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diff --git a/73/CH10/EX10.3/Example10_3.sci b/73/CH10/EX10.3/Example10_3.sci new file mode 100755 index 000000000..790ffa541 --- /dev/null +++ b/73/CH10/EX10.3/Example10_3.sci @@ -0,0 +1,36 @@ +//Chapter 10_Special Purpose Amplifiers
+//Caption : Class B Power Amplifier
+//Example10.3: A class-B audio power amplifier has a supply voltage of abs(Vcc)=15V. The closed loop gain Av=50 and the amplifier has to deliver 10W of power into an 8 ohm load.Find:
+//a)the peak output voltage swing
+//b)the peak output current swing
+//c)the input signal required(rms)
+//d)the total power from the power supply
+//e)the power dissipated in the amplifier
+//f)the power conversion efficiency
+clear;
+clc;
+//a)Solution:
+Po=10;//power in Watt
+Rl=8;//load resistance in Ohm;
+Vorms=sqrt(Po*Rl);// since output power Po=Vorms^2/Rl
+Vom=sqrt(2)*Vorms;//peak output voltage swing
+disp('V',abs(Vom),'The peak output Voltage swing:')
+//b)Solution:
+Iom=Vom/Rl;
+disp('A',abs(Iom),'The peak output current swing is:')
+//c)Solution:
+Av=50;//closed loop gain
+Vsrms=Vorms/Av;
+disp('V',Vsrms,'The input rms signal required is:')
+//d)Solution:
+Vcc=15;//absolute value of poer supply in volt
+Pin=2*Vcc*Iom/%pi;// since Iorms*2^(1/2)=Iom
+disp('W',Pin,'The total power from power supply is:')
+//e)Solution:
+Pd=(2/%pi)*Vcc*sqrt(2*Po/Rl)-Po;
+disp('W',Pd,'The power dissipated in the amplifier is:')
+//f)Solution:
+n=(Po/Pin)*100;
+disp('%',n,'The power conversion efficiency is:')
+//Note:
+//Vcc,Vom and Iom can be of either polarity but here only absolute value is considered and calculated
\ No newline at end of file diff --git a/73/CH10/EX10.4/Example10_4.sci b/73/CH10/EX10.4/Example10_4.sci new file mode 100755 index 000000000..275f0b55a --- /dev/null +++ b/73/CH10/EX10.4/Example10_4.sci @@ -0,0 +1,15 @@ +//Chapter 10_Special Purpose Amplifier
+//Caption : Power Output
+//Example10.4: For the amplifier of Example10.3, find the power output level at which the power dissipation will bw maximum and the maximum power dissipation.
+//Solution:
+clear;
+clc;
+Vcc=15;//power supply in volt
+Rl=8;//load resistance in ohm
+//since Pd=2*Vcc/%pi*sqrt(2*Po/Rl)-Po
+//to determine the value of Po at which Pd is maximum we differentiate above equation and equate to zero,we find Po as
+Po=2*Vcc^2/((%pi)^2*Rl);
+// there fore maximum power dissipated is
+Pdmax=2*Vcc/%pi*sqrt(2*Po/Rl)-Po;
+disp('W',Po,'The power output level for maximum power dissipation is:')
+disp('W',Pdmax,'Maximum power dissipation for corresponding output power level is;')
\ No newline at end of file diff --git a/73/CH10/EX10.8/Example10_8.sci b/73/CH10/EX10.8/Example10_8.sci new file mode 100755 index 000000000..3966390f3 --- /dev/null +++ b/73/CH10/EX10.8/Example10_8.sci @@ -0,0 +1,21 @@ +//Chapter 10_Special Purpose Amplifiers
+//Caption : LM4250 Parameters
+//Example10.8: The micropower programmable Op-Amp LM 4250 is supplied by 3 v sourse(absolute value) source.Determine the value of set resistor for Iset=0.1 uA if Rset is connected to (a)Vee and (b)ground. (c)determine the quiescent supply current and the quiescent power dissipation.
+clear;
+clc;
+//a)Solution:
+Vcc=3;//power supply in Volt
+Vee=-Vcc;//negative power supply in Volt
+Iset=0.1*10^-6;//bias setting current in A;
+Rset=(Vcc+abs(Vee)-0.5)/Iset;
+disp('mega Ohm',Rset/10^6,'The bias setting current resistor for Vee=-10 V is:')
+//b)Solution:
+clear Vee;
+Vee=0;//since Rset is connected to ground
+Rset=(Vcc+abs(Vee)-0.5)/Iset;
+disp('mega Ohm',Rset/10^6,'The bias setting current resistor for Vee=0 V is:')
+//c)Solution:
+Qcurrent=5*Iset;
+Qpower=(Vcc+3)*Qcurrent;//where abs(Vee)=3 V
+disp('uA',Qcurrent*10^6,'The quiescent current supply is:')
+disp('uW',Qpower*10^6,'The quiescent power dissipated is:')
\ No newline at end of file diff --git a/73/CH10/EX10.9/Example10_9.sci b/73/CH10/EX10.9/Example10_9.sci new file mode 100755 index 000000000..358cb7d07 --- /dev/null +++ b/73/CH10/EX10.9/Example10_9.sci @@ -0,0 +1,43 @@ +//Chapter 10_Special Purpose Amplifiers
+//Caption : Common Emitter Amplifier Parameters
+//Example10.9: A single common emitter amplifier has following device and circuit parameters: Rb=60 Ohm,Rs=40 Ohm,Cu=1.5 pF,Cl=1 pF,ft=1.6 GHz at Ic=2.5 mA quiescent current.Determine each of the following for two values of Rl: 30 Ohm and 100 Ohm. a)f1 b)F2 (c)BW (d)Avmid (e)avmid*Bw.
+clear;
+clc;
+Ft=1.6*10^9;//reduced unity gain frequency in Hz
+Ic=2.5*10^-3;//collector current in A
+Vt=25*10^-3;//threshold voltage at room temperature
+gm=Ic/Vt;//transconductance
+Cu=1.5*10^-12;
+Cl=1*10^-12;
+Rs=40;
+Rb=60;
+C2=gm/(2*%pi*Ft)-Cu
+for i=1:2,
+ if i==1 then
+ Rl=30;//load resistance
+ F1=1/(2*%pi*(Rs+Rb)*(C2+Cu*(1+gm*Rl)));// first break frequency
+ F2=1/(2*%pi*Rl*(Cu+Cl));//second break frequency
+ BW=F1;//since single common emitter amplifier so n=1 thus BW=F1*sqrt(2^(1/n)-1),i.e.,BW=F1
+ Avmid=-gm*Rl;//mid frequency gain
+ GBW=Avmid*BW;// gain-bandwidth product
+ disp('********For Rl=30 Ohm********')
+ disp('MHz',F1/10^6,'first break frequency is:')
+ disp('MHz',F2/10^6,'second break frequency is:')
+ disp('MHz',BW/10^6,'Bandwidth is:')
+ disp(abs(Avmid),'mid frequency gain is:')
+ disp('MHz',abs(GBW)/10^6,'gain-bandwidth product is:')
+ else
+ Rl=100;//load resistance in ohm
+ F1=1/(2*%pi*(Rs+Rb)*(C2+Cu*(1+gm*Rl)));// first break frequency
+ F2=1/(2*%pi*Rl*(Cu+Cl));//second break frequency
+ BW=F1;//since single common emitter amplifier so n=1 thus BW=F1*sqrt(2^(1/n)-1),i.e.,BW=F1
+ Avmid=-gm*Rl;//mid frequency gain
+ GBW=Avmid*BW;// gain-bandwidth product
+ disp('********For Rl=100 Ohm********')
+ disp('MHz',F1/10^6,'first break frequency is:')
+ disp('MHz',F2/10^6,'second break frequency is:')
+ disp('MHz',BW/10^6,'Bandwidth is:')
+ disp(abs(Avmid),'mid frequency gain is:')
+ disp('MHz',abs(GBW)/10^6,'gain-bandwidth product is:')
+ end
+end
\ No newline at end of file diff --git a/73/CH11/EX11.11/Example11_11.sci b/73/CH11/EX11.11/Example11_11.sci new file mode 100755 index 000000000..48613d3d5 --- /dev/null +++ b/73/CH11/EX11.11/Example11_11.sci @@ -0,0 +1,17 @@ +//Chapter 11_Nonlinear Circuit Application
+//Caption :Design Peak Detector
+//Example11.11: Design a positive peak detector using type uA 760 comparator that can respond to a 100 mV(pp),5 MHz sinusoidal input signal.The device has following specifications. Response time=25 ns, propagation time=12 ns, and Input bias current=8uA.
+//Solution:
+clear;
+clc;
+Vp=50*10^-3;//since peak-peak voltage is 100 mV
+f=5*10^6;
+T=200*10^-9;
+t=15*10^-9//since rise time(t) should be greater than propagation delay(12ns)
+deltaVc=Vp*(1-cos(4*t/T*90*(%pi)/180));
+Ib=8*10^-6;//input bias current
+C=Ib/(deltaVc/T);
+disp('mV',deltaVc*10^3,'voltage change is:')
+disp('pF',C*10^12,'capacitor value is:')
+//Note:
+// the Exact value as calculated is taken to calculate C, so C=293.59555 pF. If approx value of deltaVc is taken as 5 mV then C=320 pF
\ No newline at end of file diff --git a/73/CH11/EX11.4/Example11_4.sci b/73/CH11/EX11.4/Example11_4.sci new file mode 100755 index 000000000..1199f670b --- /dev/null +++ b/73/CH11/EX11.4/Example11_4.sci @@ -0,0 +1,10 @@ +//Chapter 11_Nonlinear Circuit Application
+//Caption :Time taken
+//Example11.4: b)Type 741 Op-amp is used as a comparator and its slew rate is 0.5V/us.How long will it change from +10 V to -10v?
