ESE319 Introduction to Microelectronics

High Frequency BJT Model &

Cascode BJT Amplifier

Kenneth R. Laker, update 01Oct14 KRL 1

 ESE319 Introduction to Microelectronics

Gain of 10 Amplifier – Non-ideal Transistor

RC
R1
C in
V CC
RS
R2 RE
vs Gain starts dropping at > 1MHz.

Why!
Because of internal transistor
capacitances that we have ignored
in our low frequency and mid-band models.

Kenneth R. Laker, update 01Oct14 KRL 2

 ESE319 Introduction to Microelectronics

Sketch of Typical Voltage Gain Response

for a CE Amplifier
∣Av∣dB
Low High
Frequency Midband
ALL capacitances are neglected, i.e. Frequency
Band External C's s.c. Band
Internal C's o.c.
Due to external
blocking and by- 3 dB Due to BJT parasitic
pass capacitors. capacitors Cπ and Cµ.
Internal C's o.c. External C's s.c.
20 log10∣Av∣ dB

f  Hz
fL f H (log scale)
BW = f H − f L≈ f H GBP=∣ Av−mid∣ BW
Kenneth R. Laker, update 01Oct14 KRL 3
 ESE319 Introduction to Microelectronics

High Frequency Small-signal Model (Fwd. Act.)

rx C

C

vbe

Two capacitors and a resistor added to mid-band small signal model.

● base-emitter capacitor, C
µ
● base-collector capacitor, Cπ
● resistor, rx, representing the base terminal resistance (rx << rπ);
ignored in hand calculations.
Kenneth R. Laker, update 01Oct14 KRL 4
 ESE319 Introduction to Microelectronics

High Frequency Small-signal Model (Fwd. Act.)

SPICE/MultiSim
rx C CJC = Cµ0
CJE = Cje0
C TF = τF
RB = rx
vbe

C =C de C je≈C de 2 C je 0 ≈C de
C0 C je0
C = C je = IC
V CB
m
V EB
m
C de=g m  F =  F Note C= i dt
1  1  VT dv
0cb 0eb C ≈2 C 0
Cde = base-charging (diffusion) cap
Non-linear, voltage controlled
m = junction grading coefficient 0.2 to 0.5) τF = forward-base transit time
Kenneth R. Laker, update 01Oct14 KRL 5
 ESE319 Introduction to Microelectronics

High Frequency Small-signal Model (IC)

Cπ
Cµ

CCS = collector-substrate depletion capacitance, ignored in hand calcs.

Kenneth R. Laker, update 01Oct14 KRL 6
 ESE319 Introduction to Microelectronics
High Frequency Small-signal Model
The transistor parasitic capacitances have a strong effect on circuit high
frequency performance!
● Cπ attenuates base signal, decreasing vbe since XCπ → 0 (short circuit)
at very high-frequencies.
●

● As we will see later; Cµ is principal cause of gain loss at high-frequencies.

At the base base-collector Cµ looks like a capacitor of value k Cµ between
base-emitter, where k > 1 and may be >> 1.
rx C

C

● This phenomenon is called the Miller Effect.

Kenneth R. Laker, update 01Oct14 KRL 7
 ESE319 Introduction to Microelectronics

Prototype Common Emitter Circuit

V CC
C
RC RS b c
R S C in vo vo
+
v be
RB vs RB r C g m v be RC
vs -
V B RE C byp
e

At high frequencies High frequency

large blocking and bypass small-signal ac model
capacitors are “short circuits”

Kenneth R. Laker, update 01Oct14 KRL 8

 ESE319 Introduction to Microelectronics

Multisim Simulation
C
2 pF
RS
50  b c vo

RB v be
vs r C g m v be RC Mid-band gain @ 100 kHz
50 k  2.5 k  12 pF 40 mS v be 5.1 k 
e

