hw1 diode bridge rectifiers-rev0

8/2/2019 HW1 Diode Bridge Rectifiers-Rev0

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I. Nikolakakos

MASDAR INSTITUTE OF SCIENCE AND TECHNOLOGY

Spring 2012

EPE 504: Power Electronics

Assignment 1Diode Bridge Rectifiers

Dr. Vinod Khadkikar

Due Date: February 6th, 2012

Submitted By:

Iraklis Nikolakakos


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I. Nikolakakos Page 2 of 28

Solutions:

Problem 1:


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Problem 1:


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Problem 2:

The circuit is shown below (screen-shot from the OrCAD design environment w.r.t toAppendix B, Computer Tools ). The diode is considered ideal, i.e. in the Model EditorWindow the following parameters are given:

.model Dbreak D Is=1e-14 Cjo=.1pF Rs=.0001 n=0.0001

Matlab is used for the calculations in the following sections, and the corresponding codeis attached in Appendix A.

a) Expression for load current

For the parameters given,Z = 33.6836 , = 1.1093 rad, = L/R = 2.0106Vm/Z = 10.0765 AHence, the equation for current becomes:

A for (1)Beta is found from Eq.

(2)

Using a numerical root-finding program, = 4.3537 rad, or 249.45o

b) Average current

[ ] (3)c) Power absorbed by the resistor


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d) Power factor

e) PSpice Simulation & Comparison with Matlab resultsIn sum, the output of matlab file is:

Problem-2(Hart,3.6):

a) An expression for load current: it is developed by hand. Here only the parameter beta is calculated beta =4.3537 rad

b) Average Current: Iav =4.8653 A

c) Power absorbed by the Resistor:PR =701.4939 W

d) Power factor: pf =0.4274

e) Simulation: Analyzed in the solution

PSpice simulation details are shown in the graph below (although T=16.666 ms, thesimulation was set to run for 20 ms so that we can see the beginning of the newcycle):

Our simulation in PSpice will reflect the complete set of parameters to understand thiscircuit: i.e. Vs, Io, Vo, V R, V L, and V d. The resulting grpahs are shown in Figure 2-1.


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The simulation results verify the Matlab results, as it can be seen in the followingtable:

SN Symbol Function (Pspice) Matlab PSpice % Dif. Figure

1 zoom-in 4.356132 4.3537 -0.06% 2-3,2-4

2 Iav avg(I(D2)) 6.8386 6.8386 0.00% 2-2,2-3

3 I rms rms(I(D2)) 4.8653 4.8651 0.00% 2-2,2-3

4 P R avg(W(R1)) 701.494 701.419 0.01% 2-2,2-3

5 pf 0.4274 0.42741 0.00% 2-2,2-3

Table 2-1: Definition of functions for the calculation of I av, I rms ,PR, and pf.

Comparing the values of the parameters 2 to 5 (i.e. from Iav to pf) we observe thatthey are very close, practically identical. Some minor discrepancies may be attributedto the resolution of the definition of the cursor point mode and the definition of theideal diode, since in the PSpice model we have set Is=1e-14, Cjo=.1pF, Rs=.0001,n=0.0001, while for our calculationsthe response of the diode has been assumedabsolutely ideal (Rs=0, n=0, Cjo=0pF).

A slightly stronger discrepancy is observed in the case of calculation, i.e. (0.6%),although it is also of minor importance considering its absolute value. This isattributed to the same reason behind the small peak that is observed in Figure 2 -4:the delay in switching between the ON and OFF states is due to the time required tochange the amount of excess minority carriers stored in the quasi-neutral regions. Thiseffect is modelled through the parameter C D according to which the voltage across thepn junction depletion region cannot be changed instantaneously. This effect ismodelled (although as a weak one) in PSpice, while in Matlab it is neglected.


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Io

Vs

/2

VL

VR

Vo

VD


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Figure 2-2: Observation of parameters and definition of functions for the calculation of I av, I rms,PR, and pf. Screenshot from PSpice.

Figure 2-3: Observation of parameters and definition of functions for the calculation of I av, I rms,PR, and pf @ t=16.668 ms, i.e. at the end of the period or 2 rad. Screenshot from PSpice.


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Figure 2-4: Observation of parameters and definition of functions for the calculation of . Screenshot from PSpice: Main and zoom-in windows.

Figure 2-5: Observation of parameters and definition of functions for the calculation of @ t , i.e. when the diode turns off. Using the information from thecursor point mode we obtain t = 11.555 ms => = 4.356132 . Screenshot from PSpice.