+//b)Solution:
+clear;
+clc;
+deltaVo=10-(-10);
+SlewRate=0.5*10^-6;
+t=deltaVo/SlewRate;
+disp('us',t/10^6,'time taken by the output voltage to change from +10 V to -10 V is:')
\ No newline at end of file diff --git a/73/CH11/EX11.5/Example11_5.sci b/73/CH11/EX11.5/Example11_5.sci new file mode 100755 index 000000000..fa7332981 --- /dev/null +++ b/73/CH11/EX11.5/Example11_5.sci @@ -0,0 +1,12 @@ +//Chapter 11_Nonlinear Circuit Application
+//Caption : Rise Time
+//Example11.5: The upper 3-dB frequency of an Op-Amp is 1MHz.Calculate the rise time of the output. If the upper 3-dB frequency of the Op-Amp is increased to 50 MHz by reducing the gain such that gain bandwidth product remains constant,then find out the new rise time.Discuss the effect of increasing bandwidth on accuracy of comparator.
+//Solution:
+clear;
+clc;
+F3dB=1*10^6;//upper 3-dB frequency of Op-Amp
+Tr=0.35/F3dB;//from definition of rise time
+disp('n sec',Tr*10^9,'Rise time of the output is:')
+F3dB1=50*10^6;
+Tr1=0.35/F3dB1;
+disp('n sec',Tr1*10^9,'Rise time of the output is:')
\ No newline at end of file diff --git a/73/CH12/EX12.11/Examle12_11.sci b/73/CH12/EX12.11/Examle12_11.sci new file mode 100755 index 000000000..8ea47b34a --- /dev/null +++ b/73/CH12/EX12.11/Examle12_11.sci @@ -0,0 +1,15 @@ +//Chapter 12_Signal Generators
+//Caption : Design
+//Example12.11: A 555 one shot circuit with Vcc=16 V is to have a 2 ms output pulse width.Design a suitable Circuit.Ithres=0.25 uA(max.) from data sheet of the device.
+//Solution:
+clear;
+clc;
+Ithres=0.25*10^-6;
+T=2*10^-3//output pulse width
+Vcc=16;//power supply to 555
+//The value of minimum capacitor charging current Ic should be much greater than the threshold Current Ithres
+Icmin=1000*Ithres;//since Icmin>>Ithres
+Ra=Vcc/(3*Icmin);
+C=T/(1.1*Ra);
+disp('kilo Ohm',Ra/10^3,'resitance design is:')
+disp('uF',C*10^6,'Capacitor design is:')
\ No newline at end of file diff --git a/73/CH12/EX12.12/Example12_12.sci b/73/CH12/EX12.12/Example12_12.sci new file mode 100755 index 000000000..45374978b --- /dev/null +++ b/73/CH12/EX12.12/Example12_12.sci @@ -0,0 +1,31 @@ +//Chapter 12_Signal Generators
+//Caption : Generating pulse by 555 timer
+//Example12.12:(a)Design a 555 astable multivibrator to generate an output pulse with pulse repetition frequency(PRF)=4 kHz and a duty cycle of 60%.Given Vcc=15V.
+//(b)Analyse the circuit designed in part (a) to determine the actual PRF and duty cycle. Given Ithres=25 uA(max.) for timer 555.
+clear;
+clc;
+//a)Solution:
+d=60*10^-2;//duty cycle given
+PRF=4*10^3;
+Vcc=15;//power supply
+T=1/PRF;//where T=Tc+Td
+Tc=d*T;
+Td=T-Tc;
+Ithres=25*10^-6;
+Icmin=1*10^-3;//since Icmin>>Ithres,so assuming Icmin=1 mA
+R=Vcc/(3*Icmin);//where R=Ra+Rb
+C=Tc/(0.7*R);
+Rb=Td/(0.7*C);
+Ra=R-Rb;
+disp('kilo Ohm',Ra/10^3,'Designed resistor(Ra) for 555 timer in astable mode is:')
+disp('kilo Ohm',Rb/10^3,'Designed resistor(Rb) for 555 timer in astable mode is:')
+disp('uF',C*10^6,'Designed Capacitor for 555 timer in astable mode is:')
+//b)Solution:
+//from equation of charging
+Tc1=0.7*R*C;
+Td1=0.7*Rb*C;
+T1=Tc1+Td1;
+PRFa=1/T1;
+da=Tc1/(Tc1+Td1)*100;
+disp('kHz',PRFa/10^3,'actual Pulse Repetition Frequency is:')
+disp('%',da,'actual duty cycle is:')
\ No newline at end of file diff --git a/73/CH12/EX12.20/Example12_20.sci b/73/CH12/EX12.20/Example12_20.sci new file mode 100755 index 000000000..2342f0206 --- /dev/null +++ b/73/CH12/EX12.20/Example12_20.sci @@ -0,0 +1,24 @@ +//Chapter 12_Signal Generators
+//Caption : Waveform Generator
+//Example12.20: Design a waveform generator using type 8038 IC.The frequency of Oscillation is 5 kHz and the duty cycles is 50%.From data sheet,typical values for the device at Vcc=5 V are as follws:
+//Voh=3.6 V; Vol=0.2 V; Ill=-1.6 mA and Ilh=40 uA.
+//Solution:
+clear;
+clc;
+Fo=5*10^3;
+//for 50% duty cycle Tp=Tn
+Vcc=5;//in volt
+Vol=0.2;//in Volt
+Voh=3.6;//in volt
+Ill=-1.6*10^-3;
+Ilh=40*10^-6;
+Tp=1/(2*Fo);
+C=0.01;//assuming the Capacitor value in uF for optimum design
+Ra=Tp/(1.66*C);
+Rb=2*Ra*Tp/(1.66*Ra*C+Tp);
+R2min=(Vcc-Vol)/(2*10^-3-abs(Ill));//since Ill is negative
+R2max=(Vcc-Voh)/(1*10^-6+Ilh);//since Ilh is positive
+disp('kilo Ohm',Ra*10^3,'designed value of Ra is:')
+disp('kilo Ohm',Rb*10^3,'designed value of Rb is:')
+disp('kilo Ohm',R2min/10^3,'minimum pull-up resistor is:')
+disp('kilo Ohm',R2max/10^3,'maximum pull-up resistor is:')
\ No newline at end of file diff --git a/73/CH12/EX12.6/Example12_6.sci b/73/CH12/EX12.6/Example12_6.sci new file mode 100755 index 000000000..33a5dc289 --- /dev/null +++ b/73/CH12/EX12.6/Example12_6.sci @@ -0,0 +1,18 @@ +//Chapter 12_Signal Generators
+//Caption : 555 Timer
+//Example12.6: Calculate (a)Tc (b)Td, and (c)the free running frequency for the timer 555 connected in astable mode.Given Ra=6.8kilo Ohm; Rb=3.3 kilo Ohm; C=0.1 uF. What is the duty cycle ,d,of the circuit?
+//Solution:
+clear;
+clc;
+Ra=6.8*10^3;
+Rb=3.3*10^3;
+C=0.1*10^-6;
+// Using equation for astable multivibrator we have
+Tc=0.69*(Ra+Rb)*C;//charging time
+Td=0.69*Rb*C;//discharging time
+f=1.44/((Ra+2*Rb)*C);//free running frequency
+d=Rb/(Ra+2*Rb);//duty cycle
+disp('ms',Tc*10^3,'charging time of 555 timer in astable mode is:')
+disp('ms',Td*10^3,'discharging time of 555 timer in astable mode is:')
+disp('kHz',f/10^3,'free running frequency of 555 timer in astable mode is:')
+disp(d,'duty cycle of 555 timer in astable mode is:')
\ No newline at end of file diff --git a/73/CH13/EX13.14/Example13_14.sci b/73/CH13/EX13.14/Example13_14.sci new file mode 100755 index 000000000..bc6a0c638 --- /dev/null +++ b/73/CH13/EX13.14/Example13_14.sci @@ -0,0 +1,25 @@ +//Chapter 13_Voltage Regulators
+//Caption : Inductor and Capacitor
+//Example13.14: A switching voltage regulator operates at a switching frequency of 30kHz and is to supply a load current Io of 1 A at a dc output voltage Vo of +10V.The dc input voltage is Vin=20V and the output(peak-peak) ripple factor is not to exceed 0.05%. Assume Rl=10 Ohm.
+//a)Find the value of the filter inductor L such that the maximum change or ripple in the current through the inductor will not exceed 40% of the average or dc current.
+//b)Find the value of the outpur capacitor CL for L1=100 uH and for L2=500 uH.
+clear;
+clc;
+//a)Solution:
+Rl=10;
+D=0.5;
+T=2.5;
+fs=30*10^3;
+L=Rl*T*(1-D)/fs;
+disp('uH',L/10^-6,'filter inductor L to ensure maximum ripple in the current through the inductor will not exceed 40% of the dc current is:')
+//b)Solution:
+L1=100*10^-6;
+RF=0.05*10^2;//output (peak-peak)ripple factor maximum limit
+//for ripple factor condition we have
+CL1=1/(15*fs^2*L1*RF);
+disp('********For L=100 uH********')
+disp('uF',CL1*10^10,'output capacitor is:')
+disp('********For L=500 uH********')
+L2=500*10^-6;
+CL2=1/(15*fs^2*L2*RF);
+disp('uF',CL2*10^10,'output capacitor is:')
\ No newline at end of file diff --git a/73/CH13/EX13.3/Example13_3.sci b/73/CH13/EX13.3/Example13_3.sci new file mode 100755 index 000000000..6265e02b2 --- /dev/null +++ b/73/CH13/EX13.3/Example13_3.sci @@ -0,0 +1,13 @@ +//Chapter 13_Voltage Regulators
+//Caption : Maximum Efficiency and Power
+//Example13.3: Calculate the maximum efficiency and associated power dissipation for the 5 V MC7805 series regulator.The input ripple is 10 V and the load current is 1 A. The output is between 4.75 to 5.25 for &v<=Vin<=20 V.
+//Solution:
+clear;
+clc;
+Vo=5;
+Vin=17;//since for MC7805 a maximum of 7.5 V is added to the ripple.Since 10 V ropple is given so Vin=10+7=17 V
+Il=1;//load current in ampere
+n=Vo/Vin*100;//series pass reguator overall efficiency
+Pd=(Vin-Vo)*Il;
+disp('%',n,'maximum efficiency for 5V MC7805 series regulator is:')
+disp('W',Pd,'power dissipation for the 5V MC7805 series regulator is:')
\ No newline at end of file diff --git a/73/CH15/EX15.2/Example15_2.sci b/73/CH15/EX15.2/Example15_2.sci new file mode 100755 index 000000000..76b4f73e0 --- /dev/null +++ b/73/CH15/EX15.2/Example15_2.sci @@ -0,0 +1,30 @@ +//Chapter 15_Phase Locked Loops
+//Caption : Output Signal Frequency
+//Example 15.2: A PLL has a Ko of 2*%pi(1kHz)/V, a Kv of 500 per sec, and a free running frequency of 500Hz.
+//a) For a constant input signal frequency of 250 Hz and 1kHz.find vf.
+clear;
+clc;
+//a)Solution:
+Ko=2*%pi*10^3;// VCO gain in kHz/V
+Kv=500;//loop bandwith in per second
+Wc=500;//Free running frequency of VCO in PLL or 2*%pi*500
+//Wi=angular input signal frequency in Hz
+//Wo=angular output signal frequency in Hz
+//since vf=(Wo(t)-Wc)/Ko
+//under locked condition Wo=Wi, so vf=(Wo-Wc)/Ko
+for i=1:2,
+ if i==1 then
+ Wo=250;
+ //or
+ Fo=2*%pi*Wo;//in Hz
+vf=(Fo-2*%pi*500)/Ko;
+disp('********For input signal frequency W=250 Hz********')
+disp('V',vf,'output signal voltage of PLL for Wo=250 Hz is:')
+else
+ Wo=1000;
+ Fo=2*%pi*Wo;//in Hz
+ vf=(Fo-2*%pi*500)/Ko;
+ disp('********For input signal frequency Wo=1 kHz********')
+ disp('V',vf,'output signal voltage of PLL for Wo=1kHz is:')
+end
+end
\ No newline at end of file diff --git a/73/CH15/EX15.3/Example15_3.sci b/73/CH15/EX15.3/Example15_3.sci new file mode 100755 index 000000000..3168822ab --- /dev/null +++ b/73/CH15/EX15.3/Example15_3.sci @@ -0,0 +1,21 @@ +//Chapter 15_Phase Locked Loops
+//Caption : VCO and Phase detector
+//Example15.3: A PLL has free running frequency Wc=500 kHz,bandwith of low pass filter=10kHz. Suppose an input signal of frequency 600kHz is applied.Will the loop acquire lock? What is VCO output frequency? The phase detector produces sum and difference frequency components.
+//Solution:
+clear;
+clc;
+BW=10;//bandwidth of low pass filter in kHz
+Fi=600;//input frequency in kHz
+Fc=500;//free running frequency in kHz
+// Output from phase detector is
+Sum=Fi+Fc;
+Difference=Fi-Fc;
+disp('kHz',Sum,'sum frequency component of phase detector in kHz')
+disp('kHz',Difference,'difference frequency component of phase detector in kHz')
+if Sum>BW then
+ if Difference>BW then
+ disp('Both Sum and Difference frequency components are outside the passbandof low-pass filter')
+ disp('Loop will not acquire lock')
+ disp('VCO frequency will be its free running frequency')
+end
+end
\ No newline at end of file diff --git a/73/CH15/EX15.4/Example15_4.sci b/73/CH15/EX15.4/Example15_4.sci new file mode 100755 index 000000000..43e6c2317 --- /dev/null +++ b/73/CH15/EX15.4/Example15_4.sci @@ -0,0 +1,17 @@ +//Chapter 15_Phase Locked Loops
+//Caption : Second Order Butterworth Filter
+//Example 15.4: A Synthesizer using PLL has Kv=5*%pi rad/s. What value of low-pass filter bandwidth should be used so that the closed-loop system approximates a second-order Butterworth filter?
+//Solution:
+clear;
+clc;
+//For Butterworth filter the damping ratio(Dr) is
+Dr=0.707;
+Kv=5*%pi;
+Wl=Kv*(2*Dr)^2;//since (Wl/Kv)^2=2*Dr
+disp('rad/sec',Wl,'low pass filter bandwidth')
+// BW for closed loop system is
+BW=sqrt(Kv*Wl);// since BW=Wn, where Wn=natural frequency,BW=bandwidth of closed loop system
+Wn=real(BW);
+t=2.2/Wn;
+disp('rad/sec',BW,'bandwidth of closed loop system is:')
+disp('sec',t,'corresponding system rise time is:')
\ No newline at end of file diff --git a/73/CH15/EX15.5/Example15_5.sci b/73/CH15/EX15.5/Example15_5.sci new file mode 100755 index 000000000..f058f2be0 --- /dev/null +++ b/73/CH15/EX15.5/Example15_5.sci @@ -0,0 +1,22 @@ +//Chapter 15_Phase Locked Loops
+//Caption : Lock Range
+//Example15.5: A PLL has a VCO with Ko=25kHz/V and Fc=50kHz.The amplifier gain is A=2 and the phase detector has a maximum output voltage swing of +0.7V and -0.7V. Find the lock range of the PLL.Assume filter gain equal to unity.
+//Solution:
+clear;
+clc;
+k1=2*0.7/%pi;//positive maximum gain value of phase detector
+k2=-k1;//negative maximum gain value of phase detector
+A=2;// amplifier gain
+Ko=25;// VCO gain in kHz
+//positive maximum output voltage swing of phase detector is
+V1=k1*%pi/2;
+//Negative maximum output voltage swing of phase detector is
+V2=k2*%pi/2;
+Vf1=k1*A*%pi/2;//Positive maximum control voltage available to drive VCO
+Vf2=k2*A*%pi/2;//negative maximum control voltage available to drive VCO
+//maximum VCO frequency swing that can be obtained is
+Fh=Ko*Vf1;//positive maximum VCO frequency swing
+Fl=Ko*Vf2;// Negative maximum VCO frequency swing
+// so lock range of PLL is
+f=Fh-Fl;
+disp('kHz',f,'The lock range of the PLL is:')
\ No newline at end of file diff --git a/73/CH16/EX16.12/Example16_12.sci b/73/CH16/EX16.12/Example16_12.sci new file mode 100755 index 000000000..4aea01049 --- /dev/null +++ b/73/CH16/EX16.12/Example16_12.sci @@ -0,0 +1,27 @@ +//Chapter 16_Bipolar and MOS Digital Gate Circuits
+//Caption : NMOS operating region
+//Example 16.12: A NMOS transistor with K=20uA/V^2 and Vth=1.5V is operated at Vgs=5V and Ids=100uA. Determine the region of the operation on I-V characteristics and find Vds.
+//Solution:
+clear;
+clc;
+K=20*10^-6;
+Vgs=5;
+Vth=1.5;
+Ids=100*10^-6;
+Id=(K/2)*(Vgs-Vth)^2;
+disp('uA',Id/10^-6,'drain current in saturation region')
+if Id>Ids then
+ disp('region of operation of NMOS transistor on I-V characteristics is LINEAR REGION')
+end
+//since NMOS lies in LINEAR REGION so Ids=(K/2)*(2*(Vgs-Vth)*Vds-Vds^2); thus substituting the values we have
+//100*10^-6==(20*10^-6/2)*(2*(5-1.5)*Vds-Vds^2);
+//so Vds^2-7*Vds+10=0; equivalent to quadrattic equation of form aX^2+b*X+c=0
+Vds=poly(0,'Vds');
+p=Vds^2-7*Vds+10;//equation whose roots has to be found
+z=roots(p);
+z=real(z)
+ if (z(1)<(Vgs-Vth)) then
+ disp('Volt',z(1),'drain to source voltage(Vds) in this Linear Region is:')
+elseif (z(2)<(Vgs-Vth)) then
+ disp('Volt',z(2),'drain to source voltage(Vds) in this Linear Region is:')
+end
\ No newline at end of file diff --git a/73/CH16/EX16.13/Example16_13.sci b/73/CH16/EX16.13/Example16_13.sci new file mode 100755 index 000000000..a3eb7bc88 --- /dev/null +++ b/73/CH16/EX16.13/Example16_13.sci @@ -0,0 +1,15 @@ +//Chapter 16_Bipolar and MOS Digital Gate Circuits
+//Caption : Power Dissipation
+// Example16.13: Calculate the maximum power dissipated by saturated load NMOS inverter for following given values: Vdd=5V;Vth=1.5V; device transconductance parameter for load device Kl=23.34*10^-6 A/V^2.Assume Vo=0V in low state.