Av−mid =−g m R C =−204 46 dB @ 180 o 

Gain @ 8.69 MHz

Kenneth R. Laker, update 01Oct14 KRL 9

 ESE319 Introduction to Microelectronics

Introducing the Miller Effect

C
RS b c vo

vs RB r C v be g m v be RC

The feedback connection of C  between base-collector causes it to appear

in the amplifier like a large capacitor 1− K C  between the base-emitter
terminals. This phenomenon is called the “Miller effect” and capacitor
multiplier “1 – K ” acting on C  equals the CE amplifier mid-band gain, i.e.
K = A v−mid =−g m R C
NOTE: CB and CC amplifiers do not suffer from the Miller effect, since in
these amplifiers, one side of C  is connected directly to ground.
Kenneth R. Laker, update 01Oct14 KRL 10
 ESE319 Introduction to Microelectronics

Miller's Theorem 1
Z=
C j 2  f C
RS b c I Z −I
vo
+ +
i v -i
vs RB r C be g m v be RC <=> V be Vo
Av−mid V be
- -
e
R B∥r  ≫ R S Vo
=A v−mid =−g m RC
V be
v o vo
Av−mid = ≈ =−g m RC
v s v be

Kenneth R. Laker, update 01Oct14 KRL 11

 ESE319 Introduction to Microelectronics

Miller's Theorem
I =I1 Z I 2 =−I I 1= I I 2 =−I
+ + + +
V1 Av V 1 V2 <=> V1 Z1 Z2 V 2 =Av V 1
- - - -

V 1−V 2 V 1− Av V 1 V1 V1 V1 Z
Z 1= = =
I = I 1=
Z
=
Z
=
Z
=> I1 I 1− Av
1− Av 1
Z 1=
1 j 2  f C  1− Av 
V 2− V 2
V 2 −V 1 Av V2 V2 V2 Z
−I =I 2= = = Z 2= = = ≈ Z if A >> 1
Z Z Z => I 2 −I 1 v
1 1−
1− Av
Av Ignored in
V2 V2 1
Z 2= = ≈ practical
I 2 −I j 2 f C  circuits

Kenneth R. Laker, update 01Oct14 KRL 12

 ESE319 Introduction to Microelectronics

Common Emitter Miller Effect Analysis

Determine effect of C :
IC Using phasor notation:
C 

RS b V be c I R =−g m V be I C
Vo C 

IR or
V be V o =I R RC =−g m V be I C  RC
C

Vs RB r C g m V be RC C 

where
e 1
IC =V be−V o

j 2 f C

Note: The current through C  I C =  V be g m V be RC − I C RC  j 2 f C 

 

depends only on V be!  1g m RC  j2  f C 

IC = V be

 1 j2  f R C C  
Kenneth R. Laker, update 01Oct14 KRL 13
 ESE319 Introduction to Microelectronics

Common Emitter Miller Effect Analysis II

IC
C 

From slide 13:

RS b V be c Vo

 1g m RC  j2  f C  v be
IR C

IC = V be Vs RB r C g m V be RC

 1 j2  f R C C  
e

 1g m RC  j2  f C 
IC = V be ≈  1 g m R C  j2 f C  V be = j2  f C eq V be

 1 j2  f R C C  
2 f RC C ≪1

Miller Capacitance Ceq: C eq=1 g m RC C =1− Av C 

Kenneth R. Laker, update 01Oct14 KRL 14
 ESE319 Introduction to Microelectronics

IC IC
C  C 

RS
b V be c Vo
RS b V be c Vo
IR C
IR C
Vs RB r C g m V be RC Vs RB r g m V be
C RC
C C eq
e e

C eq =1− Av C =1 g m RC C 

Kenneth R. Laker, update 01Oct14 KRL 15

 ESE319 Introduction to Microelectronics

Common Emitter Miller Effect Analysis III

C eq = 1 g m RC  C 

For our example circuit (Cµ = 2 pF):

