Miller Effect Cascode BJT Amplifier

ESE319 Introduction to Microelectronics
Miller Effect
Cascode BJT Amplifier
2008 Kenneth R. Laker,update 21Oct09 KRL
1

ESE319 Introduction to Microelectronics
Prototype Common Emitter Circuit
Ignore “low frequency”
capacitors
High frequency model
2008 Kenneth R. Laker,update 21Oct09 KRL
2

ESE319 Introduction to Microelectronics
Multisim Simulation
∣A v ∣=g m
R C
=40 mS ∗5.1k =20446.2 dB
Mid-band gain
Half-gain point
2008 Kenneth R. Laker,update 21Oct09 KRL
3

ESE319 Introduction to Microelectronics
Introducing the Miller Effect
The feedback connection of C between base and collector
causes it to appear to the amplifier like a large capacitor 1−K C
has been inserted between the base and emitter terminals. This
phenomenon is called the “Miller effect” and the capacitive multiplier
“1 – K ” acting on C
equals the common emitter amplifier mid-band
gain, i.e. K =−g m R C .
Common base and common collector amplifiers do not suffer
from the Miller effect, since in these amplifiers, one side of
is connected directly to ground.
C
2008 Kenneth R. Laker,update 21Oct09 KRL
4

ESE319 Introduction to Microelectronics
High Frequency CC and CB Models
B
C
ground
B
C
E
E
Common Collector
C is in parallel with R B
.
Common Base
C is in parallel with R C
.
2008 Kenneth R. Laker,update 21Oct09 KRL
5

I Z
+ +
- -
ESE319 Introduction to Microelectronics
Miller's Theorem
I 1 =I I 2 =I
+ +
Z 1
Z 2
V 1
V 2 =K V 1
- -
V 1 V 2 =K V 1
if K >> 1
I = V 1−V 2
Z
= V 1−K V 1
= V 1
Z Z
1−K
=>
Z 1 = V 1
= V 1
I 1 I = Z
1−K
Z 1
≈
1
j 2 f C
1−K
=>
Z 2 = V 2
= V 2
−I 2 −I =−
Z =
Z
1
K −1 1− 1 ≈Z
K
Let's examine Miller's Theorem as it applies to the HF model
for the BJT CE amplifier.
2008 Kenneth R. Laker,update 21Oct09 KRL
6

ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis
Determine effect of C :
B
V
C
V o
I C
I RC
I RC
=−g m
V
I C
Using phasor notation:
or
Note: The current through
depends only on !
V
E
g m
V
C
V o
=−g m
V
I C R C
where
I C
=V
−V o sC
V o
2008 Kenneth R. Laker,update 21Oct09 KRL
7

ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis II
From slide 7:
I C
=V
g m V
R C −I C
R C s C
Collect terms for and :
I C
V
1s R C C I C
=1g m R C sC
V
I C
= 1g m R C sC
1s R C C
V
= s 1g m R C C
1s R C C
V
Miller Capacitance C eq
: C eq
=1−K C
=1g m
R C
2008 Kenneth R. Laker,update 21Oct09 KRL
8

ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis III
C eq =1g m R C C
For our example circuit:
1g m
R C
=10.040⋅5100=205
C eq
=205⋅2 pF≈410 pF
2008 Kenneth R. Laker,update 21Oct09 KRL
9

ESE319 Introduction to Microelectronics
Apply Miller's Theorem to BJT CE Amplifier
For the BJT CE Amplifier: Z = 1
j C
Z 1
=
Z
1−K
1
=> Z 1
=
and Z
j 1g m R C C 2
=
C eq =1g m R C C
and K=−g m R C
Z 2
=
Z
1− 1 K
1
j 1 1 ≈ 1
j C
C
g m
R
C
Miller's Theorem => important simplification to the HF BJT CE Model
2008 Kenneth R. Laker,update 21Oct09 KRL
10