Zoom in window


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I. Nikolakakos

Problem 3:


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Problem 4:

- Simulate the previous circuit for L=40mH and L=100 H- Discuss the difference in the behavior of the two circuits- Plot the necessary waveforms- Observe steady state conditions

Figure 4-1: The simulated OrCAD model for L1=40mH and 100 H

Since the target of this problem is to study the steady state condition of the circuit we firstneed to identify an appropriate window to run the simulation. First we run the simulation

from t=0 to t=120 ms (more than 7 cycles) and we plot I R1 to see the transient effects. Weclearly observe that from 60ms onwards the steady state is reached. For simplicity in timecalculations, we will run our simulation in the window [100ms,150ms], which is equivalentto [6 cycle, 9 cycle].

Figure 4-1: Transient response of the circuit for L=40 mH (i o).


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Figure 4-2: Simulated waveforms

The simulation refers to two cases: One with Ld=40 mH and the other with Ld=100 uH.

40 mH Case:

From the waveform of the output current we note that this is a case of Continuous current:when the circuit is energized, the load current reaches the steady-state after a few periodsand then never returns to zero. For continuous-current operation, one pair of diodes isalways conducting, and the voltage across the load is a full-wave rectified sine wave. Duringthe positive cycle it is D1 and D2 that are conducting, while during the negative cycle it isD3 & D4 that are ON. The energy stored in L is high enough so that current always flows.

100 uH Case:

From the waveform of the output current we note that this is a case of discontinuous current:the load current returns to zero during every period. The value of L is not high enough sothat all the energy stored in the inductor is dissipated during every cycle. Again, in thepositive half-cycle diodes D1 and D2 are ON, while in the negative D3 and D4.When thesource voltage becomes less than Vdc the energy stored in the inductive element is less thanwhat it is required to keep the current flowing during the rest of the halfcycle. So the currentdrops to zero and a similar sequence start from the beginning in the next cycle.

Cd = 100 H

Cd = 40 mH


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Problem 5:

a) Circuit simulation without the dc side filter

Figure 5-1: The simulated OrCAD model

The simulated OrCAD model is shown in Fig. 5-1. Due to the small value of the inductance,transient effects in this circuit are negligible. However, to allow easier comparison betweenthis and the following results, we will simulate the waveforms in the time window: [5T,7T],where T=20 ms, and thus the window is defined as [100ms,140ms]. The maximum step size

is again set 0.01ms. The option to have a Fourier Analysis for the calculation of theharmonics is also selected. The center frequency is 50 Hz and the first 10 harmonics areconsidered.

The simulated waveforms are shown in Fig. 5-2 and for clarity a zoom-in window has alsobeen provided.

Zoom in window

vs v's is id vo


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Figure 5-2: Simulated waveforms.

The results from the Fourier analysis of i s and are given below ( is taken after L s and

Rs):FOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)

DC COMPONENT = 1.823971E-04

HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORM.NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 5.000E+01 8.484E+00 1.000E+00 -9.023E-01 0.000E+002 1.000E+02 3.512E-04 4.139E-05 6.081E+01 6.262E+013 1.500E+02 3.414E-04 4.024E-05 -1.344E+02 -1.317E+024 2.000E+02 3.191E-04 3.762E-05 3.102E+01 3.462E+015 2.500E+02 2.982E-04 3.515E-05 -1.638E+02 -1.593E+026 3.000E+02 2.685E-04 3.165E-05 9.869E-01 6.400E+007 3.500E+02 2.396E-04 2.824E-05 1.661E+02 1.724E+028 4.000E+02 2.061E-04 2.429E-05 -3.002E+01 -2.280E+019 4.500E+02 1.727E-04 2.036E-05 1.344E+02 1.425E+02

10 5.000E+02 1.393E-04 1.642E-05 -6.325E+01 -5.423E+01

TOTAL HARMONIC DISTORTION = 9.516903E-03 PERCENT

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(N04149,N04079)



1 5.000E+01 1.697E+02 1.000E+00 -9.023E-01 0.000E+002 1.000E+02 7.023E-03 4.139E-05 6.082E+01 6.262E+013 1.500E+02 6.780E-03 3.996E-05 -1.340E+02 -1.313E+024 2.000E+02 6.383E-03 3.762E-05 3.102E+01 3.462E+015 2.500E+02 5.924E-03 3.491E-05 -1.637E+02 -1.591E+026 3.000E+02 5.371E-03 3.165E-05 9.888E-01 6.402E+007 3.500E+02 4.763E-03 2.807E-05 1.660E+02 1.723E+028 4.000E+02 4.122E-03 2.429E-05 -3.002E+01 -2.280E+019 4.500E+02 3.437E-03 2.025E-05 1.342E+02 1.423E+02

10 5.000E+02 2.786E-03 1.642E-05 -6.325E+01 -5.423E+01

TOTAL HARMONIC DISTORTION = 9.488930E-03 PERCENT

From the above data we conclude that THD is negligible for both i s and . Based on theformulas and the above phase value of i s:

(5-1)

(5-2)we conclude that DF= 0.999955 1 and pf (1) = DF x cos(0+0.9023 o) 0.9999.DPF=cos(0.9023 o) 0.999876.