1g m RC =10.040⋅5100=205

C eq =205⋅2 pF ≈410 pF

Kenneth R. Laker, update 01Oct14 KRL 16

 ESE319 Introduction to Microelectronics
Simplified HF Model
C
RS b V be IC  c Vo

Vs RB r v be g m V be '
C ro RC R L R L =r o∥ R C∥R L 

e
C
R B∥r  '
'
V =V s
s
RS b V be IC c Vo
R B∥r  R S


'
R S =r∥ RB∥R S  '
RB r v be g m V be
Vs C '
RL

e
Thevenin Equiv.
Kenneth R. Laker, update 01Oct14 KRL 17
 ESE319 Introduction to Microelectronics

Simplified HF Model
C
'
R
'
IC R L =r o∥ R C∥R L 
S b  c Vo '
V be R S =r∥ RB∥R S 
'
R B∥r 
'
Vs RB r C g m V be ' V s =V s
RL R B∥r  R S

e Miller's Theorem
'
RS b c Vo
IC V be


g m V be
RB Cr 
'
Vs C eq '
RL


C eq =1g m RC C  e
C tot =C C eq C tot
Kenneth R. Laker, update 01Oct14 KRL 18
 ESE319 Introduction to Microelectronics

Simplified HF Model
'
RS
b c Vo '
IC 
V be R L =r o∥ R C∥R L 
g m V be '
RB Cr  R S =r∥ RB∥R S 
'
V s C eq '
RL


'
RB∥r 
e V s =V s ≈V s
R B∥r R S
C tot
C tot =C 1g m R C  C 

V be =
1/ j2  f C tot
1/ j2  f C tot R S '
V s' ∣∣
Vo
Vs
 dB

' '
Vo −g m R L −g m R L
A v  f = ≈ =
V s 1 j 2  f C tot R'S f
1 j
fH
' 1
Av−mid =−g m R L and H f =
2  C tot R'S
Kenneth R. Laker, update 01Oct14 KRL 19
 ESE319 Introduction to Microelectronics

Multisim Simulation
C
2 pF
RS
50  b c vo

RB r v be g m v be
vs C RC
50 k  2.5 k  12 pF 40 mS v be 5.1 k  Mid-band gain @ 100 kHz
e

Av−mid =−g m R C =−204 46 dB @ 180 o 

C tot =12 pF 205⋅2 pF ≈422 pF
1 1
f H ≈ f −3dB= '
≈ ≈7.54 MHz Gain @ 8.69 MHz
2C tot RS 2 422 pF 50

Kenneth R. Laker, update 01Oct14 KRL 20

 ESE319 Introduction to Microelectronics
Frequency-dependent “beta”
0V j2π f CµVb IC = (gm – j2π f Cµ)Vb
b

gm V b
short-circuit
current

The relationship ic = βib does not apply at high frequencies f ≥ fH!

Using the relationship – ic = f (Vb ) – find the new relationship
between ib and ic. For ib (using phasor notation (Ix=Ix (jf) & Vx=Vx (jf))
for frequency domain analysis):
@ node B': b I =
1
r  
 j2  f C  C   V b where r x ≈0 (ignore rx)

Kenneth R. Laker, update 01Oct14 KRL 21

 ESE319 Introduction to Microelectronics
Frequency-dependent “beta”
j2π f CµVb IC = (gm – j2π f Cµ)Vb
Vb
∞
gm V b

I b=
1
r 
 j2  f C  C   V b @ node C: I c = g m − j2  f C  V b
(ignore ro)
Leads to a new relationship between the Ib and Ic:
Ic g m − j2  f C 
β(jf) → hfe  jf = =
Ib 1
 j2  f C C  
r
Kenneth R. Laker, update 01Oct14 KRL 22
 ESE319 Introduction to Microelectronics

Frequency Response of “beta”

g m − j2  f C 
 jf = IC VT
1 g m= r  =
 j2 f C C   VT IC
r
multiplying N&D by rπ C 1
For f = f low s.t. 2  f low ≪1 ≤
 g m − j2  f C   r  gm 10
 jf =
1 j2  f C  C   r  1
and 2  f low C  C   r  ≪1 ≤
factor N to isolate gm 10
C 
1− j2  f  g m r  jf =g m r  =
gm
 jf =
1 j2  f C  C   r 

