# Chapter 8:Multivibrators And Switching Regulators¶

## Example 8.1,Page number 426¶

In [2]:
#Variable declaration
C=0.1                      #capacitance(uF)
R1=10                      #resistance(k ohms)
R2=2.3                     #resistance(k ohms)
Vcc=12.                    #supply voltage(V)
Rl=10**3.                  #resistance(k ohms)

#Calculations
#Part a
f=1/(0.693*C*(R2+R1/2))    #frequency(Hz)

#Part b
D=(1+(R2/R1))/(1+2*(R2/R1))*100          #duty cycle

#Part c
#(i)
T1=0.693*C*(R1+R2)                        #time period through R1(ms)
T2=0.693*R2*C                             #time period through R2(ms)
Pavg=(Vcc/Rl)**2*(T1/(T1+T2))             #average power dissipated during current sourcing(mW)

#Part d
Pavg1=(T2/(T1+T2))*(Vcc/Rl)**2           #average power dissipated during current sinking(mW)

#Results
print"print",round(f,2),"kHz"
print"duty cycle is",round(D,2),"%"
print"average power dissipated in current sourcing is",round((Pavg/1E-3),3),"mW"
print"average power dissipated in current sinking is",round(Pavg1/1e-3,3),"mW"

print 1.98 kHz
duty cycle is 84.25 %
average power dissipated in current sourcing is 0.121 mW
average power dissipated in current sinking is 0.023 mW


## Example 8.2,Page number 426¶

In [1]:
import math

#Variable declaration
t=1                               #time constant
e=1.8                             #e=R1/R2 min=1.8
e1=9.                              #e1=R1/R2 max=9

#Calculations
Betamin=1/(1+e)                 #current gain minimum
Betamax=1/(1+e1)                #current gain maximum
Tmax=2*t*math.log((1+Betamin)/(1-Betamin))
Tmin=2*t*math.log((1+Betamax)/(1-Betamax))
fmin=1/Tmax                     #minimum freq(Hz)
fmax=1/Tmin                     #maximum freq(k Hz)

#Results
print"fmin is",round(fmin/1E-3),"Hz and fmax is",round(fmax,1),"KHz"

fmin is 669.0 Hz and fmax is 2.5 KHz


## Example 8.3,Page number 427¶

In [1]:
import math

#Variable declaration
C=0.01                     #capacitance(uF)
R2=15                      #resistance(k ohms)
Va2=4                      #voltage(V)
Vcc=15.                    #supply voltage(V)
R1=33                      #resistance(k ohms)

#Calculations
Va1=0.67*Vcc               #voltage(V)
Vamax=Va1+Va2              #Va maximum(V)
Vamin=Va1-Va2              #Va minimum(V)
T1max=C*(R1+R2)*(math.log((1-(Vamax/(2*Vcc)))/(1-(Vamax/Vcc))))   #time period(ms)
T1min=C*(R1+R2)*(math.log((1-(Vamin/(2*Vcc)))/(1-(Vamin/Vcc))))   #time period(ms)
T2=0.693*R2*C
fmax=1/(T1min+T2)                    #maximum frequency(K Hz)
fmin=1/(T1max+T2)                    #miniimum frequency(K Hz)

#Results
print"minimum freq is",round(fmin,2),"(solution given in the textbook is incorrect)"
print"maximum freq is",round(fmax,2),"(solution given in the textbook is incorrect)"

minimum freq is 0.89 (solution given in the textbook is incorrect)
maximum freq is 4.1 (solution given in the textbook is incorrect)


## Example 8.4,Page number 433¶

In [2]:
#Variable declaration
Vi=25                         #input voltage(V)
Vsmax=30                      #supply voltage max(V)
Vomin=Vl=12                   #output minimum voltage or load voltage(V)
Io=15.                        #output current(mA)
Iq=3.                         #quinscent current of regulator(mA)
Vo=20.                        #output voltage(V)

#Calculations
#Part a
#(i)
Vimax=Vsmax                     #maximum permissible voltage(V)
Ro=0                            #for Vomin=beta=0
#(ii)
Vomax=Vi-2
betaVomax=Vomax-Vomin                   #output voltage(V)
R2max=(R1*betaVomax)/(Vomax-betaVomax)  #R2max(k ohms)
#(iii)
R3=betaVomax/Io                         #R3(k ohms)

#Part b
Vt=(Iq*betaVomax)/Io                   #common terminal fall(V)
Vomin1=Vl+Vt                           #voltage output minimum(V)

#Part c
betaVo=Vo-Vl                          #output voltage(V)
beta=betaVo/Vo                        #current gain
R2=(R1*betaVo)/(Vo-betaVo)            #R2(ohms)