+// Solution:
+clear;
+clc;
+Vdd=5;// drain voltage of NMOS inverter in Volt
+Vth=1.5;// threshold voltage of NMOS inverter in Volt
+Kl=23.34*10^-6;// transconductance Parameter for load device
+// since maximum power can be obtained if maximum device current flows whish is when Vo=low i.e.,0 V. So,for saturation region of operation we have Id=Kl*(Vgs-Vth)^2/2;
+// for saturated load inverter Vgs=Vds and
+//Vds=Vdd in low output condition,so Id=Kl*(Vdd-Vth)^2/2
+Id=23.34*10^-6*(Vdd-Vth)^2/2;
+Pmax=Id*Vdd;
+disp('mW',Pmax/10^-3,'maximum power dissipated by saturated load NMOS inverter is:')
\ No newline at end of file diff --git a/73/CH16/EX16.14/Example16_14.sci b/73/CH16/EX16.14/Example16_14.sci new file mode 100755 index 000000000..96e5d4937 --- /dev/null +++ b/73/CH16/EX16.14/Example16_14.sci @@ -0,0 +1,12 @@ +//Chapter 16_Bipolar and MOS Digital Gate Circuits
+//Caption : AC Power
+//Example16.14: Calculate the ac power dissipated by a CMOS inverter which drives a 20pF load.Given f=1MHz and Vdd=10V.
+//Solution:
+clear;
+clc;
+Ct=20*10^-12;// load capacitor in Farad
+Vdd=10;//drain voltage supply in Volt
+f=1*10^6;//frequency at which output voltage changes
+//since P=Ct*Vdd^2*f
+P=20*10^-12*(10)^2*10^6;
+disp('W',P,'ac power dissipated by a CMOS inverter is:')
\ No newline at end of file diff --git a/73/CH16/EX16.2/Example16_2.sci b/73/CH16/EX16.2/Example16_2.sci new file mode 100755 index 000000000..68bc44bc6 --- /dev/null +++ b/73/CH16/EX16.2/Example16_2.sci @@ -0,0 +1,33 @@ +//Chapter 16_Bipolar and MOS Digital Gate Circuits
+//Caption : Noise Margin
+//Example 16.2: An RTL gate has the worst case voltages listed below:
+// Temp(degree C) Voh(V) Vih(V) Vil(V) Vol(V)
+// -55 1.014 1.01 0.718 0.710
+// 25 0.844 0.815 0.565 0.300
+// 125 0.673 0.67 0.325 0.320
+//Calculte the worst case NMl and NMh noise margins.
+//Solution:
+clear;
+clc;
+T=[-55;25;125];// temperatures in degree celsius given in table
+for j=1:3,
+if j==1 then
+ disp('Noise margins for T=-55 degree celsius are:')
+ NMl=0.718-0.710;// since NMl=Vil-Vol
+ NMh=1.014-1.01;// since NMh=Vih-Voh
+ disp('Volt',NMl,'lower limit of noise margin at -55 degree celsius is:')
+ disp('volt',NMh,'upper limit of noise limit at -55 degree celsius is:')
+elseif j==2 then
+ disp('Noise margin for T=25 degree celsius are:')
+ NMl=0.565-0.300;
+ NMh=0.844-0.815;
+ disp('Volt',NMl,'lower limit of noise margin at 25 degree celsius is:')
+ disp('Volt',NMh,'upper limit of noise margin at 25 degree celsius is:')
+elseif j==3 then
+ disp('Noise margin for T=125 degree celsius are:')
+ NMl=0.325-0.320;
+ NMh=0.673-0.670;
+ disp('Volt',NMl,'lower limit of noise margin at 125 degree celsius is:')
+ disp('Volt',NMh,'uppwr limit of noise margin at 125 degree celsius is:')
+end
+end
\ No newline at end of file diff --git a/73/CH16/EX16.3/Example16_3.sci b/73/CH16/EX16.3/Example16_3.sci new file mode 100755 index 000000000..debdd5976 --- /dev/null +++ b/73/CH16/EX16.3/Example16_3.sci @@ -0,0 +1,18 @@ +//Chapter 16_Bipolar and MOS Digital Gate Circuits
+//Caption : Fanouts
+//Example 16.3: A TTL gate is guartnteed to sink 10mA without exceeding ann output voltage Vol=0.4V and to source 5mA without dropping below Voh=2.4V. If Tih=100uA at 2.4V and Iil=1mA at 0.4V,calculate the low-state and high-state fan-outs.
+// Solution:
+clear;
+clc;
+// for TTL gate
+// fanout at low output is= collector saturation current of output transitor/load current of the driven gate.
+// fanout for high output is=source current in driving gatte/input current of load gate
+// from question given
+Ic3=10*10^-3;// collector saturatioon current at output transistor
+Ie=1*10^-3;// load current of driven gate
+Ie4=5*10^-3;// source current in driving gate
+Ic1=100*10^-6;// input current of load gate
+Fl=Ic3/Ie;
+disp(Fl,'fan out at low output state is:')
+Fh=Ie4/Ic1;
+disp(Fh,'fan out at high output state is:')
\ No newline at end of file diff --git a/73/CH17/EX17.2/Example17_2.sci b/73/CH17/EX17.2/Example17_2.sci new file mode 100755 index 000000000..eb96e37f2 --- /dev/null +++ b/73/CH17/EX17.2/Example17_2.sci @@ -0,0 +1,12 @@ +//Chapter 17_Light Emitting Diodes and Liquid Crystal Displays
+//Caption :Viewing distance
+//Example17.2: Find out the viewinng distance d for a seven segmant LED display for a character height of 1cm and a height angle of 3 meters.
+//Solution:
+clear;
+clc;
+//d:viewing distance
+h=1*10^-2;//height of character in cm
+O=3;//height angle in meters
+// equivaqlent to height angle of 3 meters
+d=h/tan(0.167*%pi/180);//where 3 meters height angle is equivalent to 0.167 degrees.
+disp('meters',d,'viewing distance is:')
\ No newline at end of file diff --git a/73/CH2/EX2.1/Example2_1.sci b/73/CH2/EX2.1/Example2_1.sci new file mode 100755 index 000000000..851381b67 --- /dev/null +++ b/73/CH2/EX2.1/Example2_1.sci @@ -0,0 +1,21 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Resistance
+//Example2.1: a) A resistor has an aspect ratio of 20:1 and sheet resistance of 200 ohm/square. Find out the value of resistance.
+// b) Find out the number of squares contained in a 2kiloohm resistor whose sheet resistance is 200 ohm per square.
+//Solution: a)
+clear;
+clc;
+function y= myfunction(x,z)//y:resistance, x: sheet resistance=200ohm/square, z: aspect ratio=20:1
+y = x*z //since,resiatance=sheet resistance
+disp('resistance is=")
+disp('ohm',y)
+endfunction
+
+// Solution: b)
+// we have to find number of square which is to find aspect ratio.
+function s=myfunction1(r,p)//r:resistor=2kohm(or 2000ohm), p:sheet resistance=200 Ohm/square, s: number of square(aspect ratio)
+s=r/p//since, number of square=(resistor/sheet resistance)
+disp('number of squares are=")// include ";"at the time of calling the function
+disp('squares',s)
+endfunction
+//myfunction(200,20/1);myfunction1(2000,200);
\ No newline at end of file diff --git a/73/CH2/EX2.10/Example2_10.sci b/73/CH2/EX2.10/Example2_10.sci new file mode 100755 index 000000000..23a3bc5a7 --- /dev/null +++ b/73/CH2/EX2.10/Example2_10.sci @@ -0,0 +1,12 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Absolute Coefficient
+//Example2.10:A thin film resistor measures 150ohmat 25degree celcius and 151.5ohm at 100degree celcius. Calculate its absolute coefficient of resistance in parts per million(ppm) per degre celcius.
+//Solution:
+clear;
+clc;
+function TCR=absresistor(Rt1,Rt2,T1,T2)// TCR:absolute temperature coefficient of resistance, Rt1:resistance at 100 degree celcius=150ohm, Rt2:resistance at 25degree celcius=151.5ohm, T1:temperature=100degree celcius, T2:temperature=25degree celcius
+ TCR=(Rt2-Rt1)*10^6/(Rt1*(T1-T2))
+ disp('absolute coefficient of resistance is=")// include ";" at the time of calling the function at last
+ disp('ppm/degree Celsius',TCR)// ppm: part per million
+endfunction
+//absresistor(150,151.5,100,25);
\ No newline at end of file diff --git a/73/CH2/EX2.11/Example2_11.sci b/73/CH2/EX2.11/Example2_11.sci new file mode 100755 index 000000000..8c2e03a83 --- /dev/null +++ b/73/CH2/EX2.11/Example2_11.sci @@ -0,0 +1,14 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Ratio
+//Example2.11: Two thin resistor are measured at 50 degree celcius and 100degree celsius and are found to have the following values:
+//Temperatur(degree C) Ra(ohm) Rb(ohm)
+// 50 50 100
+// 100 51 102.1
+//Calcullate the ratio TCR in ppm/degree celcius.