Another way to obtain the same result is through calculating the power factor from PSpicesimulation window(pf (2)), based on the function:


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i.e. (RMS(V1(Rs)-V2(Rs))+RMS(V1(Rload)-V2(Rload)))/RMS(V(V1:+)-V(V1:-)), with, factored out, or

(AVG(W(Rload))+AVG(W(Rs)))/(RMS(V(V1:+)-V(V1:-))*RMS(I(Rs)))

If we assume that Rs is part of the source and therefore should not be considered in thecalculation of the power factor, then:

and the function ( AVG( W(Rload)))/(RMS(V(V1:+)-V(V1:-))* RMS(I(Rs))) calculated att=120 ms gives:

pf (3)=0.9998.


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b) Circuit with Cd= 200 F, 500 F, 1000 F, and 2000 F

Figure 5-3: The simulated OrCAD model. Cd varies according to the requirements.

This circuit has stronger transients as the capacitor Cd has higher capacity. For 200 F, thetransients are almost negligible and the system reaches the PSS almost by the end of the firstcycle. However, for Cd=2000 F the transients become significant, but they still fade awayin less than 3-4 cycles. In any case, the window [100ms,140ms] is appropriate for ouranalysis and comparison. The above are reflected in the following figure:

Figure 5-4: Transients and the Periodic Steady State

The simulated waveforms are shown in Fig. 5-5:

Cd = 2000 F

Cd = 200 F


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Figure 5-5: Simulated waveforms (steady state)

Cd = 200 F

Cd = 500 F

Cd = 1000 F

Cd = 2000 F

Cd = 2000 F & Ls


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Regarding Fourier analysis, i s and THD, DPF, and PF for the cases under study we have:

200 FFOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)



1 5.000E+01 1.092E+01 1.000E+00 2.807E+01 0.000E+002 1.000E+02 6.887E-06 6.307E-07 -1.235E+02 -1.796E+023 1.500E+02 4.249E+00 3.891E-01 -7.997E+01 -1.642E+024 2.000E+02 1.702E-05 1.559E-06 6.991E+01 -4.238E+015 2.500E+02 4.210E+00 3.856E-01 -9.360E+01 -2.340E+026 3.000E+02 2.626E-05 2.405E-06 -1.319E+02 -3.003E+027 3.500E+02 5.417E+00 4.960E-01 1.502E+02 -4.629E+018 4.000E+02 5.004E-05 4.583E-06 4.401E+01 -1.806E+029 4.500E+02 2.430E+00 2.225E-01 2.321E+01 -2.294E+02

10 5.000E+02 4.726E-05 4.328E-06 -1.164E+02 -3.971E+02

TOTAL HARMONIC DISTORTION = 7.717835E+01 PERCENT



HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMNO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 5.000E+01 1.713E+02 1.000E+00 -1.016E+00 0.000E+002 1.000E+02 5.828E-03 3.401E-05 -1.242E+02 -1.221E+023 1.500E+02 4.009E+00 2.340E-02 -1.702E+02 -1.671E+024 2.000E+02 1.288E-02 7.515E-05 -1.628E+01 -1.222E+015 2.500E+02 6.618E+00 3.863E-02 1.764E+02 1.815E+026 3.000E+02 7.934E-03 4.631E-05 1.529E+02 1.590E+027 3.500E+02 1.192E+01 6.957E-02 6.015E+01 6.726E+018 4.000E+02 1.054E-02 6.152E-05 -1.196E+02 -1.114E+029 4.500E+02 6.868E+00 4.008E-02 -6.683E+01 -5.768E+01

10 5.000E+02 9.847E-03 5.747E-05 5.626E+01 6.642E+01


, and ( ) DF=0.791646, DPF=0.882373,pf (1)= 0.69853, pf (2)= 0.697730, and pf (3)= 0.697649

