ESE319 Introduction to Microelectronics
Miller Effect
Cascode BJT Amplifier
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
1
ESE319 Introduction to Microelectronics
Prototype Common Emitter
Ignore “low frequency”
capacitors
Circuit
High frequency model
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
2
ESE319 Introduction to Microelectronics
Multisim Simulation
Mid-band gain
∣Av∣= g m RC = 40 mS∗5.1 k =20446.2 dB
Half-gain point
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 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 RC .
Common base and common collector amplifiers do not suffer
from the Miller effect, since in these amplifiers, one side of C 
is connected directly to ground.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
4
ESE319 Introduction to Microelectronics
High Frequency CC and CB Models
B
ground
B
C
E
Common Collector
C  is in parallel with R .
B
C
E
Common Base
C  is in parallel with R .
C
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
5
ESE319 Introduction to Microelectronics
+
V1
-
I
Z
Miller's Theorem
+
V 2= K V1
-
V 1− V 2 V 1− K V 1
V1
I=
=
=
Z
Z
Z
1− K
I 2= I
I 1= I
+
<=>
V1
i c2
=>
V 2 = K V1
-
-
=>
+
V1 V1
Z
Z 1= = =
I 1 I 1− K
V2 V2
Z
Z
Z 2=
=
=−
=
≈Z
−I 2 −I
1
1
−1 1−
K
K if K >> 1
Let's examine Miller's Theorem as it applies to the HF model
for the BJT CE amplifier.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
6
ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis
IC
B
Determine effect of C :

Using phasor notation:
I R =− g m V  I C
or
V o= − g m V  I C  RC
C V
o
V
IR
C
C
g mV 

where
E
Note: The current through C 
depends only on V  !

I C =  V − V o  s C 

I C = V   g m V  RC − I C RC  s C 

2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL

7
ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis II
From slide 7:
I C = V   g m V  RC − I C RC  s C 


Collect terms for I C and V  :