ESE319 Introduction to Microelectronics
Simplified HF Model
R s
C
V s
R B
r
C
g m V
r o
R C
R L
V o
V
V
C
'
R L
'
R s
I C
R L
'
=r o ∥R C ∥R L
V s ' =V s
R B ∥r
R B
∥r
R sig
'
V s
C
'
R L
V o
R s ' =r ∥ R B ∥R s
Thevenin
g m V
2008 Kenneth R. Laker,update 21Oct09 KRL
11

ESE319 Introduction to Microelectronics
Simplified HF Model
'
V s
Miller's Theorem
'
R s
I C
C in =C C eq
'
.=C
C
1g m R L
V s
'
C
C eq
'
R L
V o
(c)
C in
g m V
2008 Kenneth R. Laker,update 21Oct09 KRL
12

ESE319 Introduction to Microelectronics
Simplified HF Model
'
V s
A v f = V '
o −g m R L
≈
'
V s 1 j 2 f C in R s
f H
=
1
2C in
R s
'
∣ V o
V s∣ dB
2008 Kenneth R. Laker,update 21Oct09 KRL
13

ESE319 Introduction to Microelectronics
The Cascode Amplifier
A two transistor amplifier used to obtain simultaneously:
1. Reasonably high input impedance.
2. Reasonable voltage gain.
3. Wide bandwidth.
None of the conventional single transistor designs will meet
all of the criteria above. The cascode amplifier will meet all
of these criteria. a cascode is a combination of a common
emitter stage cascaded with a common base stage. (In “olden
days” the cascode amplifier was a cascade of grounded
cathode and grounded grid vacuum tube stages – hence the
name “cascode,” which has persisted in modern terminology.
2008 Kenneth R. Laker,update 21Oct09 KRL
14

C byp
v s
R 1
R 2
ESE319 Introduction to Microelectronics
The Cascode Circuit
R 3
i B1
i E1
i B2
i C1
R C
i C2
v-out
i c2
i b2
i c1
i e2
i
V b1
CC
R B
i E2
R E
v o
R B
=R 2
∥R 3
CE Stage
ac equivalent circuit
Comments:
1. R 1
, R 2
, R 3
, and R C
set the bias levels for both Q1 and Q2.
2. Determine R E
for the desired voltage gain.
3. C b
and C byp
are to act as “open circuits” at dc and act as “short circuits”
at all operating frequencies of interest, i.e. f f min .
i e1
CB Stage
R s
C b
v s
R s
R E
R C
R in1 = v eg1
i e1
= v cg2
i c2
=low
v o
2008 Kenneth R. Laker,update 21Oct09 KRL
15

ESE319 Introduction to Microelectronics
Cascode Mid-Band Small Signal Model
i b1
R B
r 1
i e1
i b2
r 2
i e2
i c2
i c1
g m
v be1
R in1 =low
Vout
R s
R
g C
m
v be2
v s
R E
v o
1. Show reduction in Miller effect
2. Evaluate small-signal voltage gain
OBSERVATIONS
a. The emitter current of the CB stage is
the collector current of the CE stage. (This
also holds for the dc bias current.)
i e1 =i c2
b. The base current of the CB stage is:
g m1
=g m2
=g m
r e1 =r e2 =r e
r 1
=r 2
=r
2008 Kenneth R. Laker,update 21Oct09 KRL
i b1 = i e1
1 = i c2
1
c. Hence, both stages have about same
collector current i c1
≈i c2 and same g m
, r e
, r π
.
16

ESE319 Introduction to Microelectronics
Cascode Small Signal Analysis cont.
The input resistance R in1
to the CB stage
is the small-signal “R C
” for the CE stage
i b1 = i e1
1 = i c2
1
The CE output voltage, the voltage drop
from Q2 collector to ground, is:
v cg2
=v eg1
=−r 1
i b1
=− r 1
1 i c2=− r 1
1 i e1
Therefore, the CB Stage input resistance is:
R in1
= v eg1
−i e1
= r 1
1 =r e1
A vCE−Stage
= v cg2
v sig
≈− R in1
R E
=− r e
R E
1 => C eq
=1 r e
R E
C
2C
2008 Kenneth R. Laker,update 21Oct09 KRL
17