+//Solution:
+function TCR= ratio(Rat1,Rbt1,Rat2,Rbt2,T1,T2)
+ TCR=(Rat2/Rbt2-Rat1/Rbt1)*10^6/((Rat1/Rbt1)*(T1-T2))
+ disp('ratio TCR is=")// iinclude ";" at the time of calling the function at last
+ disp('ppm/degree Celsius',TCR)//ppm: part per million
+endfunction
+//ratio(100,50,102.1,51,100,50);
\ No newline at end of file diff --git a/73/CH2/EX2.2/Example2_2.sci b/73/CH2/EX2.2/Example2_2.sci new file mode 100755 index 000000000..1d581ead3 --- /dev/null +++ b/73/CH2/EX2.2/Example2_2.sci @@ -0,0 +1,12 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption: Resistance calculation
+//Example2.2: A thick film resistor is screened with a paste off sheet resistivity 10000 ohm/square,and the resistor is defined as 0.24 cm long and 0.06 cm wide. Calculate the resistance R.
+// Solution:
+clear;
+clc;
+function R=myfunction3(p,l,w)//r:resistor, p=sheet resistance=10000 ohm/square, l:length of resistor=0.24cm, w:width of the resistor=0.06cm
+ R=p*(l/w)//since, resistance=sheet resistance*(length of resistor/width of the resistor)
+ disp('resistance of the thick film resistor is=')// at the time of calling the function include ";" after it
+ disp('ohm',R)
+endfunction
+// myfunction3(10000,0.24,0.06);
\ No newline at end of file diff --git a/73/CH2/EX2.3/Example2_3.sci b/73/CH2/EX2.3/Example2_3.sci new file mode 100755 index 000000000..1d3cb0b47 --- /dev/null +++ b/73/CH2/EX2.3/Example2_3.sci @@ -0,0 +1,12 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Sheet Resistivity
+//Example2.3: Calculate the sheet resistivity of a square of thick film resistor material with the following properties: bulk resistivity=10^-1ohm-cm and thick film thickness=10micrometer.
+//Solution:
+clear;
+clc;
+function Ps=myfunction4(p,t)// Ps:sheet resistance, p:bulk resistivity of thick film=10^-1(or 0.01), t:thickness of thick film=10micrometer(=10*10^-4=0.001)
+ Ps=p/t// since, sheet resistance of the film=bulk resistance of the film/thickness of the film
+ disp('sheet resistivity is=')// include ";" atat the time of calling the function
+ disp('ohm per square',Ps)
+endfunction
+//myfunction4(10^-1,10*10^-4);
\ No newline at end of file diff --git a/73/CH2/EX2.4/Example2_4.sci b/73/CH2/EX2.4/Example2_4.sci new file mode 100755 index 000000000..a5653d11b --- /dev/null +++ b/73/CH2/EX2.4/Example2_4.sci @@ -0,0 +1,14 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Design Capacitor
+//Example2.4: Design a circular 100pF capacitor with the thick film dielectric having dielectric film thickness=0.02mm(or 0.002cm), assume Er=100
+//Solution: We have to find the radius of crcular capacitor inorder to design it.
+clear;
+clc;
+function r1= myfunction5(c,t)
+ r1=c*t// constant=capacitor * thickness of thick film
+ Er=100//given relative permeability of thick film
+ r=sqrt(r1/(Er*%pi*8.85*10^-12))//radius of circular capacitor, Eo=8,85*10^-12(dielectric constant of free space)
+ disp('radius of circular capacitor is=')//include ";" at the time of calling the function
+ disp('meter',r)
+endfunction
+//myfunction5(100*10^-12,0.002);
\ No newline at end of file diff --git a/73/CH2/EX2.6/Example2_6.sci b/73/CH2/EX2.6/Example2_6.sci new file mode 100755 index 000000000..d68c67574 --- /dev/null +++ b/73/CH2/EX2.6/Example2_6.sci @@ -0,0 +1,13 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Capacitance
+//Example2.6: Find out the capacitance of a thick film capacitor,if the dielectric constant Er=100,dielectric film thickness=25micrometer and area A=0.0625 cm square.
+//Solution:
+clear;
+clc;
+function c=capacitance(Er,A,t)
+ c=8.8*10^-12*Er*A/(10^-12*t)// capacitance value will be 2.2*10^-10 or 220pF, Eo:dielectric constant of free space=8.8*10^-12, Er: dielectric constant of thick film= 100(given), A:area of thick film=0.0626 cm square( or 0.0625*10^-4m square), t:thickness of the thick film= 25*10^-6m)
+ // capacitance=Eo*Er*A/t
+ disp('capacitance is=')// c=2.200D-10(=2.2*10^-10)F, include ";" at last at the time of calling the function
+ disp('pF',c)//pF:pico Farad
+endfunction
+//capacitance(100,0.0625*10^-4,25*10^-6);
\ No newline at end of file diff --git a/73/CH2/EX2.8/Example2_8.sci b/73/CH2/EX2.8/Example2_8.sci new file mode 100755 index 000000000..02b0a38d7 --- /dev/null +++ b/73/CH2/EX2.8/Example2_8.sci @@ -0,0 +1,10 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Thickness
+//Example2.8: The bulk resistivity of nichrom is 120uohm-cm. Calculate the thickness T in angstroms of a film with sheet resistivity of 100ohm/square.
+// Solution:
+function T=thickness(Ps,p)// Ps: sheet resistivity of nichrom=100ohm/square, p:bulk resistivity of nichrom=120uohm-cm
+ T=p/(Ps*10^-8)// since Ps=p/T and 1 angstrom =10^-8cm, so dividing by 10^-8 here
+ disp('thickness is=')// include ";" at the time of callimg the function
+ disp('angstrom',T)
+endfunction
+//thickness(100,120*10^-6);
\ No newline at end of file diff --git a/73/CH2/EX2.9/Example2_9.sci b/73/CH2/EX2.9/Example2_9.sci new file mode 100755 index 000000000..3099422e8 --- /dev/null +++ b/73/CH2/EX2.9/Example2_9.sci @@ -0,0 +1,12 @@ +//Chapter 2_Thick Film and Thin Film Hybrid ICs
+//Caption : Length
+//Rxample2.9: Calculate the length of a 400ohm thin film resistor.Given a sheet resistivity oof 100ohm/square and a resistor width of 100um
+//Solution:
+clear;
+clc;
+function L=extent(r,w,Ps)// L:length of thin film, r:resistance of thin film=400ohm, w:width of resistor=100um, Ps:sheet resistance=100ohm/square
+ L=r*w/(10^-6*Ps)//since, r=Ps*L/w and length in micrometer so dividing by 10^-6.
+ disp('length of thin film is=')// include";" at the time of calling the function at last
+ disp('micrometer',L)
+endfunction
+//extent(400,100*10^-6,100);
\ No newline at end of file diff --git a/73/CH3/EX3.2/Example3_2.sci b/73/CH3/EX3.2/Example3_2.sci new file mode 100755 index 000000000..d50dbfdbe --- /dev/null +++ b/73/CH3/EX3.2/Example3_2.sci @@ -0,0 +1,14 @@ +//Chapter 3_Semoconductor Devices Fundamentals
+//Caption : Resistivity
+//Example3.2: A Sample of Si is doped with 10^17 phosphorus atoms/cubic cm. What is its resistivity? Given Un=700square cm/v-sec.
+//Solution:
+clear;
+clc;
+function Res=resistivity(u,n)//n:doped concentration=10^17 atoms/cubic cm, u: mobility of electrons=700square cm/v-sec.
+ q=1.6*10^-19 //q:charge
+ Res=1/(q*u*n)// since P is neglegible.
+ disp('resistivity of the si doped with n-dopant is:')// include ";" at the time of calling
+ disp('ohm-cm',Res)
+endfunction
+// after executing calling resitivity( u=700 and n=10^17)i.e., resistivity(10^17,700);
+// Result: Resistivity of the Si doped with n-dopant is : 0.089 ohm-cm(approx)
\ No newline at end of file diff --git a/73/CH3/EX3.3/Example3_3.sci b/73/CH3/EX3.3/Example3_3.sci new file mode 100755 index 000000000..4b35dc43a --- /dev/null +++ b/73/CH3/EX3.3/Example3_3.sci @@ -0,0 +1,12 @@ +//Chapter 3_Semoconductor Devices Fundamentals
+//Caption : Resistivity of Intrinsic Ge
+//Example3.3: Find the resistivity of intrinsic Ge at 300K. Given un=3900, and up=1900 cm^2/N sec. and ni=2.5*10^13 cm^-3for intrinsic Ge.
+//Solution:
+function RES=resistivity(un,up)// un:electron concentration, up:hole concentration
+ q=1.6*10^-19;//in coulumb
+ ni=2.5*10^13;//concentration in cm^-3
+ RES=1/(q*ni*(un+up))//since n=p=ni
+ disp('resistivity of intrinsic Ge is :')
+ disp('ohm-cm',RES)
+endfunction
+//resistivity(3900,1900);
\ No newline at end of file diff --git a/73/CH3/EX3.4/Example3_4.sci b/73/CH3/EX3.4/Example3_4.sci new file mode 100755 index 000000000..b526a291e --- /dev/null +++ b/73/CH3/EX3.4/Example3_4.sci @@ -0,0 +1,12 @@ +//Chapter 3_Semoconductor Devices Fundamentals
+//Caption: Hole Concentration
+//Example3.4: A semiconductor is doped with a concentration of 10^17 atoms/cm^3 of rsenic . What is the equilibrium hole concentration p at 300K. Given ni=1.5*10^10 cm^-3
+// Solution:
+clear;
+clc;
+function p=holeconcentration(ni,Nd)//ni=intrinsiic concentration=1.5*10^10 cm^-3, Nd: donar concentration; since, Nd>>ni, so Nd=n=10^17 atoms/cm^3.
