CHAPTER 05 - BJT AT HIGH FREQUENCY

Example E01 - Pg 139

In [1]:
#  Exa 5.1
#  Given data
import math
I_C = 2.;#  in mA
I_C =I_C * 10.**-3.;#  in A
V_CEQ = 20.;#  in V
h_fe = 100.;
I_BQ = 20.;#  in uA
I_BQ = I_BQ * 10.**-6.;#  in A
Beta = 100.;
f_T = 50.;#  in MHz
f_T = f_T * 10.**6.;#  in Hz
Cob = 3.;#  in pF
Cob = Cob * 10.**-12.;#  in F
h_ie = 1400.;#  in ohm
T = 300.;#  in K
#  (i) Transconductance
g_m = 11600.*(I_C/T);#  in S
g_m=g_m*10.**6.;#  in uS
print '%s %.2e' %("The transconductance in uS is",g_m);
#  (ii) Input resistance
g_m=g_m*10.**-6.;#  in S
r_be = h_fe/g_m; # in ohm 
print '%s %.2f' %("The input resistance in ohm is",r_be);
#  (iii) Capacitance
Cbc = Cob ;#  in F
Cbe = g_m/(2.*math.pi*f_T)-Cbc;#  in F 
Cbe= round(Cbe*10.**12.);#  in pF
print '%s %.2f' %("The capacitance in pF is",Cbe);
#  (iv) Base Spreading Resistance
r_bb = round(h_ie - r_be);#  in ohm
print '%s %.2f' %("The base spreading resistance in ohm is",r_bb);

The transconductance in uS is 7.73e+04
The input resistance in ohm is 1293.10
The capacitance in pF is 243.00
The base spreading resistance in ohm is 107.00

Example E02 - Pg 141

In [2]:
#  Exa 5.2
#  Given data
import math
I_C = 10.;#  in mA
I_C =I_C * 10.**-3.;#  in A
V_CE = 10.;#  in V
V_T= 26.*10.**-3.;#  in V
h_ie = 500.;#  in ohm
h_oe = 4.*10.**-5.;#  in S
h_fe  = 100.;
g_be = 1./260.;
h_re = 10.**-4.;
f_T = 50.;#  in MHz
f_T = f_T * 10.**6.;#  in Hz
T = 300.;#  in K
Cob =3.;#  in pF
Cob = Cob * 10.**-12.;#  in F
#  (i) Transconductance
g_m = I_C/V_T;#  in A/V
g_m= round(g_m*10.**3.);#  in mA/V
print '%s %.2f' %("The Transconductance in mA/V is",g_m);
#  (ii) Input resistance
g_m= g_m*10.**-3.;#  in A/V
r_be = round(h_fe/g_m);#  in ohm
print '%s %.2f' %("The input resistance in ohm is",r_be);
#  (iii) Base spreading resistance 
r_bb = h_ie - r_be;#  in ohm
print '%s %.2f' %("The base spreading resistance in ohm is",r_bb);
#  (iv) The feedback conductance 
g_bc = h_re*g_be;
print '%s %.2e' %("The feedback conductance is",g_bc);
#  (v) The output conductance 
g_ce = h_oe - (1.+h_fe)*g_bc
print '%s %.2e' %("The output conductance is : ",g_ce)
#  (vi) Capacitance
Cbe= g_m/(2.*math.pi*f_T);#  in F
Cbe= Cbe*10.**12.;#  in pF
print '%s %.2f' %("The value of C_b''e in pF is : ",Cbe)
Cc= Cob;#  in F
Cc= Cc*10.**12.
print '%s %.2f' %("The value of Cc in pF is : ",Cc)

The Transconductance in mA/V is 385.00
The input resistance in ohm is 260.00
The base spreading resistance in ohm is 240.00
The feedback conductance is 3.85e-07
The output conductance is :  1.15e-06
The value of C_b''e in pF is :  1225.49
The value of Cc in pF is :  3.00

