ENEE 303H Fall 2009 HW7 YHC
11/03/09
1.
Rsig
Cgs
G
S
Cgd
Vo
CL
Vsig
RL'
gm*vgs
gmb*vbs
B
D
0
0
T1 (s) 
vbs  vo
, in HW5 solution we have solved vo and v g with MATLAB

vsig vsig
and Mathematica,
C 

 gm
1  s gs 
'
gm 
YL  g m 
 vo

'
vsig
CL  C gs  C gs  C gd Rsig YL  g mC gd Rsig
Rsig CL C gs  C gd  C gsC gd
1 s
 s2
'
YL  g m
YL'  g m

I in 

YL'
vsig  vg
Rsig
 
 g m  s CL  C gs 


vg
, where
vsig




 





N (s)

D(s)

YL'  g m  s CL  C gs

C gs  C gd Rsig YL'


 

 g mC gd Rsig  s 2 Rsig CL C gs  C gd  C gsC gd

vo
)
vsig
( the denominator is the same as that of
----------------------------------------------------------------

1 
or from the node equation at S: g m vg  vo  sC gs vg  vo  vo  sCL  ' 
RL 



gm  sCgs vg  
vo sCL  YL'
 g m  sC gs

 vg  vo




YL'  g m  s CL  Cgs
gm  sCgs 

---------------------------------------------------------------see revised HW5 solution for zeros, poles and magnitude response.
By the way, by eq.(6.36) f H 
1
 1
1   1
1 
2  2  2   2 2  2 
  P1  P 2   Z 1 Z 2 
= 843.8 MHz
Now back to I in , input admittance = T2 ( s) 
Y
'
L
 

  
I in

vsig

N ( s)
D( s) D( s )  N ( s )


Rsig
Rsig D( s)
1
s YL' C gs  C gd  g mC gd  s 2 CL C gs  C gd  C gsC gd
 



 


 g m  s CL  C gs  C gs  C gd Rsig YL'  g mC gd Rsig  s 2 Rsig CL C gs  C gd  C gsC gd






 
 YL' C gs  C gd  g mC gd 

CL C gs  C gd  C gsC gd 
s
1

s




YL'  g m
YL' C gs  C gd  g mC gd 







C  C gs  C gs  C gd Rsig YL'  g mC gd Rsig
R C C  C  C gsC gd
1 s L
 s 2 sig L gs ' gd
'
YL  g m
YL  g m


mid-band magnitude T2 (s)  high 



YL' Cgs  Cgd   g mCgd
1
,  z  0,
CL Cgs  Cgd   CgsCgd
Rsig
Example: again use the values given by Exercise 6.34, p.639 of
Sedra/Smith.
We get fz=wz/2/pi= 5.44 GHz
Besides, another zero is at 0, this is because at DC, Cgs and
Cgd are open to Vsig.
fhp1=whp1/2/pi= 849 MHz and fhp2=whp2/2/pi= 5.18 GHz
(by solving zeros of the polynomial D(s))
fp1=wp1/2/pi= 729.5 MHz (by dominant pole approximation)
Im
s plane
-wz
-whp2
-whp1
0
Re
The following figure shows the magnitude response of T2(s). The
unit is 1/Ω.
Magnitude Response
-85.5
-86
Magnitude, dB
-86.5
-87
-87.5
-88
9
10
10
11
10
10
Frequency, Hz
12
10
2.
f0 
1
 CC 
2 L 1 2 
 C1  C2 
The assumptions we have made here include ignoring internal
capacitances and resistances of BJT. A good choice of L, C1 and
C2 should maintain this assumption still a good approximation.
Higher L value of RFC is used to make it open to AC signal. The
bypass capacitances C3, C4 are chosen to make AC signal can
bypass the resistors RB1 and RE, respectively.
10Vdc
VCC
0
RFC
0.01H
C3
0.01
C2
0.2n
L
2.5uH
RB1
100
C1
0.2n
0
Q1
V
Q2N3904
RB2
100
RE
1k
0
C4
1m
14V
12V
10V
8V
6V
4V
0s
0.2us
0.4us
0.6us
0.8us
1.0us
1.2us
1.4us
1.6us
V(Q1:c)
Time
10V
8V
6V
4V
2V
0V
0Hz
5MHz
10MHz
15MHz
20MHz
25MHz
V(Q1:c)
Frequency
From SPICE FFT result, the oscillation frequency is at 10MHz,
which corresponds to what we calculated. Besides, there exists
higher order harmonics as well.
L
vc
v b'
vb
C1
C2
RB2
gm*vb
0
At node C: g m  sC1vc 
At node B: sC 2 vb' 
v
c


v '  sLg
 vb'
 0  vc  b 2 m
sL
1 s C1 L


v  v'
vb'
v'  v
 b c  0  vb' sC2  GB 2   c b
sL
RB 2
sL

( You can set vb'  1 and solve the node equations to get the
oscillation condition g m RB 2 
vb' sC 2  GB 2  
C1
)
C2
 sLg m  s 2C1Lvb'  g m  sC1vb'

sL 1  s 2C1L
1  s 2C1L



 

vb' sC2  GB2  1  s 2C1L  vb' s3C1C2 L  s 2C1LGB 2  sC2  GB 2   gm  sC1vb'
vb' 
 gm
s C1C2 L  s C1LGB 2  sC1  C2   GB 2
3
2
To oscillate, s    j  j0 and 02 
vb' 
C1  C2
LC1C2
 gm
gm

1
 C1  C2 
 C1 
GB 2  j0 C1  C2   GB 2  GB 2
 j0 C1  C2   
 C2 
 C2 
so g m RB 2 
C1
C2