 1 s RC C   I C =  1 g m RC  s C  V

s  1 g m RC  C 
 1 g m RC  s C 
IC =
V =
V
 1 s RC C  
 1 s RC C  

Miller Capacitance Ceq: C eq =1− K  C =1 g m RC 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
8
ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis III
C eq = 1 g m RC  C 
For our example circuit:
1 g m RC =10.040⋅5100=205
C eq =205⋅2 pF ≈410 pF
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
9
ESE319 Introduction to Microelectronics
Apply Miller's Theorem to BJT
CE Amplifier
Z
Z 1=
1− K
For the BJT CE Amplifier: Z =
=>
Z 1=
1
j 1 g m RC  C 
1
j C
and
Z 2=
Z
1−
and K =− g m RC
Z 2=
1
j 1
1
C
g m RC 
1
K
≈
1
j C
C eq =1 g m RC  C 
Miller's Theorem => important simplification to the HF BJT CE Model
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
10
ESE319 Introduction to Microelectronics
Simplified HF Model
R'sig B' I C 
.≈0
Rsig
V sig
B rx
B'
++
r
RB
V
C
C
C
-
gmV
ro
E
R'sig B' I C 
V
C
-
C
-
C eq
gmV
C
(b)
RC RL V'sig
-
V'sig = V sig
gmV
'
L
R
R'sig = r ∥ RB∥Rsig  V o
V sig
+
Vo
-
−3 dB
20 log 10 AM
0
(c)
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
(d)
≈
-6 dB/octave
-20 dB/decade
f =
H
C in
C in= C C eq =C  C  1 g m R'L
R'L
+ R'L= r o∥ RC∥ RL
Vo
-
RB∥r 
R B∥r  Rsig
|Vo/Vsig| (dB)
++
V
V
R'L
(a)
Thevenin
'
sig
++
'
sig
+
V
C
− g m R'L
'
1 j  C in Rsig
1
'
2  C in Rsig
f  f (Hz log scale)
H
11
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 (based on P. V. Lopresti 2006) update 15Oct08 KRL
12
ESE319 Introduction to Microelectronics
The Cascode Circuit
i C1
i E1
i B2
i c2
v-out
i B1
CB Stage
CE Stage
i b2
i C2
i e2
i e1
i c1
i b1
Rb
i E2
RB= R2∥ R3
Rin1 =
veg1 vcg2
=
= low
i e1
i c2
ac equivalent circuit
Comments:
1. R1, R2, R3, and RC set the bias levels for both Q1 and Q2.
2. Determine RE for the desired voltage gain.
3. Cb and Cbyp are to act as “open circuits” at dc and act as “short circuits”
at all operating frequencies of interest, i.e. min.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
13
ESE319 Introduction to Microelectronics
Cascode Mid-Band Small Signal Model
i c1
Vout
i b1
i e1
i b2
Rb
i e2
i c2 Rin1= low
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:
i e1
i c2
i b1=
=
1 1
c. Hence, both stages have about same
collector current i c1≈ ic2 and same gm.
g m1= g m2= g m
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
14
ESE319 Introduction to Microelectronics
Cascode Small Signal Analysis cont.
i c1
Vout
i b1
i e1
i b2
Rb
i e2
Rin1= low
i c2
Rc
The input resistance Rin1 to the CB stage
is the small-signal “RC” for the CE stage
i e1
ic2
i b1=
=
1 1
The CE output voltage, the voltage drop
from Q2 collector to ground, is:
r 1
r 1
vcg2 = veg1=−r 1 i b1=−
i c2=−
i e1
1
1
Therefore, the CB Stage input resistance is:
veg1 r  1
Rin1=
=
=r e1
−i e1 1
vcg2
Rin1
r e1
r
AvCE− Stage =
≈−
=− 1 => C eq =1 e1  C  2 C 
vsig
RE
RE
RE
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
15
ESE319 Introduction to Microelectronics
Cascode Small Signal Analysis - cont.
i c1
i b1
i e1
i b2
Rb
i e2
i c2
Vout
Now, find the CE collector current in terms of
the input voltage Vsig: Recall i c1≈ ic2
vsig
i b2≈
Rsig∥RBr  21 RE
 vsig
 vsig
i c2= i b2 ≈
≈
Rsig∥RB r  2 1 RE 1 RE
for bias insensitivity: 1 RE ≫ Rsig∥RBr  2
vsig
i c2≈
RE
,
=>
OBSERVATIONS:
1. Voltage gain Av is about the same as a stand-along CE Amplifier.
2. HF cutoff is much higher then a CE Amplifier due to the reduced Ceq.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
16
ESE319 Introduction to Microelectronics
Approximate Cascode HF Voltage Gain
RC
−
RE
V out
Av =
≈
'
V sig
1 j  C in Rsig
where
r e1
C in =C  C eq =C 1  C C 2 C 
RE
'
Rsig = Rsig∥RB≈ Rsig
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
17
ESE319 Introduction to Microelectronics
Cascode Biasing
I1
IC1
v-out
IE1
IC2
IE2
Rin1= low
1. Choose IE1 – make it relatively large to
reduce Rin 1 = r e1=V T / I E1 to push out HF
break frequencies.
2. Choose RC for suitable voltage swing
VC1G and RE for desired gain.