+ p=ni^2/Nd
+ disp('hole concentrartion at 300K is:')
+ disp('per cubic cm',p)
+endfunction
+//holeconcentration(1.5*10^10,10^17);
\ No newline at end of file diff --git a/73/CH3/EX3.5/Example3_5.sci b/73/CH3/EX3.5/Example3_5.sci new file mode 100755 index 000000000..d58f58175 --- /dev/null +++ b/73/CH3/EX3.5/Example3_5.sci @@ -0,0 +1,11 @@ +//Chapter 3_Semoconductor Devices Fundamentals
+//Caption: Resistivity of Cu
+//Example3.5:The resistivity of metal is given by p=1/nqu, where n is number of electrons per cubic meter, u is mobility ,a nd q is electronic charge. Determine the resistivit of copper at room temperature. Given n=8.5*10^28 per cubic meter, u=3.2*10^-3 m^2/V-sec, at room temperature.
+//Solution:
+q=1.6*10^-19;
+n=8.5*10^28;
+u=3.2*10^-3;
+p=1/(n*q*u);
+disp('resistivity of the copper is :')
+disp('ohm-meter',p)
+// 2.298D-08 means 2.298*10^-8
\ No newline at end of file diff --git a/73/CH3/EX3.6/Example3_6.sci b/73/CH3/EX3.6/Example3_6.sci new file mode 100755 index 000000000..63009dba5 --- /dev/null +++ b/73/CH3/EX3.6/Example3_6.sci @@ -0,0 +1,41 @@ +//Chapter 3_Semoconductor Devices Fundamentals
+//Caption : Bipolar Transistor Parameters
+//Example3.6: Determine Cu,Ccs,Gm,C1,R1,R0 and Ru for a bipolar transisitor. Given : Ic=0.2 mA, Vcb=10V,Vcs=15V,Cuo=0.25pF,Cje=1 pF,Ccso=1.5pF,Bo=2000,Tf=0.3ns, n=2*10^-4 and Vo=0.55V for all junctions.
+// Solution:
+clear;
+clc;
+Cuo=0.25;// collector-base depletion region capacitance in pico Farad(pF) for zero bias
+Ccso=1.5 ;// collector-substrate junction capacitance in pico Farad(pF) for zero bias
+q=1.6*10^-19 ;//electron charge in coulomb
+Ic=0.2 ;//collector current in ampere(A)
+k=8.6*10^-5;//in eV/K, where 1eV=1.6*10^-19
+T=300;//absolute temperature in kelvin(K)
+Vcb=10 ;//forward bias on the junction in volt(v)
+Vcs=15 ;//collector-substrate bias in volt(V)
+Cje=1 ;//depletion region capacitance in pico Farad(pF)
+Bo=200;//small signal current gain
+Tf=0.3;//transit time in forward direction in nano seconds(nS)
+n=2*10^-4;// proportionality constant for Ro and gm
+Vo=0.55;// bias voltage in volt(V)
+Cu=Cuo/sqrt(1+(Vcb/Vo));// collector-base capacitance
+disp('Cu is:')
+disp('pF',Cu)
+Ccs=Ccso/sqrt(1+(Vcs/Vo));// collector-substrate capacitance
+disp('Ccs is:')
+disp('pF',Ccs)
+gm=q*Ic/(k*T*1.6*10^-19);// since k is in eV so converting it in Coulomb/Kelvin
+disp('gm is:')// transconductance of the bipolar transistor here
+disp('mA/V',gm)
+Cb=Tf*gm;//diffusion capacitance in pico Farad(pF)
+C1=Cb+Cje;//small signal capacitance of bipolar transistor
+disp("C1 is:")
+disp("pF",C1)
+R1=Bo/gm;// small signal input resistance of bipolar transistor
+disp('R1 is:')
+disp('kilo Ohm',R1)
+Ro=1/(n*gm);//small signal output resistance
+disp('R0 is')
+disp('kilo Ohm',Ro)
+Ru=10*Bo*Ro/10^3;//collector-base resistance
+disp('Ru is:')
+disp('Mega Ohm',Ru)
\ No newline at end of file diff --git a/73/CH5/EX5.1/Example5_1.sci b/73/CH5/EX5.1/Example5_1.sci new file mode 100755 index 000000000..c85d84612 --- /dev/null +++ b/73/CH5/EX5.1/Example5_1.sci @@ -0,0 +1,13 @@ +//Chapter 5_Monolithic Components
+//Caption : Transit Time
+//Example5.1: A lateral pnp device base width is 8 um and the diffusion cofficient for base region is 10 cm^2/sec. Calculate the base transit time and the unity gain frequency.
+//Solution:
+function T=transittime(W,D)//W:base width=8um; D:base diffusion cofficient=10 sq cm/sec.
+ T= W^2/(2*D);// since f(transit frequency response )=2*D/(W^2)
+ disp('base transit time is:')
+ disp('ns',T*10^9)// in nanoseconds(ns)
+ F=1/(2*%pi*T)// where F=unity gain frequency=1/(2*%pi*transit time)
+ disp('unity gain frequency is:')
+ disp('MHz',F/10^6)// in Mega Hertz
+endfunction
+//transittime((8*10^-6),10*10^-4);
\ No newline at end of file diff --git a/73/CH5/EX5.2/Example5_2.sci b/73/CH5/EX5.2/Example5_2.sci new file mode 100755 index 000000000..dbe69b044 --- /dev/null +++ b/73/CH5/EX5.2/Example5_2.sci @@ -0,0 +1,18 @@ +//Chapter 5_Monolithic Components
+//Caption : Unit gain frequency
+//Example5.2: a) Find Fl(unit gain frequency ) for the lateral pnp device. Assume diffusion cofficient of holes in the base of 0.5 sq cm/sec and base width of 10um.
+// b) Find the Fs(unit gain frequency) for the substrate pnp device.Assume D=20 sqcm/sec and W=8um.
+clear;
+clc;
+//a)Solution: for the lateral pnp device
+Wl=10*10^-4;//base width in micro centimeter(ucm)
+Dl=0.5;// base diffusion cofficient in sq cm/sec
+Fl=2*Dl/(Wl^2);
+disp('unit gain frequency for lateral pnp device is;' )
+disp('MHz',Fl/10^6)
+//b)Solution: for substrate pnp device
+Ws=8*10^-4;// base in ucm
+Ds=20;//base diffusion cofficient in sq cm/sec
+Fs=Ds/(Ws^2);
+disp('unit gain frequency for substrate pnp device is;')
+disp('MHz',Fs/10^6)
\ No newline at end of file diff --git a/73/CH5/EX5.3/Example5_3.sci b/73/CH5/EX5.3/Example5_3.sci new file mode 100755 index 000000000..1bbc897a2 --- /dev/null +++ b/73/CH5/EX5.3/Example5_3.sci @@ -0,0 +1,17 @@ +//Chapter 5_Monolithic Components
+//Capation : Resistance and Sheet resistance
+//Example5.3: a)A base diffusion layer length is 100um and it's width is 10um.The sheet resistance of the layer is 100 ohm/square.Calculate its resistance.
+//b) Calculate the sheet resistance of a 20um thick,5 ohm-cm ep-layer.
+//a)Solution:
+L=100;//base diffusion layer in um
+W=10;//base diffusion width in um
+Rs=100;//sheet resistance in ohm/square
+R=L*Rs/W;
+disp('resistance of base diffusion layer is:')
+disp('Ohm',R)
+//b)Solution:
+Pe=5*10^-2;//ep-layer resistivity in ucm
+t=20*10^-6;//thickness of the layer in um
+Rse=Pe/t;//sheet resitivity of ep-layer
+disp('sheet resistance of ep-layer is:')
+disp('Ohm',Rse)
\ No newline at end of file diff --git a/73/CH5/EX5.4/Example5_4.sci b/73/CH5/EX5.4/Example5_4.sci new file mode 100755 index 000000000..bf26e8614 --- /dev/null +++ b/73/CH5/EX5.4/Example5_4.sci @@ -0,0 +1,12 @@ +//Chapter 5_Monolithic Components
+//Caption : Capacitance per unit area
+//Example5.4: Determine the capacitance per unit area of the 400 armstrong gate oxide of a MOSFET device relative permittivity of silicon dioxide=3.9.
+//Solution:
+clear;
+clc;
+Eo=8.86*10^-14;//permittivity of free space in F/cm
+Er=3.9;//relative permittivity of MOSFET device
+t=0.4*10^-5;//thickness of the gate oxide in cm
+Co=Eo*Er/t;// since capoacitance(C)=permittivity(E)*area(A)/thicknes(t); so C/A=e/t
+disp('capacitance per unit area of gate oxide is:')
+disp('F/cm^2',Co)
\ No newline at end of file diff --git a/73/CH7/EX7.1/Example7_1.sci b/73/CH7/EX7.1/Example7_1.sci new file mode 100755 index 000000000..7da550bc4 --- /dev/null +++ b/73/CH7/EX7.1/Example7_1.sci @@ -0,0 +1,35 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Bipolar Differential Amplifier Parameter
+//Example7.1: The following specification are given for the dual input,balanced output bipolar diferential amplifier:
+//Rc=2.2 kOhm,Re=4/7kOhm,Rs=50 ohm,Vcc=10V,Vee=-10V and Bf=Bo=100. Assume Vbe=0.7V.
+//Determine
+//a)Icq and Vceq
+//b)Differential-mode voltage fgain , and
+//c)Input and output resistances.
+clear;
+clc;
+//a)Solution:\
+Rc=2.2*10^3;// collector resistance in one transistor in ohm
+Re=4.7*10^3;// emitte resitance of one transistor in ohm
+Rs=50;//base or source resitance of one transistor in ohm
+Vcc=10;// collector power supply in Volt
+Vee=-10;// negative emitter power supply
+Bf=100;// gain of the transistor
+Bo=Bf;
+Vbe=0.7;//base emitter voltage of one transiator
+Icq=(abs(Vee)-Vbe)/(2*Re+(Rs/Bf));
+Vceq=Vcc+Vbe-Rc*Icq;
+//b)Solution:
+gm=Icq/(25*10^-3);// where transconductance gm=Ic/Vt, Vt=25mV at room temperature, so gm =Ic/25
+Ad=-gm*Rc;// differential mode voltage gain Icq here will be taken as found above not approximated to as given book
+//c)Solution:
+r=Bo/gm;// input resistance of one transistor
+Ri=2*r;//differential mode input resistance
+Ro=Rc;//differential mode output resistance
+disp('A',Icq*10^3,'operating point collector current')
+disp('V',Vceq,'collector-to-emitter voltage is:')
+disp(abs(Ad),'Differential-mode voltage gain')
+disp('kilo Ohm',Ri/10^3,'Input Resistance')
+disp('kilo Ohm',Ro/10^3,'Output Resistance')
+// Note:
+//value of Icq is taken as 0.0009893 A or 0.9893 mA not approximated to 0.98 mA
\ No newline at end of file diff --git a/73/CH7/EX7.14/Example7_14.sci b/73/CH7/EX7.14/Example7_14.sci new file mode 100755 index 000000000..c159d9d1d --- /dev/null +++ b/73/CH7/EX7.14/Example7_14.sci @@ -0,0 +1,18 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Effect on Output Voltage
+//Example7.14: For the noninverting OP-Amp with input resistance R1 nad feedback resistance R2 find the effect on output voltage Vo because of the common mode voltage Vcm when the input voltage Vs changes by 1V. Given CMRR=70 dB.
+//Solution:
+clear;
+clc;
+CMRR=70;//Common Mode Rejection Ratio in dB
+// since CMRR=20*log(Vcm/Vdm) dB
+//so Vdm=Vcm/10^(CMRR/20)
+//since output voltage of OP-Amp is Vo=(R1+R2)*Vdm/R1=(R1+R2)*Vcm/(R1*10^(CMRR/20))
+R1=100;//assuming input resistance standard value in kilo Ohm
+R2=900;//assuming feedback resistance standard value in kilo Ohm
+Vs=1;//change in input voltage given in question
+Vcm=Vs;//since change in input voltage is applied to noninverting input and through the feedback to the inverting iput of the Op-Amp as well.
+Vo=(R1+R2)*Vcm/(R1*10^(CMRR/20))
+disp('mV',abs(Vo)*10^3,'change in output voltage due to common mode Voltage(Vcm) is:')
+//Note:
+// CMRR,Vdm,Vo may be of either polarity. Here absolute value is calculated
\ No newline at end of file diff --git a/73/CH7/EX7.15/Example7_15.sci b/73/CH7/EX7.15/Example7_15.sci new file mode 100755 index 000000000..16eac7761 --- /dev/null +++ b/73/CH7/EX7.15/Example7_15.sci @@ -0,0 +1,16 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Slew rate and Fmax
+//Example7.15: For type 741 Op-Amp following parameter are given.Quiescent collector current Ic=9.5 uA, Cc=30 pF. Peak amplitude of input voltage Vm=15V.
+//a)Determine the slew rate
+//b)Determine full power bandwidth Fmax for the slew rate as obtained from part (a).
+clear;
+clc;
+//a)Solution:
+Ic=9.5*10^-6;// operating collector current in A
+Cc=30*10^-12;// parasitic capacitance
+SlewRate=2*Ic/Cc;
+disp('V/us',SlewRate/10^6,'Slew rate is:')
+//b)Solution:
+Vm=15;//amplitude of input voltage in Volt
+Fmax=SlewRate/(2*%pi*Vm);// full power bandwidth
+disp('kHz',Fmax/10^3,'full power bandwidth Fmax for the Slew Rate obtained above is:')
\ No newline at end of file diff --git a/73/CH7/EX7.16/Example7_16.sci b/73/CH7/EX7.16/Example7_16.sci new file mode 100755 index 000000000..96d49ce90 --- /dev/null +++ b/73/CH7/EX7.16/Example7_16.sci @@ -0,0 +1,12 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Largest Amplitude
+//Example7.16: An amplifier has a 10 kHz sinewave input signal. Find the largest amplitude that the output of the amplifier can be,without distortion owing to slew rate limiting. Given slew rate=0.5V/u sec.
+//Solution:
+clear;
+clc;
+Fmax=10*10^3;//frequency of sinewave input signal in Hz
+SlewRate=0.5*10^6;//given in question in V/sec
+Vm=SlewRate/(2*%pi*Fmax);//Since Fmax=slew rate/(2*%pi*Vm)
+disp('V(peak)',Vm,'largest amplitude that the output of the amplifier can be without distortion owing to slew rate limitation is:')
+//Note:
+// calculated amplitude is 7.9577 V, which can be approximated to 8 V
\ No newline at end of file diff --git a/73/CH7/EX7.17/Example7_17.sci b/73/CH7/EX7.17/Example7_17.sci new file mode 100755 index 000000000..6b4df5f57 --- /dev/null +++ b/73/CH7/EX7.17/Example7_17.sci @@ -0,0 +1,14 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Maximum allowable frequency
+//Example7.17: When a low frequency sinusoidal waveform is applied to an input of the noninverting Op-Amp the amplifier responds linearly over an output range from -10V to +10V. If R1=R2 and the slew rate of the amplifier is 50V/u sec,what is the maximum allowable frequency of an input sinusoid if the output signal swing is to be maintained from -10V to +10V without distortion? resistance and R2 is feedback resitance.
+//Solution
+clear;
+clc;
+SlewRate=50/10^-6;//in V/sec
+Vo=10-(-10);//from question output is ranging from -10V to +10V
+Vom=Vo;//where Vom is the maximum value of Vo
+//therefore
+Fmax=SlewRate/(2*%pi*Vom);
+disp('kHz',Fmax/10^3,'maximum allowable frequency of an input sinusoidal for output swing maintained from -10V to +10v is:')
+//Note:
+// obtained maximum allowable amplitude is 397.88736 kHz which can be approximated to 400 kHz
\ No newline at end of file diff --git a/73/CH7/EX7.2/Example7_2.sci b/73/CH7/EX7.2/Example7_2.sci new file mode 100755 index 000000000..0e6e8e795 --- /dev/null +++ b/73/CH7/EX7.2/Example7_2.sci @@ -0,0 +1,16 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Rc and Re
+//Example7.2: A bipolar differentail amplifier uses a transistor having Bo=200 and biased at Icq=100 uA. Determine Rc and Re so that abs(Ad)=500 and CMRR=80 dB.
+// Solution:
+clear;
+clc;
+//CMRR in dB is expressed as 20logCMRR, so 80=20logCMRR or
+CMRR=10^(80/20);
+Icq=100*10^-6;//collector current
+Vt=25*10^-3;//standard value of threshold voltage at room temperature
+gm=Icq/Vt;
+Re=CMRR/(2*gm);//since CMRR=2*gm*Re(approx)
+Ad=500;// absolute value of differential mode voltage gain
+Rc=-Ad/gm;//Collector resistance
+disp('Mega Ohm',Re/10^6,'emitter resitance(Re) of bipolar differential amplifier is:')
+disp('Kilo Ohm',abs(Rc)/10^3,'collector resistance(Rc) of bipolar differential amplifier is:')
\ No newline at end of file diff --git a/73/CH7/EX7.4/Example7_4.sci b/73/CH7/EX7.4/Example7_4.sci new file mode 100755 index 000000000..ebf78e1b7 --- /dev/null +++ b/73/CH7/EX7.4/Example7_4.sci @@ -0,0 +1,18 @@ +//Chapter 7_Operation Amplifier Characteristics
+//Caption : Offset Voltage Change
+//Example7.4: What is the change in the offset voltage of a bipolar transistor amplifier for a difference of 10V in the collector-to-emitter voltage and Va=250 V. Assume room temperature.