Kenneth R. Laker, update 01Oct14 KRL 23

 ESE319 Introduction to Microelectronics

Frequency Response of “beta” cont.

 jf =
1− j2  f
gm
C

=
 g m r  1− j
f
fz  g m r =
 1− j
f
fz  
 f  dB

20log10 
1 j2  f C C   r 
 1 j
f
f   1 j
f
f  f
f f z

Hence, the lower break frequency or – 3dB frequency is fβ

1 gm 1 gm
f = = f z= =
2 C  C   r  2 C  C    the upper: 2 C  / g m 2 C 
where f z ≫ f 

Kenneth R. Laker, update 01Oct14 KRL 24

 ESE319 Introduction to Microelectronics

Frequency Response of “beta” cont.

Using Bode plot concepts, for the range where: f  f 
 jf =g m r  =
For the range where: f   f  f z s.t. ∣1− j f / f z∣≈1

We consider the frequency-dependent numerator term to

be 1 and focus on the response of the denominator:
f  f  f z gm r 
 jf ≈ =

 1 j
f
f  1 j
f
f 
Kenneth R. Laker, update 01Oct14 KRL 25
 ESE319 Introduction to Microelectronics

Frequency Response of “beta” cont.

gm r 
Neglecting numerator term:  jf = =

 1 j
f
f  1 j
f
f 
 f
And for f / f  >>1 (but < f / f z ): ∣ jf ∣≈ =
f
f  
f

f
Unity gain bandwidth: ∣ jf ∣=1⇒  ¿ ¿| f = f =1⇒ f T = f 
f T

T BJT unity-gain fre-

f T= = f  quency or GBP
2
Kenneth R. Laker, update 01Oct14 KRL 26
 ESE319 Introduction to Microelectronics

Frequency Response of “beta” cont.

−3
=100 r  =2500  C  =12 pF C =2 pF g m =40⋅10 S
12 −3
1 10 ⋅10 6
 = = =28.57⋅10 rps
 C C   r  122⋅2.5
  28.57 6
f = = 10 Hz=4.55 MHz f T = f  =455 MHz
2 6.28
g m 40⋅10−3⋅1012 9
 z= = Hz=20⋅10 rps
C 2
z
f z= =3.18⋅109 Hz=3180 MHz
2
Kenneth R. Laker, update 01Oct14 KRL 27
 ESE319 Introduction to Microelectronics

“beta” Bode Plot

|β(jf)| (dB)
(dB) |β(jf)| vs. Frequency

f (MHz)
fβ fT

Kenneth R. Laker, update 01Oct14 KRL 28

 ESE319 Introduction to Microelectronics

Multisim Simulation
Vb v-pi Ic
1 Ohm
Ib
Vc
1 Ohm

v-pi
mS

Insert 1 ohm resistors – we want to measure a current ratio.

Ic g m − j2  f C 
 jf = =
Ib 1
 j2  f C C  
r
Kenneth R. Laker, update 01Oct14 KRL 29
 ESE319 Introduction to Microelectronics

Simulation Results

Theory:
Low frequency |β(jf)| f T = f  =455 MHz

Unity Gain frequency about 440 MHz.

Kenneth R. Laker, update 01Oct14 KRL 30
 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 satisfy all

of the criteria above.
● The cascode amplifier will satisfy all of these criteria.

● A cascode is a CE Stage cascaded with a CB Stage.