#Results
print"a)i)max permissible supply voltage is",Vimax,"V"
print"ii)output voltage range for Vi=25V is",Vomin,"V to",Vomax,"V and R2max is",R2max,"k ohms"
print"iii)R3 is",round(R3,2),"kohms"
print"b)Vomin is",Vomin1,"V"
print"c)R2 is",round(R2,2),"ohms and R3 is",round(R3,3),"ohms"

a)i)max permissible supply voltage is 30 V
ii)output voltage range for Vi=25V is 12 V to 23 V and R2max is 18 k ohms
iii)R3 is 0.73 kohms
b)Vomin is 14.2 V
c)R2 is 13.33 ohms and R3 is 0.733 ohms


## Example 8.5,Page number 434¶

In [11]:
#Variable declaration
A=.0025                     #voltage gain
Vi=8                        #input voltage(V)
R2=1.5                      #resistance 2(k ohms)
R1=1                        #resistance 1(k ohms)

#Calculations
beta=R2/(R1+R2)               #current gain
Vo=Vl/(1-beta)                #output voltage(V)
Vo1=(A*Vi)/(1+(A*beta)-beta)  #output voltage ripple if Vi=8Vp-p

#Results
print"Vo is",Vo,"V"
print"expression of output voltage ripple",round(Vo1,2),"Vp-p"

Vo is 12.5 V
expression of output voltage ripple 0.05 Vp-p


## Example 8.6,Page number 435¶

In [3]:
#Variable declaration
Ro=7.5                               #output resistance(ohms)
hfe=50
Ve=20                                #voltage given to emitter(V)
Vbe=0.8                              #base to emitter voltage(V)
Vc=15                                #collector voltage(V)
P=12                                 #maximum power dissipation(W)
Ib1=5                                #for minimum load current Il=0,Ib=5

#Calculations
Io=(Vc/Ro)*10**3                      #output current(A)
Is=Il+5                               #supply current(mA)
Ic=Io-Is                              #collector current(A)
Ib=Ic/hfe                             #base current(mA)
Ie=Ic-Ib                              #emitter current(mA)
Pt=(Ve*Ie)-(Vc*Ic)                    #power dissipated in transistor(W)
Pl=(Ve-Vbe)*Is-Vc*Il                   #power dissipated in LR
Vimax=(P+Vc*(Ic*10**-3))/(Ie*10**-3)  #input voltage maximum
Iomin=hfe*Ib1                         #output current minimum(mA)

#Results
print"power dissipated in the transistor is",round((Pt/1E+3),2),"W and in LR is",round((Pl/1E+3),3),"W"
print"maximum permissible input voltage is",round(Vimax,2),"V"

power dissipated in the transistor is 8.83 W and in LR is 0.415 W
maximum permissible input voltage is 21.69 V
minimum load current for load voltage to remain stabalized is 250 mA


## Example 8.7,Page number 440¶

In [9]:
#Variable declaration
I=2.             #current at 12 V
V=240            #dc source(V)
d=17/50.         #duty cycle
d1=0.6           #duty cycle
eta1=0.8         #efficiency

#Calculations
Isav=(1*d)/2                  #average supply current(A)
Pav=V*Isav                    #average supply power(W)
eta=(P/Pav)*100               #regulator efficiency
Isav1=(1*d1)/2                #average supply current(A)
Po=Il*Vdc                     #power output(W)

#Results
print"regulator efficiency is",round(eta,1),"%"
print"average supply current is",Il,"A"
print"power output is",Po,"W"

regulator efficiency is 58.8 %
average supply current is 4.8 A
power output is 57.6 W


## Example 8.8,Page number 441¶

In [3]:
#Variable declaration
Vs=200                #dc source voltage(V)
eta=.85               #efficiency
f=20                  #oscillator frequency(Hz)
iSmax=2.6             #peak value of supply current(A)
pdf=0.4               #pulse duty factor

#Calculations
Isav=(Vl*Il)/(Vs*eta)   #average peak supply current(A)
iS=(2*Isav)/pdf         #supply current(A)
T=1000/f                #oscillation time period(uS)
tp=pdf*T                #transistor time(us)
d=iS/tp                 #change in iS with respect to time(A/us)
tp1=iSmax/d             #transistor time(us)
pdf1=tp1/T              #pulse duty factor
Isav1=(iSmax*pdf1)/2    #average peak supply current(A)
eta1=(P*100)/(Vs*Isav1) #efficiency

#Results
print"peak value of supply current is",round(Isav,3),"A"
print"pdf is",round(pdf,3)
print"overall efficienc is",round(eta1,1),"%"

peak value of supply current is 0.441 A
pdf is 0.4
overall efficienc is 81.6 %