ro4 ∥Rin,b5 )
= gm1 (ro2 ∥ro4 ∥Rin,b5 )
For simplicity, let VA = 100V for both pnp and npn transistors, so that ro2 = ro4 =
VA
.
ICQ1
Then
(
)
ICQ,1 1 VA
vb5
=
∥Rin,b5
vd
Vt
2 ICQ1
−gm5 (rOE ∥Rin,b6 )
vc5
=
vb5
1 + gm5 RC3
vout
≃ 1
vc5
1
Rin,b5 = rπ,5 + (101)RC3
VA
ICQ2
=
The design equations are
vout
ICQ,1
=
vd
Vt
(
1 VA
∥Rin,b5
2 ICQ1
)(
−gm5 (rOE ∥Rin,b6 )
1 + gm5 RC3
)
vout (vd = 0) = IQ6 RE4 − 15V
rπ,6 + rOE
Rout = RE4 ∥
101
Rin = 2rπ,1 = 2rπ,2
Now we can begin assigning values. Select
RE4 so that vout (vd = 0) = 0V
a large ICQ,6 so that the output stage can drive large loads
ICQ,6 = 10mA
rπ,6 =
RE4 = 15V /10mA = 1.5kΩ
100(0.025V )
= 250Ω
10mA
a small ICQ,1 so that Rin is large;
Rin = 2
100(0.025)
⇒ ICQ,1 = 10µA ⇒ Rin = 500kΩ
ICQ,1
let RC3 = 10kΩ for a large Rin,b5
Rin,b5 = rπ,5 + (101)RC3 ≃ 101(10kΩ) ≃ 1M Ω
The gain is
(
)(
−gm5 (rOE ∥151.5kΩ)
1 + gm5 10kΩ


ICQ,5 100V
− 0.025V
( ICQ,5 ∥151.5kΩ)

≃ (333.3) 
ICQ,5
1 + 0.025V 10kΩ
vout
10µA
≃
vd
0.025V
1 100V
∥1M Ω
2 10µA
)
select ICQ,5 = 50µA; then
vout
≃ (333.3)
vd
(
−281.7
50µA
1 + 0.025V
10kΩ
)
= 333.3 ×
−281.7
= 4471V /V
21
The output resistance is
Rout = RE4 ∥
100V /50µA
rπ,6 + rOE
≃ 1.5kΩ∥
= 1.4kΩ
101
101
which is a large value. A second common-collector stage could be added at the output
to further reduce this value.
Biasing
Assume ICB = ICA (same transistor area in QA and QB )
ICB = ICQ,1 + ICQ,2 = 20µA = ICA ≃
15V − 1.4V
⇒ RA = 680kΩ
RA
Use an area 5 times larger for (relative to QA ) QC,D,E so that ICQ,5 = 50µA as, desired.
Transistor QF will have to be 200 times larger than QD to achieve ICQ,6 = 10mA.
The input bias current is ICQ,1/β = ICQ,2/β = 10µA/100 = Ibias = 100nA .
Common-mode rejection ratio (CMRR)
For our selection for IQB = 20µA, the current source’s Norton resistance for the differential stage is RCS = 100V /20µA = 5M Ω.
Using the technique previously shown in the handout for differential amplifiers,
vcm
vcm
= 100 ×
≃ vcm/10M Ω
rπ,1 + (β + 1)(2RCS )
250kΩ + 101 × 10M Ω
ic1,cm = ic2,cm = β
If the active load is assumed perfect (by neglecting the base currents for Q3 and Q4 ) the
differential-stage’s output current is zero. If the base currents are taken into account,
ic4,cm =
β
ic1,cm
β+2
and the differential-stage’s output current is
(
)
β
vcm
−2 vcm
io1,cm = ic4,cm − ic2,cm =
−1
=
β+2
10M Ω
102 10M Ω
Since from this point on, the same loads and gains affect both the differential- and commonmode signals, we can compute the CMRR without actually finding the common-mode voltage
gain.
io1,dm/v
CM RR = | i
d
/vcm
o1,cm
|=
gm1 (102)10M Ω
10µA
=
510M Ω = 204000 = 106dB
2
0.025V