(Historical Note: the cascode amplifier was a cascade of grounded

cathode and grounded grid vacuum tube stages – hence the
name “cascode,” which has remained in modern terminology.)
Kenneth R. Laker, update 01Oct14 KRL 31
 ESE319 Introduction to Microelectronics

The Cascode Amplifier

CB Stage CE Stage CB Stage
i C1
R1 RC i c2
C byp Q1 vo
B1
i B1 C1 vO
v-out RRs S i b2
Q1 i e1 i c1
Q2
CE Stage i E1 E1
R2 i C2 V CC v i e2 i b1
RRs S Cin i B2 RC
C2 s
RB RE
B2
Q2
i E2
vs R3
E2
v e1 v c2
RE R B =R 2∥R3 R in1= = =low≈r e1
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. Cin and Cbyp are to act as “open circuits” at dc and act as near
“short circuits” at all operating frequencies f  f min.
Kenneth R. Laker, update 01Oct14 KRL 32
 ESE319 Introduction to Microelectronics

CE and CB Amplifier Feature Review

CE Amplifier CB Amplifier
Voltage Gain (AV) moderate (-RC/RE) High (RC/(Rs + re))

Important for
Current Gain (AI) High (β) low (about 1)
Cascode
Input Resistance High (RB||βRE) low (re)

Output Resistance High (RC||ro) High (RC||ro)

Kenneth R. Laker, update 01Oct14 KRL 33

 ESE319 Introduction to Microelectronics

Cascode Mid-Band Small Signal Model

B1
i c1 vo
CB
CB Stage i C1 Stage C1
R1 RC i b1 vvbe1
r1 1 ggm vv 1
C byp B1
i B1 C1 vO
v-out m1 be1

Q1 i e1 E1
i E1 B2
CE Stage E1 i c2 R in1 =low
R2 i C2 V CC i b2 C2
RRs S Cin i B2 C2 RRs S vvbe2 g m vv 2
RC
Q2 r2 2
m2 be2
B2 vs
i E2
vs R3
E2 E2
RE CE RB i e2 RE
Stage
=R∥R
RRB B=R 2∥R 3
2 3

ASSUME: Q1 = Q2; e.g. β's identical, and both Forward Active

Kenneth R. Laker, update 01Oct14 KRL 34

 ESE319 Introduction to Microelectronics

Cascode Small Signal Analysis

B1
i c1 vo
1. Show reduction in Miller effect
CB
2. Evaluate small-signal voltage gain
Stage i b1 C1
r1 vvbe1 ggm vv 1
1
m1 be1 OBSERVATIONS
i e1 E1 a. The emitter current of the CB Stage is
B2
C2 i c2 R in1 =low the collector current of the CE Stage. (This
RRs S i b2 RC
vv 2 g m vv 2 also holds for the dc bias currents.)
r be2
2 m2 be2
vs
E2
i e1=i c2
CE RB i e2 RE b. The base current of the CB Stage is:
Stage i e1 i c2
i b1= =
1 1
g m1=g m2=g m c. Hence, both stages have about same
r e1=r e2=r e collector current i c1≈ic2 and same gm, re, rπ.
r  1=r  2 =r  I C1≈ I C2
Kenneth R. Laker, update 01Oct14 KRL 35
 ESE319 Introduction to Microelectronics

Cascode Small Signal Analysis cont.

The input resistance Rin1 to the CB Stage is
i c1
CB B1 vo the small-signal “re1” for the CB Stage.
Stage i b1 C1 i e1 i c2
r vvbe1
1 ggm v 1 i b1= =
m be1
1 1
i e1 E1 R ≈r
B2 in 1 e1 The CE output voltage, the voltage drop
C2 i c2
RRs S i b2 RC from Q2 collector to ground, is:
vv 2
r be2 g mvv 2 r r
vs
 m be2
v c2=v e1 =−r  i b1=− i e1=− i c2
E2 1 1
CE RB i e2 RE Therefore, the CB Stage input resistance is:
Stage
v e1 r
Rin1= = =r e1
−i e1 1
v c2 R in1 re r Miller Effect
AvCE −Stage = ≈− =− 1 => C eq =1 e C 2 C 
vs RE RE RE Much reduced
Kenneth R. Laker, update 01Oct14 KRL 36
 ESE319 Introduction to Microelectronics

Cascode Small Signal Analysis - cont.

