from the example problem in lecture 07 (which is also .... from the example problem in lecture 07...

1. From the example problem in Lecture 07 (which is also Example 3-2 in Hart):

a. Figure 1: Using MATLAB, plot the current ( )i wt from 0 2π− . From theMATLAB data and on the graph, determine and NEATLY label the extinctionangle, β , the average current, avgI , and the rms current rmsI . Do these values

match the values determined in class?b. Figure 2: Using MATLAB, plot the instantaneous voltage, instantaneous current,

and instantaneous power. From the MATLAB data on the graph, determineand NEATLY label the average power. Does this value match the valuedetermined in class?

c. Using MATLAB, print out the values for β , rmsI , avgI , oP , and power factor pf .

Do these values match the values determined in class? d. Turn in:

i. Figures 1 and 2ii. Printout of the values determined in Part c

iii. Print out your MATLAB codeiv. Upload your MATLAB files to your shared folder on the Google Drive

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

w*t [rad]

i D [A

]

Diode Current for Half-Wave Rectifier with R Load and R-L Load

iRiR,rmsiRLiRL,rms

0 1 2 3 4 5 6-100

-50

0

50

100Instantaneous power for Half-Wave Rectifier with R Load and R-L Load

Vs [V

]

0 1 2 3 4 5 60

0.5

1

i D [A

]

iRiR,rmsiRLiRL,rms

0 1 2 3 4 5 60

20

40

60

80

w*t [rad]

Pow

er [W

]

p(wt),Rpavg,Rp(wt),RLpavg,RL

1/15/15 10:43 PM MATLAB Command Window 1 of 1

beta =

3.4959

iDrms =

0.4730

iDave =

0.3081

Pout =

22.3720

pf =

0.6689

>>

Problem 3.8: RL-Source Load Half-Wave Rectifier

A half-wave rectifier of Fig. 3-5a has a 240 V rms, 60 Hz ac source. The load is a series inductance, resistance, and dc source, with L = 75 mH, R = 10 Ω, and Vdc = 100 V. Determine (a) the power absorbed by the dc voltage source, (b) the power absorbed by the resistance, and (c) the power factor.

Using Eqs. 3-22 and 3-23, /

/

2 2 2 2

1 1

1

) ( ) sin( )

sin( )

( ) 10 (377(.075) 30.0377(.075)tan tan 1.23

10377(0.075) 2.83

10100sin 0.299 17.

240 2

tm dc

m dc

dc

m

V Va i t t AeZ R

V VA eZ R

Z R LL rad

RL

RV radV

ω ωτ

α ωτ

ω ω θ

α θ

ω

ωθ

ωωτ

α

−

− −

−

= − − +

= − − +

= + = + = Ω

= = =

= = =

= = = =

/2.83

2 2

1

( ) 11.3sin( 1.23) 10 21.2 ; 3.94 2263.13 . ( (100)(3.13) 313 .

) 4.81 . ( (4.81) (10) 231 .313 231)

(240

t

avg dc dc avg

rms R rms

i t t e radI A numerical integration), P V I W

b I A numerical integration) P I R WPc pfS

ωω ω β−

°

= − − + = = °= = = =

= = = =+

= = 0.472 47.2%)(4.81)

= =

Hart, Power Electronics, 1/e. © 2011 The McGraw-Hill Companies.

Problem 3.13: Freewheeling Diode

The half-wave rectifier with a freewheeling diode (Fig. 3-7a) has R = 12 Ω and L = 60 mH. The source is 120 V rms at 60 Hz. (a) From the Fourier series of the half-wave rectified sine wave that appears across the load, determine the dc component of the current. (b) Determine the amplitudes of the first four nonzero ac terms in the Fourier series. Comment on the results.

Solution 3.13

Using Eq. 3-34,

a)

00 0

120 2 5454.0 .; 4.50 .12

mV VV V I ARπ π

= = = = = =

b) n Vn Zn In

0 54.02 12.00 4.50

1 84.85 25.6 3.31

2 36.01 46.8 0.77

4 7.20 91.3 0.08 The terms beyond n = 1 are insignificant.



