# Chapter 11 : Power Devices¶

## Example number 11.1 , Page number 4420¶

In [1]:
#importing module
import math
from __future__ import division

#Variable declaration
RL=8 #in ohm
VCC=30 #in V

#Calculations
IC_max=VCC/RL
VCE_max=VCC
IC=VCC/(2*RL)
VCE=VCC-(IC*RL)
PT=VCE*IC

#Result
print("maximum collector current= %1.2f A\n" %IC_max)
print("Maximum collector-emiiter voltage= %i V\n" %VCE_max)
print("Maximum Power rating= %2.2f W" %PT)
maximum collector current= 3.75 A

Maximum collector-emiiter voltage= 30 V

Maximum Power rating= 28.12 W

## Example number 11.2 , Page number 421¶

In [3]:
#importing module
import math
from __future__ import division

#Variable declaration
VDD=25 #voltage axis intersection point in V
ID=4 #current in A

#Calculations
RD=VDD/ID
ID=VDD/(2*RD)
VDS=VDD-(ID*RD)
PT=VDS*ID

#Result
print("Drain Resistance= %1.2f ohm\n" %RD)
print("Drain current at maximum power ditribution point= %i A\n" %ID)
print("Drain-to-source voltage at maximum power dissipation point= %2.1f V\n" %VDS)
print("Maximum power dissipation= %i W" %PT)
Drain Resistance= 6.25 ohm

Drain current at maximum power ditribution point= 2 A

Drain-to-source voltage at maximum power dissipation point= 12.5 V

Maximum power dissipation= 25 W

## Example number 11.3 , Page number 422¶

In [6]:
#importing module
import math
from __future__ import division

#Variable declaration
beta1=20 #bjt gain
beta2=20 #bjt gain

#Calculations
beta0=beta1+beta2+(beta1*beta2)

#Result
print("net common-emitter current gain= %i" %beta0) #The answer in the textbook is mathematically incorrect
net common-emitter current gain= 440

## Example number 11.4 , Page number 91¶

In [7]:
#importing module
import math
from __future__ import division

#Variable declaration
TJ_max=150 #in C
Tamb=27 #in C
Rth_dp=1.7 #Thermal resistance in C/W
Rth_pa=40 #in C/W
Rth_ps=1 #in C/W
Rth_sa=4 #in C/W

#Calculations
PD1_max=(TJ_max-Tamb)/(Rth_dp+Rth_pa)
PD2_max=(TJ_max-Tamb)/(Rth_dp+Rth_sa+Rth_ps)

#Result
print("Case(a):No heat sink used :-Maximum power distribution= %1.2f W\n" %PD1_max)
print("Case(b):Heaat sink used :- Maximum power distribution= %2.2f W" %PD2_max)
Case(a):No heat sink used :-Maximum power distribution= 2.95 W

Case(b):Heaat sink used :- Maximum power distribution= 18.36 W

## Example number 11.5 , Page number 436¶

In [8]:
#importing module
import math
from __future__ import division

#Variable declaration
B=10 #current gain
IB=0.6 #in A
VBE=1 #in V
RC=10 #in ohm
VCC=100 #in Vs

#Calculations
IC=B*IB #in A
VCE=VCC-(IC*RC) #in V
VCB=VCE-VBE #in V
PT=(VCE*IC)+(VBE*IB)

#Result
print("Total power dissipation= %.1f W" %PT)
print("The BJT is working outside the SOA")
Total power dissipation= 240.6 W
The BJT is working outside the SOA

## Example number 11.6 , Page number 437¶

In [9]:
#importing module
import math
from __future__ import division

#Variable declaration
Beff=250 #effective gain
B1=25 #current gain of transistor
B2=8.65 #effective gain of Darlington-pair
iB=50*10**-3 #in A

#Calculations
iC2=iB*(Beff-B1)
iE2=(1+(1/B2))*iC2

#Result
print("Emitter current= %2.2f A" %iE2)
Emitter current= 12.55 A

## Example number 11.7 , Page number 438¶

In [11]:
#importing module
import math
from __future__ import division

#Variable declaration
VBB=24 #in V
r1=3 #in k-ohm
r2=5 #in k-ohm

#Calculations
n=r1/(r1+r2)
VP=(n*VBB)+0.7

#Result
print("peak-point voltage= %1.1f V" %VP)
peak-point voltage= 9.7 V

## Example number 11.8 , Page number 437¶

In [12]:
#importing module
import math
from __future__ import division

#Variable declaration
Rth_sink=4 #resistance in C/W
Rth_case=1.5 #in C/W
T2=200 #Temperature in C
T1=27 #Room temperature in C
P=20 #power in W

#Calculations
Rth=(T2-T1)/P
Tdev=T2
Tamb=T1
Rth_dp=Rth
Rth_ps=Rth_case #case-sink resistance
Rth_sa=Rth_sink #sink-ambient resistance
PD=(Tdev-Tamb)/(Rth_dp+Rth_ps+Rth_sa)

#Result
print("Actual power dissipation= %2.2f W" %PD) #The answers vary due to round off error
Actual power dissipation= 12.23 W
In [ ]: