Chapter 4 - High Frequency Amplifiers

Example 4.1 Page No. 4-34

In [2]:
from __future__ import division  
gm=(1/26)*10**3
print "(i) g_m = I_C / V_T =%0.2f mA/V"%gm
rbe=200/(38.46)
print "(ii) r_b''e = h_fe / g_m =%0.2f kohm"%rbe
cc=((38.46*10**-3)/(500*10**6))*10**12
print "(iii) (C_e + C_C) = g_m / 2*pi*f_T = g_m / omega_T =%0.2f pF"%-cc
cbe=76.92-3
print "Therefore,  C_b''e  = C_e =%0.2f pF"%cbe
print "(iv) We know that,"
print "f_T = h_fe*f_beta"
print "Therefore,  2*pi*f_T = h_fe*2*pi*f_beta"
print "omega_T = h_fe*omega_beta"
ob=((500*10**6)/200)*10**-3
print "omega_beta = omega_T / h_fe =%0.2f rad/s"%ob

(i) g_m = I_C / V_T =38.46 mA/V
(ii) r_b''e = h_fe / g_m =5.20 kohm
(iii) (C_e + C_C) = g_m / 2*pi*f_T = g_m / omega_T =-76.92 pF
Therefore,  C_b''e  = C_e =73.92 pF
(iv) We know that,
f_T = h_fe*f_beta
Therefore,  2*pi*f_T = h_fe*2*pi*f_beta
omega_T = h_fe*omega_beta
omega_beta = omega_T / h_fe =2500.00 rad/s

Example 4.2 Page No. 4-35

In [3]:
from math import sqrt
from __future__ import division  
ft=25*2
print "(i) f_T = |A_i|*f =%0.2f MHz"%ft
hfe=50000/200
print "(ii) h_fe(in kHz) = f_T / f_beta =%0.2f kHz"%hfe
print "(iii) |A_i| = h_fe / sqrt(1+((f/f_beta)**2)) :"
print "At  f = 10 MHz"
ai=250/sqrt(1+(((10*10**6)/(200*10**3))**2))
print "|A_i| =%0.2f"%ai
print "At  f = 100 MHz"
ai=250/sqrt(1+(((100*10**6)/(200*10**3))**2))
print "|A_i| =%0.2f"%ai

(i) f_T = |A_i|*f =50.00 MHz
(ii) h_fe(in kHz) = f_T / f_beta =250.00 kHz
(iii) |A_i| = h_fe / sqrt(1+((f/f_beta)**2)) :
At  f = 10 MHz
|A_i| =5.00
At  f = 100 MHz
|A_i| =0.50

Example 4.10 Page No. 4-39

In [5]:
from math import sqrt,pi
from __future__ import division  
print "a) The 3dB frequency for circuit gain and voltage gain is given as,"
print "(f_H)=1/(2*pi*R_eq*C_eq)"
r=(200*1000)/(200+1000)
print "where R_eq =(R_s+r_bb'')parallel to r+b''e =%0.2f ohm"%r
c=(100*10**-12)+((1+50)*3*10**-12)
print "and  C_eq =(C_b''e)+(1+(g_m*R_L)*C_b''c)= %0.2f F"%c
f=1/((2*pi*166.67*253*10**-12))
print "f_H = %0.2f Hz"%f
print "b)Voltage gain is given as,"
a=(-50*1)
print "A=(-g_m*R_L)=%0.2f"%a

a) The 3dB frequency for circuit gain and voltage gain is given as,
(f_H)=1/(2*pi*R_eq*C_eq)
where R_eq =(R_s+r_bb'')parallel to r+b''e =166.67 ohm
and  C_eq =(C_b''e)+(1+(g_m*R_L)*C_b''c)= 0.00 F
f_H = 3774350.04 Hz
b)Voltage gain is given as,
A=(-g_m*R_L)=-50.00

Example 4.12 Page No. 4-40

In [6]:
from __future__ import division  
print "f_H=1/(2*pi*R_eq*C_eq)"
print "and f_H''=2(f_H)"
print "1/(2*pi*R_eq*C_eq) = 2/(2*pi*R_eq*C_eq)"
print "R_eq'' = R_eq/2"
print "R_eq=(r_b''e)parallel to (r_bb''+R_s)"
print "= (r_b''e)=1000 ohm"
print "Therefore  R_eq'' =500 ohm"
print "Therefore  500=((r_b''e)*(r_bb''+R_s))/((r_b''e)+(r_bb'')+R_s)"
print " = 1000(100+R_s)/(1000+100+R_s)"
r=(4.5*10**5)/500
print "R_s = %0.2f ohm"%r

f_H=1/(2*pi*R_eq*C_eq)
and f_H''=2(f_H)
1/(2*pi*R_eq*C_eq) = 2/(2*pi*R_eq*C_eq)
R_eq'' = R_eq/2
R_eq=(r_b''e)parallel to (r_bb''+R_s)
= (r_b''e)=1000 ohm
Therefore  R_eq'' =500 ohm
Therefore  500=((r_b''e)*(r_bb''+R_s))/((r_b''e)+(r_bb'')+R_s)
 = 1000(100+R_s)/(1000+100+R_s)
R_s = 900.00 ohm