>> CircuitData

CircuitData =

wt: [1x400 double]Angle: [1x1 struct]

Current: [1x1 struct] Voltage: [1x1 struct]

Power: [1x1 struct]

>> CircuitData.Angle

ans =

AlphaMin: 0.2991 DelayAngle: 0

Beta: 3.9368

>> CircuitData.Current

ans =

iout: [1x400 double] Iavg: 3.1256 Irms: 4.8009

>> CircuitData.Voltage

ans =

vsource: [1x400 double]vout: [1x400 double]vR: [1x400 double]

vD_forward: [1x400 double] vD_reverse: [1x400 double]

vL: [1x400 double]Vdc: 100

>> CircuitData.Power

ans =

pSource: [1x400 double]pR: [1x400 double]pD: [1x400 double]

pVdc: [1x400 double]pL: [1x400 double]

Ptotal: 543.0392 PaverageR: 230.4837 PaverageD: 0

PVdc: 312.5555


PL: -0.0015pf: 0.4713

efficiency: 0.5756

>>

0 1 2 3 4 5 6

-300

-200

-100

0

100

200

300

v sour

ce [V

], i ou

t [A]

Voltage Source and Output Current vs. Time

vsourceiout

0 1 2 3 4 5 6-500

0

500

1000

1500

2000

2500

Pow

er [W

]

Instantaneous Source Power Delivered

wt [rad]

psourcePtot,avg

0 1 2 3 4 5 60

20

40

60

80

v R [V

], i ou

t [A]

Voltage Across Load Resistance and Output Current vs. Time

vRiout

0 1 2 3 4 5 60

200

400

600

800

Pow

er [W

]

Instantaneous Power Absorbed by the Load Resistance

wt [rad]

pRPR,avg

0 1 2 3 4 5 60

20

40

60

80

100

v dc [V

], i ou

t [A]

DC Voltage Source and Output Current vs. Time

VdcioutIavg

0 1 2 3 4 5 60

200

400

600

800

Pow

er [W

]

Instantaneous Power Absorbed by the DC Voltage Source

wt [rad]

pVdcPVdc,avg

0 1 2 3 4 5 6

-300

-200

-100

0

100

v L [V],

i out [A

]

Voltage Across Inductor and Output Current vs. Time

vLiout

0 1 2 3 4 5 6

-1000

-500

0

500

1000

Pow

er [W

]

Instantaneous Power Absorbed/Delivered by the Inductor

wt [rad]

pLPL,avg

1/27/15 9:27 PM C:\Users\J...\EE344_HW_05_Spring_2015_v2.m 1 of 4

clear;clc;

% example 3-2 in Hart% CircuitParameters.Vm=100; %[V] Amplitude of sinusoidal input% CircuitParameters.R=100; % [Ohms] Resistance% CircuitParameters.L=0.1; % [H] Inductance% CircuitParameters.f=60; % [Hz] Frequency% CircuitParameters.Vf=0; %[V] Non-ideal diode dc voltage drop (forward bias)% CircuitParameters.Rf=0; % [Ohms] Non-ideal diode resistance (forward bias)% CircuitParameters.Vdc=0; % [V] dc voltage source in series [V]% CircuitParameters.alphad=0; % [rad] delay angle (=0 if unctontrolled)% CircuitParameters.numpoints=400; % number of points in the output waveforms

% example 3-5 in Hart% CircuitParameters.Vm=120*sqrt(2); %[V] Amplitude of sinusoidal input% CircuitParameters.R=2; % [Ohms] Resistance% CircuitParameters.L=0.02; % [H] Inductance% CircuitParameters.f=60; % [Hz] Frequency% CircuitParameters.Vf=0; %[V] Non-ideal diode dc voltage drop (forward bias)% CircuitParameters.Rf=0; % [Ohms] Non-ideal diode resistance (forward bias)% CircuitParameters.Vdc=100; % [V] dc voltage source in series [V]% CircuitParameters.alphad=0; % [rad] delay angle (=0 if uncontrolled)% CircuitParameters.numpoints=400; % number of points in the output waveforms