3. Choose bias resistor string such that
its current I1 is about 0.1 of the collector
current IC1.
4. Given RE, IE2 and VBE2 = 0.7 V calculate
R3.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
18
ESE319 Introduction to Microelectronics
Cascode Biasing - cont.
I1
VB1G
VB2G
v-out
VC2G
VCE2 =V C2G −V Re= V C2G−V B2G−0.7 V 
. =V B1G −V BE1−V B2G V BE2
.≈ V B1G−0.7 V −V B2G0.7 V
. =V B1G −V B2G
Since the CE-Stage gain is very small:
a. The collector swing of Q2 will be small.
b. The Q2 collector bias VC2G= VB1G - 0.7 V.
5. Set V B1G−V B2G ≈1 V ⇒ VCE2≈1 V
This will limit VCB2 VCB2=V CE2−V BE2=0.3 V
which will keep Q2 forward active.
6. Next determine R2. Its drop VR2 = 1 V
V B1G −V B2G
with the known current.
R2 =
I1
VCC −V B1G
7. Then calculate R1. R1 =
I1
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
19
ESE319 Introduction to Microelectronics
Cascode Bias Example
VCC-ICRE-1.7
v-out
ICRC
VCE1=ICRC–1–ICRE
1.0
=12 V
VCE2=1
ICRE+0.7
Cascode circuit
=12 V
ICRE
I E2≈ I C2= I E1≈ I C1 ⇒ I C1≈ I E2
Typical Bias Conditions
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
20
ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
VCC-ICRE-1.7
1. Choose IE1 – make it a bit high to lower re1
or r  1. Try IE1 = 5 mA => r e1 =0.025 V / I E =5  .
ICRC
VCE1=ICRC–1–ICRE
1.0
VCE2=1
ICRE+0.7
ICRE
=12 V
2. Set desired gain magnitude. For example
if |AV| = 10, then RC/RE = 10.
3. Since the CE stage gain is very small,
VCE2 can be small. Use VCE2 = VB1G – VB2G = 1 V.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
21
ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
VCC-ICRE-1.7
ICRC
VCE1=ICRC–1–ICRE
1.0
VCE2=1
ICRE+0.7
ICRE
V CC =12
I C =5 mA.
RC
∣Av∣= R =10
E
Determine RC for a 5 V drop across RC.
5V
=1000 
−3
5⋅10 A
RC RC
RE =
= =100 
∣Av∣ 10
RC =
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
22
ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
I1
V CC =12
VCC-ICRE-1.7
ICRC
VCE1=ICRC–1–ICRE
1.0
VCE2=1
ICRE+0.7
ICRE
RC =1 k  I C =5 mA. RE =100
Make current through the string of bias
resistors I1 = 1 mA.
VCC
12
R1  R2 R3=
=
=12 k 
−3
I 1 1⋅10
We now calculate the bias voltages:
V CC − I C RE −1.7 V =12 V −0.5 V −1.7 V =9.8 V
V B1B2=1.0 V
V B2G= I C RE 0.7=5⋅10−3⋅1000.7=1.2 V
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
23
ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
V B2G=10−3 R3=1.2 V
v-out
R3=1.2 k 
V B1G−V B2G=1⋅10−3 R2=1.0 V
R2=1 k 
R1 =12000−1200−1000=9.8 k 
R1 =10 k 
V CC =12 RC =1 k 
V B2G=1.2 V
I C =5 mA. RE =100 V B1G−V B2G=1.0 V
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
24
ESE319 Introduction to Microelectronics
Multisim Results – Bias Example
IC1
V CC −7.3290.9710.339 12 V −8.639 V
=
=3.36 mA
Check IC1: I C1=
RC
1000 
IC1 about 3.4 mA. That's a little low.
Increase R3 to 1.5 k Ohms and re-simulate.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
25
ESE319 Introduction to Microelectronics
Improved Biasing
10 uF
10 uF
V CC −5.0480.9410.551 12 V −6.540 V
Check IC1: I C1=
=
=5.46 mA
RC
1000 
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 (based on P. V. Lopresti 2006) update 15Oct08 KRL
26
ESE319 Introduction to Microelectronics
Single Frequency Gain
10 uF
v-out
10 uF
Gain |Av| of about 8.75 at 100 kHz. - OK for rough calculations. Some attenuation from low CB input impedance (RB = R2||R3)and some from 5 Ohm re.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
27
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 (based on P. V. Lopresti 2006) update 15Oct08 KRL
28
ESE319 Introduction to Microelectronics
Scope Plot – Near 5 V Swing on Output
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
29
ESE319 Introduction to Microelectronics
Determine Bypass Capacitors
Low Frequency f ≤ fmin
v-out
From CE stage determine Cb
Cb ≥
10
10
≈
F
2  f min RB∥r bg 2 f min RB∥1 RE
RB= R2∥R3
From CB stage determine Cbyp
10
C byp≥
F
2  f min  1 RE  r 1 
where RS -> RE
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL
30