Example 4.16 Page No. 4-44

In [7]:
from math import sqrt,pi
from __future__ import division  
print "Hybrid-pi Equivalent is as shown in fig.4.29"
print "(i) Mid frequency voltage gain :"
print "V_o / V_s = -h_fe*R_L / R_s+h_ie"
hie=(100+1000)*10**-3
print "h_ie = r_bb'' + r_b''e =%0.2f kohm"%hie
hfe=0.2*1000
print "h_fe = g_m * r_b''e =%0.2f"%hfe
vo=-200/2
print "Therefore,  V_o / V_s =%0.2f"%vo
fb=(1/(2*pi*1000*(204*10**-12)))*10**-3
print "(ii) f_beta = 1 / 2*pi*r_b''e*(C_e+C_C) =%0.2f kHz"%fb
print "f_beta = %0.2f kHz"%fb
ft=(200*780)*10**-3
print "(iii) f_T = h_fe * f_beta =%0.2f kHz"%ft

Hybrid-pi Equivalent is as shown in fig.4.29
(i) Mid frequency voltage gain :
V_o / V_s = -h_fe*R_L / R_s+h_ie
h_ie = r_bb'' + r_b''e =1.10 kohm
h_fe = g_m * r_b''e =200.00
Therefore,  V_o / V_s =-100.00
(ii) f_beta = 1 / 2*pi*r_b''e*(C_e+C_C) =780.17 kHz
f_beta = 780.17 kHz
(iii) f_T = h_fe * f_beta =156.00 kHz

Example 4.17 Page No. 4-44

In [8]:
from math import sqrt,pi
from __future__ import division  
print "(i) We know that,"
print "  f_H = 1 / 2*pi*R_eq*C_eq"
print "where  R_eq = (R_s+r_bb'')*r_b''e / R_s+r_bb''+r_b''e"
print "and  C_eq = C_e + C_C*[1+g_m*R_L]"
rbe=100/100
print "  r_b''e(in k-ohm) = h_fo / g_m = %0.2f kohm"%rbe
print "C_eq = C_e + C_C*[1+g_m*R_L] = C_e + C_C[1+100*10**-3*500]"
print "     = C_e + 51 pF"
ce=((100*10**-3)/(2*pi*(400*10**6)))*10**12
print "C_e = g_m / 2*pi*f_T =%0.2f pF"%ce
ceq=39.79+51
print "Therefore,  C_eq =%0.2f pF"%ceq
req=1/(2*pi*5*90.79*10**-6)
print "R_eq = 1 / 2*pi*f_H*C_eq =%0.2f ohm"%req
print "Therefore, 350.6 = (R_s+100)*1000 / R_s+1100"
rs=(285.66*10**3)/649.4
print "Therefore,  R_s = %0.2f ohm"%rs
print "(ii) The mid-band voltage gain V_o/V_s is given as"
print "  V_o/V_s = -h_fe*R_L / R_s+h_ie"
hie=(100+1000)*10**-3
print "where  h_ie = r_bb'' + r_b''e =%0.2f K"%hie
vo=(-100*500)/(439.88+1100)
print "Therefore,  V_o/V_s =%0.2f"%vo

(i) We know that,
  f_H = 1 / 2*pi*R_eq*C_eq
where  R_eq = (R_s+r_bb'')*r_b''e / R_s+r_bb''+r_b''e
and  C_eq = C_e + C_C*[1+g_m*R_L]
  r_b''e(in k-ohm) = h_fo / g_m = 1.00 kohm
C_eq = C_e + C_C*[1+g_m*R_L] = C_e + C_C[1+100*10**-3*500]
     = C_e + 51 pF
C_e = g_m / 2*pi*f_T =39.79 pF
Therefore,  C_eq =90.79 pF
R_eq = 1 / 2*pi*f_H*C_eq =350.60 ohm
Therefore, 350.6 = (R_s+100)*1000 / R_s+1100
Therefore,  R_s = 439.88 ohm
(ii) The mid-band voltage gain V_o/V_s is given as
  V_o/V_s = -h_fe*R_L / R_s+h_ie
where  h_ie = r_bb'' + r_b''e =1.10 K
Therefore,  V_o/V_s =-32.47

Example 4.20 Page No. 4-47

In [10]:
from math import sqrt,pi
from __future__ import division  
print "Assume that the output time-constant is negligible as compared to the time consedtant. When this is the case"
print "A_vs = V_o/V_s = -g_m*R''_L*G''_s / G''_s+g_b''e+sC"
gs=6.66*10**-3
print "where  G''_s = 1 / (R_s||R_b)+r_bb'' =%0.2f"%gs
gbe=1/1000
print "g_b''e = 1 / r_b''e =%0.2f"%gbe
rl=(0.5/1.5)*10**3
print "R''_L = R_L || R_C =%0.2f ohm"%rl
print "  sC = admittance of C"
c=100+(3*(1+(50*333.33*10**-3)))
print "where  C = C_e + C_C*(1+g_m*R''_L) =%0.2f"%c
print "At 10 kHz,"
sc=2*pi*10*153*10**-9
print "sC = 2*pi*f*C =%0.2f"%sc
print "Therefore, At 10kHz signal frequency"
avs=(-50*333.33*6.66*10**-6)/((6.66*10**-3)+(10**-3)+(9.613*10**-6))
print "A_vs = V_o / V_s =%0.2f"%avs

Assume that the output time-constant is negligible as compared to the time consedtant. When this is the case
A_vs = V_o/V_s = -g_m*R''_L*G''_s / G''_s+g_b''e+sC
where  G''_s = 1 / (R_s||R_b)+r_bb'' =0.01
g_b''e = 1 / r_b''e =0.00
R''_L = R_L || R_C =333.33 ohm
  sC = admittance of C
where  C = C_e + C_C*(1+g_m*R''_L) =153.00
At 10 kHz,
sC = 2*pi*f*C =0.00
Therefore, At 10kHz signal frequency
A_vs = V_o / V_s =-14.47