% Problem 3.8 in Hart:CircuitParameters.Vm=240*sqrt(2); %[V] Amplitude of sinusoidal inputCircuitParameters.R=10; % [Ohms] ResistanceCircuitParameters.L=0.075; % [H] InductanceCircuitParameters.f=60; % [Hz] FrequencyCircuitParameters.Vf=0; %[V] Non-ideal diode dc voltage drop (forward bias)CircuitParameters.Rf=0; % [Ohms] Non-ideal diode resistance (forward bias)CircuitParameters.Vdc=100; % [V] dc voltage source in series [V]CircuitParameters.alphad=0; % [rad] delay angle (=0 if uncontrolled)CircuitParameters.numpoints=400; % number of points in the output waveforms

% example 3-12 in Hart:% CircuitParameters.Vm=120*sqrt(2); %[V] Amplitude of sinusoidal input% CircuitParameters.R=2; % [Ohms] Resistance% CircuitParameters.L=0.02; % [H] Inductance% CircuitParameters.f=60; % [Hz] Frequency% CircuitParameters.Vf=0; %[V] Non-ideal diode dc voltage drop (forward bias)% CircuitParameters.Rf=0; % [Ohms] Non-ideal diode resistance (forward bias)% CircuitParameters.Vdc=100; % [V] dc voltage source in series [V]% CircuitParameters.alphad=45*(pi/180); % [rad] delay angle (=0 if uncontrolled)% CircuitParameters.numpoints=400; % number of points in the output waveforms

%% Find values for the half-wave rectifierCircuitData=HWRectifier(CircuitParameters);wt=CircuitData.wt; % [rad] x-axis


%% plot resultsfigure(1)subplot(2,1,1);plot(wt,CircuitData.Voltage.vsource,'r',... wt,CircuitData.Current.iout,'b');grid;ylabel('v_source [V], i_out [A]');title('Voltage Source and Output Current vs. Time');legend('v_source','i_out');axis([0,2*pi,... min(min(CircuitData.Voltage.vsource),min(CircuitData.Current.iout)),... max(max(CircuitData.Voltage.vsource),max(CircuitData.Current.iout))]);subplot(2,1,2);plot(wt,CircuitData.Power.pSource,'r',... wt,CircuitData.Power.Ptotal,'b');grid;ylabel('Power [W]');title('Instantaneous Source Power Delivered');xlabel('wt [rad]');legend('p_source','P_tot,avg');axis([0,2*pi,min(CircuitData.Power.pSource),max(CircuitData.Power.pSource)]); figure(2)subplot(2,1,1);plot(wt,CircuitData.Voltage.vR,'r',... wt,CircuitData.Current.iout,'b');grid;ylabel('v_R [V], i_out [A]');title('Voltage Across Load Resistance and Output Current vs. Time');legend('v_R','i_out');axis([0,2*pi,... min(min(CircuitData.Voltage.vR),min(CircuitData.Current.iout)),... max(max(CircuitData.Voltage.vR),max(CircuitData.Current.iout))]);subplot(2,1,2);plot(wt,CircuitData.Power.pR,'r',... wt,CircuitData.Power.PaverageR,'b');grid;ylabel('Power [W]');title('Instantaneous Power Absorbed by the Load Resistance');xlabel('wt [rad]');legend('p_R','P_R,avg');axis([0,2*pi,min(CircuitData.Power.pR),max(CircuitData.Power.pR)]); figure(3)subplot(2,1,1);plot(wt,CircuitData.Voltage.vD_forward,'r',... wt,CircuitData.Current.iout,'b');grid;ylabel('v_D [V], i_out [A]');title('Voltage Drop Across Non-ideal Diode and Output Current vs. Time');legend('v_D','i_out');


axis([0,2*pi,... min(min(CircuitData.Voltage.vD_forward),min(CircuitData.Current.iout)),... max(max(CircuitData.Voltage.vD_forward),max(CircuitData.Current.iout))]);subplot(2,1,2);plot(wt,CircuitData.Power.pD,'r',... wt,CircuitData.Power.PaverageD,'b');grid;ylabel('Power [W]');title('Instantaneous Power Loss in Non-ideal Diode');xlabel('wt [rad]');legend('p_D','P_D,avg');%axis([0,2*pi,min(CircuitData.Power.pD),max(CircuitData.Power.pD)]);axis([0,2*pi,-10,10]);

figure(4)subplot(2,1,1);plot(wt,CircuitData.Voltage.Vdc,'r',... wt,CircuitData.Current.iout,'b',... wt,CircuitData.Current.Iavg,'g');grid;ylabel('v_dc [V], i_out [A]');title('DC Voltage Source and Output Current vs. Time');legend('V_dc','i_out','I_avg');axis([0,2*pi,... min(min(CircuitData.Voltage.Vdc),min(CircuitData.Current.iout)),... max(max(CircuitData.Voltage.Vdc),max(CircuitData.Current.iout))]);subplot(2,1,2);plot(wt,CircuitData.Power.pVdc,'r',... wt,CircuitData.Power.PVdc,'b');grid;ylabel('Power [W]');title('Instantaneous Power Absorbed by the DC Voltage Source');xlabel('wt [rad]');legend('p_Vdc','P_Vdc,avg');axis([0,2*pi,min(CircuitData.Power.pVdc),max(CircuitData.Power.pVdc)]);

figure(5)subplot(2,1,1);plot(wt,CircuitData.Voltage.vL,'r',... wt,CircuitData.Current.iout,'b');grid;ylabel('v_L [V], i_out [A]');title('Voltage Across Inductor and Output Current vs. Time');legend('v_L','i_out');axis([0,2*pi,... min(min(CircuitData.Voltage.vL),min(CircuitData.Current.iout)),... max(max(CircuitData.Voltage.vL),max(CircuitData.Current.iout))]);subplot(2,1,2);plot(wt,CircuitData.Power.pL,'r',... wt,CircuitData.Power.PL,'b');grid;


ylabel('Power [W]');title('Instantaneous Power Absorbed/Delivered by the Inductor');xlabel('wt [rad]');legend('p_L','P_L,avg');axis([0,2*pi,min(CircuitData.Power.pL),max(CircuitData.Power.pL)]);

1/27/15 9:27 PM C:\Users\John\Google Driv...\HWRectifier.m 1 of 4

function CircuitData = HWRectifier(CircuitParameters)%HWRectifier.m generates the Half-wave Rectifier waveforms for one period%(0-2*pi rad). The types of HWRs it can handle are R, RL, RL-Source, %controlled and uncontrolled.%%Author: John D. Stevens%Date: 22 January 2015%%Input arguments:% CircuitParameters.Vm - [V] Amplitude of sinusoidal input% CircuitParameters.R - [Ohms] Resistance% CircuitParameters.L - [H] Inductance% CircuitParameters.f - [Hz] Frequency of the input voltage source% CircuitParameters.Vf - Non-ideal diode dc voltage drop (forward bias)% CircuitParameters.Rf - [Ohms] Non-ideal diode resistance (forward bias)% CircuitParameters.Vdc - [V] dc voltage source in series [V]% CircuitParameters.alphad - [rad] delay angle (=0 if unctontrolled)% CircuitParameters.numpoints - number of points in the output waveforms%Output:% CircuitData.wt - [rad] x-axis% CircuitData.Angle.AlphaMin - [rad] angle at which io > 0% CircuitData.Angle.DelayAngle - [rad] delay angle% CircuitData.Angle.Beta - [rad] extinction angle% CircuitData.Current.iout - [A] output current% CircuitData.Current.Iavg - [A] avg output current% CircuitData.Current.Irms - [A] rms output current% CircuitData.Voltage.vsource - [V] Input (source) voltage% CircuitData.Voltage.vout - [V] Output voltage% CircuitData.Voltage.vR - [V] Voltage across the load resistance R% CircuitData.Voltage.vD_forward - [V] Voltage across the non-ideal diode resistance (forward bias)% CircuitData.Voltage.vD_reverse - [V] Voltage across the diode (reverse bias)% CircuitData.Voltage.vL=vL - [V] Voltage across the inductor% CircuitData.Voltage.Vdc=Vdc - [V] dc source voltage% CircuitData.Power.pSource - [W] instantaneoud source power% CircuitData.Power.pR=pR - [W] instantaneous power absorbed by the load resistance R% CircuitData.Power.pD=pD - [W] instantaneous power absorbed by the non-ideal diode (forward bias)% CircuitData.Power.pVdc - [W] instantaneous power absorbed by the dc voltage% CircuitData.Power.pL - [W] instantaneous power across the inductor% CircuitData.Power.Ptotal - [W] average total power supplied by the source% CircuitData.Power.PaverageR - [W] average power absorbed by the load resistor R% CircuitData.Power.PaverageD - [W] average power loss in the non-ideal diode (forward bias)% CircuitData.Power.PVdc - [W] power absorbed by the dc voltage source% CircuitData.Power.PL - [W] power absorbed/delivered in/by the inductor (should be zero)% CircuitData.Power.pf - power factor% CircuitData.Power.efficiency - efficiency (depends on what is the output)