ESE319 Introduction to Microelectronics
Cascode Small Signal Analysis - cont.
Now, find the CE collector current in terms of
the input voltage v s
: Recall i c1
≈i c2
i b2 ≈
v s
R s ∥R B r 2 1 R E
i c2 =i b2 ≈
v s
≈
v s
R s ∥R B r 2
1 R E 1 R E
for bias insensitivity:
1 R E
≫R s
∥R B
r
v s
OBSERVATIONS:
1. Voltage gain A v
is about the same as a stand-along CE Amplifier.
2. HF cutoff is much higher then a CE Amplifier due to the reduced C eq
.
2008 Kenneth R. Laker,update 21Oct09 KRL
18

ESE319 Introduction to Microelectronics
Approximate Cascode HF Voltage Gain
A v
f = V − R C
o f
V s f ≈ R E
'
1 j 2 f C in R s
where
C in
=C
C eq
=C
1 r e
C
R
C
2C
E
R ' s =R s ∥R B ≈R s
2008 Kenneth R. Laker,update 21Oct09 KRL
19

ESE319 Introduction to Microelectronics
I 1
I C1
I E1
Cascode Biasing
R in1 =low
I C2
I E2
1. Choose I E1
– make it relatively large to
reduce R in1 =r e1 =V T / I E1 to push out HF
break frequencies.
2. Choose R C
for suitable voltage swing
V C1G
and R E
for desired gain.
3. Choose bias resistor string such that
its current I 1
is about 0.1 of the collector
current I C1
.
4. Given R E
, I E2
and V BE2
= 0.7 V calculate
R 3
.
2008 Kenneth R. Laker,update 21Oct09 KRL
20

I 11
V B1
ESE319 Introduction to Microelectronics
Cascode Biasing - cont.
Since the CE-Stage gain is very small:
a. The collector swing of Q2 will be small.
b. The Q2 collector bias V C2
= V B1
- 0.7 V.
5. Set
V B1 −V B2 ≈1V ⇒V CE2 ≈1V
V B2
V C2
This will limit V CB2
V CB2 =V CE2 −V BE2 =0.3V
which will keep Q2 forward active.
V CE2 =V C2 −V R e =V C2 −V B2 −0.7 V
.=V B1 −V BE1 −V B2 V BE2
.≈V B1 −0.7V −V B2 0.7V
.=V B1 −V B2
2008 Kenneth R. Laker,update 21Oct09 KRL
6. Next determine R 2
. Its drop V R2
= 1 V
with the known current.
7. Then calculate R 1
.
R 2 = V B1−V B2
I 1
R 1 = V CC−V B1
I 1
21

ESE319 Introduction to Microelectronics
Cascode Bias Example
V CC
-I C
R E
-1.7
R C
I C
R C
=12 V
1.0
I C
R E
+0.7
V CE1 =V =I C
R CC C
–1–I −I C R EC
−1−I C R E
=12 V
V CE2
=1
R E
I C
R E
Cascode circuit
I E2 ≈ I C2 =I E1 ≈ I C1 ⇒ I C1 ≈I E2
Typical Bias Conditions
2008 Kenneth R. Laker,update 21Oct09 KRL
22

ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
V CE1 =V CC −I C R C −1−I C R E
1. Choose I E1
– make it a bit high to lower r e
or . Try I E1
= 5 mA => .
r
r e =0.025V / I E =5
2. Set desired gain magnitude. For example
if A V
= -10, then R C
/R E
= 10.
3. Since the CE stage gain is very small,
V CE2
can be small. Use V CE2
= V B1
– V B2
= 1 V.
2008 Kenneth R. Laker,update 21Oct09 KRL
23

ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
V CC
=12
I C
=5 mA.
∣A v ∣= R C
R E
=10
V CE1 =V CC −I C R C −1−I C R E
Determine R C
for a 5 V drop across R C
.
R C
=
5V
5⋅10 −3 A =1000
R E
= R C
∣A v ∣ = R C
10 =100
2008 Kenneth R. Laker,update 21Oct09 KRL
24

ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
I 1
=12V
V CC
=12
I C
=5 mA.
R C
=1 k R E
=100
R 1
V CC
-I C
R E
-1.7
R 2
R 3
1.0
I C
R E
+0.7
R C
R E
I C
R C
V CE1 =V =I C
R CCC −I –1–I CC R CE
−1−I C R E
V CE2
=1
I C
R E
Make current through the string of bias
resistors I 1
= 1 mA.
R 1
R 2
R 3
= V CC
= 12
I 1 1⋅10 −3=12k
We now calculate the bias voltages:
V CC
−I C
R E
−1.7 V =12 V −0.5V −1.7 V =9.8V
V B1B2
=V B1
−V B2
=1.0V
V B2
=I C
R E
0.7=5⋅10 −3 ⋅1000.7=1.2V
2008 Kenneth R. Laker,update 21Oct09 KRL
25

ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
V B2
=10 −3 R 3
=1.2V
V B1
V B2
R 3
=1.2 k
V B1
−V B2
=1⋅10 −3 R 2
=1.0V
R 2
=1 k
R 1
=12000−1200−1000=9.8k
V CC
=12 R C
=1 k
I C
=5 mA. R E
=100
V B2
=1.2 V
V B1
−V B2
=1.0V
R 1
=10 k
2008 Kenneth R. Laker,update 21Oct09 KRL
26

ESE319 Introduction to Microelectronics
Multisim Results – Bias Example
R 1
I C1
R C
C byp
v o
R s
C b
R 2
v s
R 3
R E
Check I C1
:
I C1
about 3.4 mA. That's a little low.
Increase R 3
to 1.5 k Ohms and re-simulate.
2008 Kenneth R. Laker,update 21Oct09 KRL
I C1
= V CC−7.3290.9710.339
=
R C
12V −8.639 V
1000
=3.36 mA
27

ESE319 Introduction to Microelectronics
Improved Biasing
R 1
R C
C byp
R s
10 uF
C b
R 2
v s
10 uF
R 3
R E
Check I C1
: I C1
= V CC−5.0480.9410.551
That's better! Now measure the gain at a mid-band frequency
with some “large” coupling capacitors, say 10 µF inserted.
2008 Kenneth R. Laker,update 21Oct09 KRL
R C
12V −6.540 V
=
1000
=5.46 mA
28

ESE319 Introduction to Microelectronics
Single Frequency Gain
10 uF
C byp
R 1
R C
v-out
v o
R s
v s
R 2
R 3
10 uF
C b
R E
Gain |A v
| of about 8.75 at 100 kHz - OK for rough calculations. Some attenuation
from low CB input impedance (R B
= R 2
||R 3
)and some from 5Ω r e
.
2008 Kenneth R. Laker,update 21Oct09 KRL
29

ESE319 Introduction to Microelectronics
Bode Plot for the Amplifier
18.84 dB
Low frequency break point with 10 µF. capacitors
18.84 dB
High frequency break point – internal capacitances only
2008 Kenneth R. Laker,update 21Oct09 KRL
30

ESE319 Introduction to Microelectronics
Scope Plot – Near 5 V Swing on Output
2008 Kenneth R. Laker,update 21Oct09 KRL
31

ESE319 Introduction to Microelectronics
Determine Bypass Capacitors
Low Frequency f ≤ f min
v-out
From CE stage determine C b
C b
≥
10
2 f min R B ∥r bg
≈
R B
=R 2
∥R 3
10
2 f min R B ∥1 R E
F
From CB stage determine C byp
10
C byp
≥
2 f min 1 R E
r F
where R S
-> R E
2008 Kenneth R. Laker,update 21Oct09 KRL
32