i c1 Now, find the CE collector current in terms of
i b1
the input voltage vs: Recall i c1≈ic2
v be1 gmmvvbe1
be1 vs
i b2≈
i e1 R S∥R B r 1 R E
RS
i b2 i c2 Rin 1≈r e1  vs  vs
vvbe2be2 g mvv be2 i c2 = ib2 ≈ ≈
m be2 R S∥R B r   1 R E  1 R E
i e2
for bias insensitivity: 1 R E ≫ R S∥R B r 

i c1≈ie1 =i c2≈i e2 vvss v o −RC

i c2 ≈ v o≈−i c2 R C Av = ≈
RE vs 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.
Kenneth R. Laker, update 01Oct14 KRL 37
 ESE319 Introduction to Microelectronics

RS
'
b
IC V be
c Vo
∣∣
Vo
Vs
 dB

g m V be
RB Cr 
'
Vs C eq '
RL


e
C tot

Common Emitter Stage

C tot =C 1g m R C C 
1
Cascode Stage f H= '
⇒ f H Cascode  ≫ f H CE 
2 C tot RS
re
C tot =C 1 C 
RE
.2 C 
Kenneth R. Laker, update 01Oct14 KRL 38
 ESE319 Introduction to Microelectronics
Cascode Biasing
1. Choose IE1 – make it relatively large to
reduce R in1=r e =V T / I E1 to push out HF
o.c. IC1 break frequencies.
CB I1 vO 2. Choose RC for suitable voltage swing
Q1
R in1 =low VC1 and RE for desired gain.
IE1
RS C in IC2
Q2 3. Choose bias resistor string such that
IE2
o.c. its current I1 is about 0.1 of the collector
current IC1.

1 2
4. Given RE, IE2 and VBE2 = 0.7 V calc. R3.
 I E2 =I C2=I E1= I C1 ⇒ I C1= I E2≈ I E2

5. Need to also determine R1 & R2.

Kenneth R. Laker, update 01Oct14 KRL 39

 ESE319 Introduction to Microelectronics

Cascode Biasing - cont.

II11 Since the CE-Stage gain is very small:
VB1 vO a. The collector swing of Q2 will be small.
Q1 b. The Q2 collector bias VC2= VB1 - 0.7 V.
Rin 1≈r e1
VB2 VC2 6. Set V B1−V B2 =1V ⇒V CE2 =1 V
Q2
Since VCE2 = 1 V > VCE2sat Q2 forward active.

7. Next determine R2. Its drop VR2 = 1 V

V CE2=V C2 −V R e =V C2−V B2 −0.7V  with the known current.
.=V B1−0.7V −V R e V B1−V B2
.=V B1−0.7 V −V B2 0.7V R2=
I1
.=V B1−V B2=1V

Kenneth R. Laker, update 01Oct14 KRL 40

 ESE319 Introduction to Microelectronics

Cascode Biasing - cont.

V B1−V B2 1 V
R2= =
I1 I1
V B2
V B1 8. Then calculate R3. R 3=
Q1 I1
Rin 1≈r e1
where V B2 =0.7V I E R E
Q2
V CC
Note: R 1R 2 R 3=
I1

9. Then calculate R1.

V CC
R 1= −R 2−R 3
0.1 I C

Kenneth R. Laker, update 01Oct14 KRL 41

 ESE319 Introduction to Microelectronics

Cascode Bias Summary

SPECIFIED: Av, VCC, VC1 (CB collector voltage);
SPECIFIED: fH => Ctot(re) => IE (or IC) => re.
DETERMINE: RC, RE, R1, R2 and R3. V C1 RC
STEP1: R C = RE=
SET: V B1−V B2 =1V ⇒ V CE2 =1V IC ∣A v∣
V B1−V B2 1V
STEP2: R 2 = =
V B1 I1 0.1 I C
Q1
Rin 1≈r e1 V B2 0.7 V I E R E
STEP3: R 3= =
Q2 I1 0.1 I C
V CC V CC
R 1R 2 R 3= =
I 1 0.1 I C
V
STEP4: R 1= CC −R 2−R 3
I C2= I E1≈ I C1≈ I E2 =I C 0.1 I C
Kenneth R. Laker, update 01Oct14 KRL 42
 ESE319 Introduction to Microelectronics