%% establish parametersVm=CircuitParameters.Vm; %[V] Amplitude of sinusoidal input


R=CircuitParameters.R; % [Ohms] ResistanceL=CircuitParameters.L; % [H] Inductancef=CircuitParameters.f; % [Hz] FrequencyVf=CircuitParameters.Vf; %[V] Non-ideal diode dc voltage drop (forward bias)Rf=CircuitParameters.Rf; % [Ohms] Non-ideal diode resistance (forward bias)Vdc=CircuitParameters.Vdc; % [V] dc voltage source in series [V]alphad=CircuitParameters.alphad; % [rad] delay angle (=0 if uncontrolled)N=CircuitParameters.numpoints; % number of points in the output waveforms

w=2*pi*f; % [rad/sec] assign angular frequencyRt=R+Rf; % [Ohms] total circuit resistanceVt=Vf+Vdc; % [V] total series dc voltage in the circuitZmag=sqrt(Rt^2+w^2*L^2); % [Ohms] impedance magnitudeZang=atan(w*L/Rt); % [rad] impedance angletau=L/Rt; % [sec] time constantwt=linspace(0,2*pi,N); % [rad] (x-axis) span one full input cyclealpha_min=asin(Vt/Vm); % [rad] alpha angle%determine if rectifier is controlledif alphad>=alpha_min %it also has to be less than pi alpha=alphad; %use the delay angle for alphaelse alpha=alpha_min; %use the minimum angle alpha (as though it is uncontrolled)endA=(-(Vm/Zmag)*sin(alpha-Zang)+(Vt/Rt))*exp(alpha/(w*tau));

%% find instantaneous current for a half-wave rectifier for an RL-Source Loadio=(Vm/Zmag)*sin(wt-Zang)-(Vt/Rt)+A*exp(-wt/(w*tau)); % [A]

% find beta (extinction angle)if L==0, beta=pi; else % indices for all values of wt greater than pi yy=find(wt>pi); %find zero crossing of wt after pi (this is beta) beta=wt(yy(find(abs(io(yy))==min(abs(io(yy)))))); % [rad]end

io(find(io<0))=0.0; %set all points in io after beta equal to zeroio(find(wt<alpha))=0.0; %set all points to the left of alpha equal to zeroio_avg=mean(io); % [A] average (dc) output currentio_rms=rms(io); % [A] rms output current

%% find instantaneous voltage waveformsvs=Vm*sin(wt); % [V] input voltagevo=vs; %temporaryvo(find(wt>beta))=0.0; %set all points in vo after beta equal to zerovR=io*R; % [V] Voltage across the load resistance (R)vD_rev=vs; %temporary (need to zero out points to the right of beta (below)vD_rev(find(wt<=beta))=0.0; %set all points in vo before beta equal to zerovD_for=io*Rf+Vf; % [V] Voltage across the non-ideal diode resistance (forward bias)

