Chapter 7:MESFET and Related Devices¶

Example 7.1 Page 231¶

In [1]:
#initialisation of variable
from math import *
S=-4.4*10**15;#slope
q=1.6*10**-19;#charge
E=11.9*8.85*10**-14;#constant
Vb=.42;#Voltage

#calculation
N=2/(q*E)/-S;#concentration
V=.0259*log(2.86*10**19/(2.7*10**15));#voltage
B=Vb+V;#barrier height

#result
print"donor concentration is",round(N,2),"cm^-3"
print"barrier height is",round(B,2),"V"

donor concentration is 2.69753510033e+15 cm^-3
barrier height is 0.66 V


Example 7.2 Page 233¶

In [2]:
#initialisation of variable
from math import *
Nd=10**16;#concentration
t=10**-6;#time
Js=6.5*10**-5;#A/cm^2
Nc=2.86*10**19;
q=1.6*10**-19;#charge
Dp=10;
E=4.096*10**-10;#constant
N=9.65*10**9;

#calculation
Bn=.0259*log(110*300**2/Js);#barrier height
Vn=.0259*log(Nc/Nd);
V=Bn-Vn;#built-in potential
W=(2*E*V/q/Nd)**.5;#depletion layer width
Lp=(Dp*t)**.5;
Jp=q*Dp*N**2/Lp/Nd;#minority carrier current density
R=Js/Jp;#ratio

#result
print"built-in potential is",round(V,2),"V"
print"depletion layer width is",round(W/10**8,14),"cm"
print"ratio of Js to Jpo is",round(R,2)

built-in potential is 0.46 V
depletion layer width is 4.86e-12 cm
ratio of Js to Jpo is 13795541.35


Example 7.3 Page 236¶

In [3]:
#initialisation of variable
from math import *
A=10**-5;#area
R=10**-6;#resistance
Nd=5*10**19;
V=.8;#Voltage
M=.26*9.1*10**-31;#mass
I=1;#A
E=1.05*10**-12;#constant
h=1.05*10**-34;#constant

#calculation
Rc=R/A;#ohmic resistance
C=2*(M*E)**.5/h*10;
Io=1/Rc*(Nd*10**6)**.5/C*exp(C*V/(Nd*10**6)**.5);#current
Vd=V-Nd**.5/C*log(Io/I)*1000;#voltage drop

#result
print"voltage drop is",round(Vd*1000,2),"mV"
print"the answer differs slightly from the book due to rounding off in the book"

voltage drop is 21.94 mV
the answer differs slightly from the book due to rounding off in the book


Example 7.4 Page 242¶

In [13]:
#initialisation of variable
from math import *
N=2*10**15;
T=300;#K
H=.89;#barrier height
D=12.4;
q=1.6*10**-19;#charge
E=8.85*10**-14;#constant
d=.6*10**-4;#thickness

#calculation
Vp=q*N*d**2/(2*D*E);#pinch-off voltage
Vn=.026*log(4.7*10**17/N);
Vb=H-Vn;#built-in potential

#result
print"pinch-off voltage is",round(Vp,3),"V"
print"built-in potential is",round(Vb,2),"V"

pinch-off voltage is 0.525 V
built-in potential is 0.75 V


Example 7.5 Page 248¶

In [13]:
#initialisation of variable
from math import *
N=2*10**18;
q=1.6*10**-19;#charge
E=8.85*10**-14;#constant
d=40*10**-7;#thickness
H=.85;#barrier height
D=12.3;#dielectric constant
V=.23;#Voltage

#calculation
Vp=q*N*d**2/(2*D*E);#volt
Vt=H-V-Vp;#threshold voltage
Ns=D*E/(q*(d+11*10**-7))*1.73;#electron gas concentration

#result
print"pinch-off voltage is",round(Vp,2),"V"
print"two-dimensional electron gas concentration is",round(Ns,2),"cm^-2"

pinch-off voltage is 2.35 V
two-dimensional electron gas concentration is 2.30783272059e+12 cm^-2