+//Solution:
+clear;
+clc;
+Vt=25*10^-3;// threshold voltage at room temperature in Volt
+Va=250;//early voltage of the bipolar transistor in volt
+deltaVce=1;//let us assume 1V of change in Vce(collector-to-emitter voltage)
+deltaVos1=Vt*(-deltaVce/Va);
+disp('mV',abs(deltaVos1)*10^3,'change in offset voltage for 1 V change in Vce is:')
+for i=1:1,
+ if i==1 then
+ deltaVce=10;// in volt
+ deltaVos=deltaVce*deltaVos1;
+ disp('mV',abs(deltaVos)*10^3,'change in offset voltage of bipolar transistor for 10V collector-to-emitter voltage(Vce) difference is:')
+ end
+end
\ No newline at end of file diff --git a/73/CH7/EX7.5/Example7_5.sci b/73/CH7/EX7.5/Example7_5.sci new file mode 100755 index 000000000..7514aebb4 --- /dev/null +++ b/73/CH7/EX7.5/Example7_5.sci @@ -0,0 +1,13 @@ +//Chapter 7_Operational Amplifier Characteristics
+//Caption : Temperature Coefficient
+//Example7.5: Determine the temperature coefficient of the input offset voltage for the bipolar differential amplifier having Vos=1.5 mV. What is the percentage change in the Vos per degree temperature change.
+//Solution:
+clear;
+clc;
+// temperature cofficient of the input offset voltage for the bipolar differential amplifier Vos is=dVos/dT=Vos/T;
+Vos=1.5*10^-3;//input offset voltage for bipolar differential transistor amplifier
+T=300;// assuming room temperature
+TC=Vos/T;// temperature cofficient of Vos
+//percentage change in the Vos per degree temperature change will be given by as follow:
+PC=(TC/Vos)*100;// percentage change(PC) in the Vos per degree temperature change
+disp('%per degree celcius',PC,'percentage change in the Vos per degree temperature change is:')
\ No newline at end of file diff --git a/73/CH8/EX8.1/Example8_1.sci b/73/CH8/EX8.1/Example8_1.sci new file mode 100755 index 000000000..aabd482c5 --- /dev/null +++ b/73/CH8/EX8.1/Example8_1.sci @@ -0,0 +1,12 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Device Temperature
+//Example8.1:The Heat generated by a linear IC,uA 741 is 200 mW.If the thermal resistance is 150 degree Celsius/Watt and the ambient temperature is 25 degree celsius.calculate the device temperature.
+//Solution:
+clear;
+clc;
+Pd=200*10^-3;//heat generated
+Rt=150;//thermal resistance
+Ta=25;//ambient temperature in degree celsius
+//assuming thermal equilibrium conditon
+Td=Pd*Rt+Ta;
+disp('degree celsius',Td,'The device temperature is:')
\ No newline at end of file diff --git a/73/CH8/EX8.2/Example8_2.sci b/73/CH8/EX8.2/Example8_2.sci new file mode 100755 index 000000000..f1f04ba365 --- /dev/null +++ b/73/CH8/EX8.2/Example8_2.sci @@ -0,0 +1,12 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Device Temperature
+//Example8.2: For the device in Example8.1, Pdmax=500 mW. Determine the device temperature after equilibrium is attained for an ambient temperature of 75 degree celsius and if the device is subjected to maximum heat generation.Maximum allowable device temperature is 150 degree Celsius.
+//Solution:
+clear;
+clc;
+Pmax=500*10^-3;
+Pd=Pmax;//since device is subjected to maximum heat generation
+Rt=150;//thermal resitance
+Ta=75;//ambient temperature
+Td=Pd*Rt+Ta;
+disp('degree celsius',Td,'device temperature is:')
\ No newline at end of file diff --git a/73/CH8/EX8.3/Example8_3.sci b/73/CH8/EX8.3/Example8_3.sci new file mode 100755 index 000000000..99300304a --- /dev/null +++ b/73/CH8/EX8.3/Example8_3.sci @@ -0,0 +1,11 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Device Temperature
+//Example8.3:a)The ambient temperature of the device of Example8.2 rises above 90 degree celsius. What is the new value of Td if it still generates 500 mW?
+//a)Solution:
+clear;
+clc;
+Pd=500*10^-3;
+Rt=150;//thermal resistance
+Ta=90;//ambient temperature
+Td=Pd*Rt+Ta;
+disp('degree celsius',Td,'New value of device temperature is:')
\ No newline at end of file diff --git a/73/CH8/EX8.4/Example8_4.sci b/73/CH8/EX8.4/Example8_4.sci new file mode 100755 index 000000000..e5bdccce2 --- /dev/null +++ b/73/CH8/EX8.4/Example8_4.sci @@ -0,0 +1,11 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Device Temperature
+//Example8.4: Forced air cooling provided for the device in Example8.3 lowers the ambient temperature at 60 degree celsius.What is temperature of the device?
+//Solution:
+clear;
+clc;
+Pd=500*10^-3;
+Rt=150;//thermal resistance
+Ta=60;//ambient temperature
+Td=Pd*Rt+Ta;
+disp('degree celsius',Td,'Temperature of the device is:')
\ No newline at end of file diff --git a/73/CH8/EX8.7/Example8_7.sci b/73/CH8/EX8.7/Example8_7.sci new file mode 100755 index 000000000..07e0b86d9 --- /dev/null +++ b/73/CH8/EX8.7/Example8_7.sci @@ -0,0 +1,15 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Output Voltage
+//Example8.7: In the summing amplifier(inverting mode) the signals to be combined are V1=3V, V2=2v, and V3=1V.The input resistor are R1=R2=R3=3 kilo ohm.The feeddback resistor Rf=1 kilo ohm. Consider ideal Op-Amp,determine Vo.
+//Solution:
+clear;
+clc;
+V1=3;//input signal
+V2=2;//input signal
+V3=1;//input signal
+Rf=1*10^3;//feedback resitor
+R1=3*10^3;//input resistor in ohm
+R2=R1;//input resistor in ohm
+R3=R2;//input resistor in ohm
+Vo=-(Rf/R1*V1+Rf/R2*V2+Rf/R3*V3);
+disp('V',Vo,'Output Voltage of summing amplifier is:')
\ No newline at end of file diff --git a/73/CH8/EX8.8/Example8_8.sci b/73/CH8/EX8.8/Example8_8.sci new file mode 100755 index 000000000..f7a66d872 --- /dev/null +++ b/73/CH8/EX8.8/Example8_8.sci @@ -0,0 +1,18 @@ +//Chapter 8_Applications of Operational Amplifier
+//Caption : Vp and Vo
+//Example8.8: In the circuit of non-inverting summing Op-Amp, V1=+2V, V2=-4V, V3=+5V. input resistors for all the three input signal are same and are equal to 1 kilo Ohm.The feedback resistor Rf is 2 kilo ohm. Determine the voltage Vp at the noninverting pin of the Op-Amp and the output Vo. Assume ideal Op=Amp.
+//Solution:
+clear;
+clc;
+Rf=2*10^3;//feedback resistor
+R1=1*10^3;
+R2=R1;
+R3=R2;
+V1=2;
+V2=-4;
+V3=5;
+n=3;//no of inputs
+Vp=(Rf/R1*V1+Rf/R2*V2+Rf/R3*V3)/n;
+Vo=(1+Rf/R1)*Vp;
+disp('V',Vp,'voltage at noninverting pin is:')
+disp('V',Vo,'output voltage voltage of noninverting summing Op-Amp is:')
\ No newline at end of file diff --git a/73/CH9/EX9.12/Example9_12.sci b/73/CH9/EX9.12/Example9_12.sci new file mode 100755 index 000000000..97d5605c0 --- /dev/null +++ b/73/CH9/EX9.12/Example9_12.sci @@ -0,0 +1,18 @@ +//Chapter 9_Active Filters
+//Caption : Unity gain frequency and Capacitor determination
+//Example9.12: a)Determine the unity gain frequency,Fo,of a switched capacitor integrator having following specifications:Fclk=1 kHz, C1=1 pF,and C2=15.9 pF
+//b)What is the value of capacitor for an RC integrator having R=1.6 mega Ohm and Fo as obtained in part(a).
+//a)Solution:
+clear;
+clc;
+C1=1*10^-12;//source capacitor in F
+C2=15.9*10^-12;//feedback capacitor
+Fclk=1*10^3;//clock frequency or switching frequency
+Fo=1*(C1/C2)*Fclk/(2*%pi);
+disp('Hz',Fo,'unity gain frequency is:')
+//b)Solution:
+R=1.6*10^6;//resistor of RC integrator in Ohm
+C=1/(2*%pi*Fo*R);
+disp('nF',C*10^9,'for Rc integrator value of capacitor needed is:')
+// Note:
+// Obtained results are approximated to nearest values,thus Fo=10 Hz and C=10 nF
\ No newline at end of file diff --git a/73/CH9/EX9.6/Example9_6.sci b/73/CH9/EX9.6/Example9_6.sci new file mode 100755 index 000000000..13f2c7e6c --- /dev/null +++ b/73/CH9/EX9.6/Example9_6.sci @@ -0,0 +1,13 @@ +//Chapter 9_Active Filters
+//Caption : Determine Q Fl and Fh
+//Example9.6:A certain two-pole band pass filter response is required with a centre frequency of 2 kHz and a 3 dB bandwidth of 400 Hz. Determine Q,Fl and Fh.
+//Solution:
+clear;
+clc;
+Fo=2*10^3;//centre frequency in Hz
+BW=400;//3 dB bandwidth
+Q=Fo/BW;// Q-factor of band pass filter
+Fl=Fo*sqrt(1+1/(4*Q^2))-Fo/(2*Q);
+Fh=Fo*sqrt(1+1/(4*Q^2))+Fo/(2*Q);
+disp('Hz',Fl,'lower cutt off frequency is:')
+disp('Hz',Fh,'Higher cutt off frequency is:')
\ No newline at end of file |