Cascode Bias Example

R1 VCC-ICRE-
RC I R
C C
V C1
1.7
Q1 Q1
VVCE1
CE1
=V RC−I
=ICCC CR
–1–I R−V CE2 −I C R E
C CE
1.0
Q2 =12 V Q2 V =1 =12 V
CE2

ICRE+0.
RE
7 ICR
E

I E2≈ I C2= I E1≈ I C1 ⇒ I C1≈ I E2

Cascode Amp Typical Bias Conditions

Kenneth R. Laker, update 01Oct14 KRL 43

 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

1. Choose IE1 – to set re.

Let re = 5Ω => I E1=0.025 V /r e =5 mA.
Q1
V CE1=V CC − I C RC −1− I C R E

2. Set desired gain magnitude. For example

Q2 if AV = -10, then RC/RE = 10.

3. Since the CE stage gain is very small,

VCE2 can be small, i.e. VCE2 = VB1 – VB2 = 1 V.

Kenneth R. Laker, update 01Oct14 KRL 44

 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

Specs: RC
V CC =12 V V C1=7 V r e =5 ∣Av∣= R =10
E
=100
Q1
V CE1=V CC − I C RC −1− I C R E

Determine RC for VC1 = 7V .

Q2
I E1=0.025 V / r e =5 mA
V CC −V C1 5V
RC = −3
= −3
=1000 
5⋅10 A 5⋅10 A
RC RC
RE= = =100 
∣Av∣ 10
Kenneth R. Laker, update 01Oct14 KRL 45
 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

V CC =12 RC =1 k  r e =5 R E =100 
I1
=100
R1 RC I R
VCC-ICRE-1.7 C C Make current through the string of bias
Q1
resistors I1 = 0.1 IC = 0.5 mA.
CE1=VC CC
VVCE1 =I RC−I–1–ICCREC −1− I C R E
R2 1.0
V CC 12
R1 R 2 R3= = −4
=24 k 
=12V I 1 5⋅10
Q2 VCE2=1
ICRE+0.7 Calculate the bias voltages (base side of Q1, Q2):
R3 RE I R
C E
V R1=V CC −1.7V −I C R E =12 V −1.7V −0.5 V =9.8 V
V R2=V B1−V B2=1V
V R3=V B2 =0.7 I C R E =0.75⋅10−3⋅100=1.2 V
Kenneth R. Laker, update 01Oct14 KRL 46
 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

V B2 =5⋅10−4 R3=1.2V
CB V B1
R3 =2.4 k 
Q1
V B1−V B2=5⋅10−4 R 2=1.0V
RRSS C in V B2 R 2=2 k 
Q2
Recall: R1 R 2 R3=24 k 
R1 =24000−2.400−2000=19.6 k 
Recall for ac: R B =R 2∥R3
R B =2 k ∥2.4 k =1.1 k 
V CC =12 , RC =1 k  , V B2 =1.2 V ,
I C =5 mA , R E =100  , V B1−V B2=1.0V
Kenneth R. Laker, update 01Oct14 KRL 47
 ESE319 Introduction to Microelectronics
Completed Design
=100 1
r e =5 ⇒ I C =5 mA f H=
2  C tot R'S
V C1=7 V CB V B1 re
Q1 C tot =C 1 C 
RC RE
∣Av∣= R =10 RRSS C in V B2 .=C 1.05 C 
E
Q2
If Cπ = 12 pF
Cµ = 2 pF
C tot =14.1 pF
R1 =19.6 k 
RC =1 k  f Hcascode=225.8 MHz
R 2=2 k 
R3 =2.4 k  R E =100  For CE with |Av| = 10
f HCE =94 MHz
NOTE: R B=R 2∥R 3=1.09 k ≪ R E =10 k 
Kenneth R. Laker, update 01Oct14 KRL 48