Example E03 - Pg 142

In [3]:
#  Exa 5.3
#  Given data
import math
W = 10.**-6.;#  in m
I_E =2.;#  in mA
I_E = I_E * 10.**-3.;#  in A
V_T = 26.;#  in mV
V_T = V_T * 10.**-3.;#  in V
D_B = 47.*10.**-4.;
# g_m = abs(I_C)/V_T = abs(I_E)/V_T;
#  The emitter diffusion capacitance, Cbe = g_m*((W**2)/(2*D_B));
Cbe = I_E/V_T*W**2./(2.*D_B);#  F
Cbe= Cbe*10.**12.;#  in pF
print '%s %.2f' %("The emitter diffusion capacitance in pF is",Cbe);
Cbe= Cbe*10.**-12.;#  in F
g_m = abs(I_E)/V_T;
#  The transition frequency 
f_T = g_m/(2*math.pi*Cbe);#  in Hz
f_T = f_T * 10.**-6.;#  in MHz
print '%s %.2f' %("The transition frequency in MHz is",f_T);

#  Note: The answer in the book is not accurate.

The emitter diffusion capacitance in pF is 8.18
The transition frequency in MHz is 1496.06

Example E04 - Pg 145

In [4]:
#  Exa 5.4
import math
I_CQ = 5.;#  in mA
I_CQ = I_CQ * 10.**-3.;#  in A
V_VEQ = 10.;#  in V
h_ie = 600.;#  in ohm
h_fe = 100.;
C_C = 3.;#  in pF
C_C = C_C * 10.**-12.;#  in F
Ai = 10.;#  Ai(f)
f = 10.;#  in MHz
#  Ai = h_fe/( sqrt( 1 + ((f/f_Beta)**2) ) );
f_Beta = f/(math.sqrt( ((h_fe/Ai)**2.) - 1. ));#  in MHz
print '%s %.2f' %("The Beta cut off frequency in MHz is",f_Beta);
f_T = h_fe*f_Beta;#  in MHz
print '%s %.2f' %("The gain bandwidth product in MHz is",f_T);
g_m = 0.1923;
Ce = g_m/(2*math.pi*f_T*10.**6.);#  in F
print '%s %.2f' %("The value of Ce in F is",Ce);
Cbe= Ce;#  in F
print '%s %.2e' %("The value of C_b''e in pF is : ",Cbe*10.**12)
r_be = h_fe/g_m;#  in ohm
print '%s %.2f' %("The value of r_b''e in ohm is",r_be);
r_bb = h_ie - r_be;#  in ohm
print '%s %.2f' %("The value of r_bb'' in ohm is",r_bb);

The Beta cut off frequency in MHz is 1.01
The gain bandwidth product in MHz is 100.50
The value of Ce in F is 0.00
The value of C_b''e in pF is :  3.05e+02
The value of r_b''e in ohm is 520.02
The value of r_bb'' in ohm is 79.98

Example E05 - Pg 146

In [5]:
#  Exa 5.5
#  Given data
import math 
f_T = 400.;#  in MHz
D_Beta = 13.;#  in cm**2/sec
# Ce = (g_m*(W**2))/(2*D_B), so
# f_T = (g_m/(2*%pi))*( (2*D_B)/(g_m*(W**2)) ) = D_B/(%pi*(W**2));
W = math.sqrt( D_Beta/(math.pi*f_T*10.**6.) );#  in cm
W = W * 10.**4.;#  in um
print '%s %.2f' %("The base width of silicon transistor in um is",W);

The base width of silicon transistor in um is 1.02

Example E06 - Pg 147

In [6]:
#Exa 5.6 clc;
#  Given data
import math 
D_B = 47.;#  in cm**2/sec
I_C = 2.;#  in mA
I_C = I_C * 10.**-3.;#  in A
V_CEQ = 15.;#  in V
W = 1.;#  in um
W = W * 10.**-4.;#  in cm
V_T = 0.026;#  in V
g_m =I_C/(abs(V_T));#  in ohm
Ce = (g_m*(W**2.))/(2.*D_B);#  in F
Ce = Ce * 10.**12.;#  in pF
print '%s %.2f' %("The value of Ce in pF is",Ce);
f_T = g_m/(2.*math.pi*Ce*10.**-12.);#  in Hz
f_T = f_T * 10.**-6.;#  in MHz
print '%s %.2f' %("The value of f_T in MHz is",f_T);

The value of Ce in pF is 8.18
The value of f_T in MHz is 1496.06