John

Highlight


vL=vo-vR-vD_for-Vdc; % [V] Voltage across the inductor %% find instantaneous powerps=vs.*io; % [W] instantaneous source power deliveredpR=vR.*io; % [W] instantaneous power absorbed in loadpD=vD_for.*io; % [W] instantaneoud power absorbed in the non-ideal diode (forward bias)pVdc=Vdc*io; % [W] instantaneous power absorbed in the dc voltage sourcepL=vL.*io; % [W] instantaneous inductor powern (should be zero) %% find average power, power factor, and efficiencyPR=io_rms^2*R; % [W] total power out to load resistance RPD=io_rms^2*Rf+io_avg*Vf; % [W] total power loss from the non-ideal diode (forward bias)PVdc=io_avg*Vdc; % [W] power absorbed by the dc voltage sourcePL=mean(pL); % [W] average power absorbed/delivered in/by the inductor (should be zero)Ptotal=PR+PD+PVdc; % [W] total power supplied by the sourceStotal=(Vm/sqrt(2))*io_rms; % [VA] apparent power (source)pf=Ptotal/Stotal; % power factorNu=PVdc/Ptotal; % efficiency (depends on what is the output). Is it the R or the Vdc? %% Assign outputCircuitData.wt=wt; % [rad] x-axisCircuitData.Angle.AlphaMin=alpha_min; % [rad] angle at which io > 0CircuitData.Angle.DelayAngle=alphad; % [rad] delay angleCircuitData.Angle.Beta=beta; % [rad] extinction angleCircuitData.Current.iout=io; % [A] output currentCircuitData.Current.Iavg=io_avg; % [A] avg output currentCircuitData.Current.Irms=io_rms; % [A] rms output currentCircuitData.Voltage.vsource=vs; % [V] Input (source) voltageCircuitData.Voltage.vout=vo; % [V] Output voltageCircuitData.Voltage.vR=vR; % [V] Voltage across the load resistance RCircuitData.Voltage.vD_forward=vD_for; % [V] Voltage across the non-ideal diode resistance (forward bias)CircuitData.Voltage.vD_reverse=vD_rev; % [V] Voltage across the diode (reverse bias)CircuitData.Voltage.vL=vL; % [V] Voltage across the inductorCircuitData.Voltage.Vdc=Vdc; % [V] dc source voltageCircuitData.Power.pSource=ps; % [W] instantaneoud source powerCircuitData.Power.pR=pR; % [W] instantaneous power absorbed by the load resistance RCircuitData.Power.pD=pD; % [W] instantaneous power absorbed by the non-ideal diode (forward bias)CircuitData.Power.pVdc=pVdc; % [W] instantaneous power absorbed by the dc voltageCircuitData.Power.pL=pL; % [W] instantaneous power across the inductorCircuitData.Power.Ptotal=Ptotal; %[W] average total power supplied by the sourceCircuitData.Power.PaverageR=PR; % [W] average power absorbed by the load resistor RCircuitData.Power.PaverageD=PD; % [W] average power lossed in the non-ideal diode (forward bias)CircuitData.Power.PVdc=PVdc; % [W] power absorbed by the dc voltage sourceCircuitData.Power.PL=PL; % [W] power absorbed/delivered in/by the inductor (should be zero)CircuitData.Power.pf=pf; % power factorCircuitData.Power.efficiency=Nu; % efficiency (depends on what is the output)


end


Problem 3.24

For the controlled half-wave rectifier with resistive load, the source is 120 V rms at 60 Hz. The resistance is 100 Ω, and the delay angle α is 45°.

(a) Determine the average voltage across the resistor.

120 2(1 cos ) (1 cos 45 ) 46.12 2

mo

VV Vαπ π

= + = + ° =

(b) Determine the power absorbed by the resistor.

2

2

sin 212 2

120 2 0.785 sin(2(0.785))1 80.9 .2 280.9 65.5100

rms

mrms

VPRVV

V

P W

α απ π

π π

=

= − +

= − + =

= =

(c) Determine the power factor as seen by the source.

,80.9(120) 97.1100

65.5 0.674 67.4%97.1

s rms rmsS V I VA

PpfS

= = =

= = = =


A controlled half-wave rectifier has a series resistance, inductance, and dc voltage source with R = 2 Ω, L = 75 mH, and Vdc = 48 V. The source is 120 V rms at 60 Hz. The delay angle is 50°. Determine

(a) An expression for load current

/14.1( ) 5.99sin( 1.50) 24.0 29.3 ., 0.873 4.24ti t t e A t radωω ω ω−= − − + ≤ ≤

(b) The power absorbed by the dc voltage source

1 ( ) ( ) 1.912

(1.91)(48) 91.6

o

dc o dc

I i t d t A

P I V W

β

α

ω ωπ

= =

= = =

∫

(c) The power absorbed by the resistor.

2

2 2

1 ( ) ( ) 2.932

(2.93) 2 17.1

rms

R rms

I i t d t A

P I R W

β

α

ω ωπ

= =

= = =

∫

Problem 3.34

Use MATLAB to determine the delay angle required such that the dc source in Prob. 3-33 absorbs 50 W.

81odα ≈

Problem 3.33

from the example problem in lecture 07 (which is also .... from the example problem in lecture 07...

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