500 FFOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)

DC COMPONENT = -5.645374E-06


1 5.000E+01 1.489E+01 1.000E+00 1.593E+01 0.000E+002 1.000E+02 7.297E-06 4.902E-07 1.471E+01 -1.714E+01

3 1.500E+02 1.241E+01 8.334E-01 -1.325E+02 -1.803E+024 2.000E+02 3.083E-05 2.071E-06 1.655E+02 1.018E+025 2.500E+02 8.397E+00 5.641E-01 7.777E+01 -1.855E+006 3.000E+02 4.706E-05 3.161E-06 7.855E-01 -9.476E+017 3.500E+02 4.282E+00 2.877E-01 -7.542E+01 -1.869E+028 4.000E+02 3.979E-05 2.673E-06 -1.686E+02 -2.960E+029 4.500E+02 1.308E+00 8.790E-02 1.152E+02 -2.811E+01

10 5.000E+02 2.383E-05 1.601E-06 -1.124E+01 -1.705E+02




HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZEDNO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 5.000E+01 1.710E+02 1.000E+00 -1.510E+00 0.000E+002 1.000E+02 1.622E-02 9.485E-05 6.200E+01 6.502E+01

3 1.500E+02 1.170E+01 6.843E-02 1.374E+02 1.419E+024 2.000E+02 1.281E-02 7.486E-05 -1.223E+02 -1.163E+025 2.500E+02 1.320E+01 7.716E-02 -1.229E+01 -4.745E+006 3.000E+02 1.315E-02 7.688E-05 1.414E+01 2.320E+017 3.500E+02 9.415E+00 5.504E-02 -1.655E+02 -1.550E+028 4.000E+02 1.501E-02 8.777E-05 -1.742E+02 -1.621E+029 4.500E+02 3.695E+00 2.160E-02 2.503E+01 3.862E+01

10 5.000E+02 1.235E-02 7.223E-05 -1.936E+01 -4.261E+00


, and ( ) DF=0.689537, DPF=0.961598,pf (1)= 0.66306, pf (2)= 0.66208, and pf (3)= 0.6619501000 F

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)



1 5.000E+01 1.580E+01 1.000E+00 -5.168E+00 0.000E+002 1.000E+02 4.930E-03 3.120E-04 -6.841E+01 -5.807E+013 1.500E+02 1.258E+01 7.960E-01 1.638E+02 1.793E+024 2.000E+02 4.087E-03 2.587E-04 1.281E+02 1.487E+025 2.500E+02 7.653E+00 4.844E-01 -3.003E+01 -4.192E+00

6 3.000E+02 2.383E-03 1.508E-04 -3.799E+01 -6.980E+007 3.500E+02 3.161E+00 2.001E-01 1.260E+02 1.622E+028 4.000E+02 6.308E-04 3.992E-05 -1.724E+02 -1.310E+02





6 3.000E+02 4.838E-03 2.857E-05 -1.455E+02 -1.355E+027 3.500E+02 6.945E+00 4.102E-02 3.593E+01 4.766E+018 4.000E+02 2.088E-03 1.233E-05 5.209E+00 1.862E+01


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9 4.500E+02 9.992E-01 6.324E-02 -1.303E+02 -8.378E+0110 5.000E+02 1.006E-03 6.365E-05 1.036E+02 1.553E+02


9 4.500E+02 2.824E+00 1.668E-02 1.392E+02 1.543E+0210 5.000E+02 2.251E-03 1.330E-05 6.828E+01 8.504E+01


, and ( ) DF=0.72314, DPF=0.995935,

pf (1)= 0.7202, pf (2)= 0.71849, and pf (3)= 0.718362000 F

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)



1 5.000E+01 1.535E+01 1.000E+00 -1.395E+01 0.000E+002 1.000E+02 4.691E-01 3.056E-02 6.946E+01 9.735E+013 1.500E+02 1.182E+01 7.701E-01 1.372E+02 1.790E+024 2.000E+02 2.584E-01 1.683E-02 -1.320E+02 -7.621E+015 2.500E+02 6.652E+00 4.333E-01 -7.603E+01 -6.307E+00

6 3.000E+02 5.739E-02 3.739E-03 1.983E+01 1.035E+027 3.500E+02 2.394E+00 1.559E-01 5.300E+01 1.506E+028 4.000E+02 4.121E-02 2.684E-03 1.627E+01 1.278E+029 4.500E+02 1.133E+00 7.383E-02 1.243E+02 2.498E+02

10 5.000E+02 3.507E-02 2.285E-03 1.627E+02 3.021E+02






6 3.000E+02 1.136E-01 6.738E-04 -5.600E+01 -4.642E+017 3.500E+02 5.265E+00 3.123E-02 -3.725E+01 -2.607E+018 4.000E+02 7.562E-02 4.485E-04 -7.055E+01 -5.777E+019 4.500E+02 3.215E+00 1.907E-02 3.388E+01 4.825E+01

10 5.000E+02 9.720E-02 5.765E-04 8.669E+01 1.027E+02


, and ( ) DF= 0.742917, DPF=0.970606pf (1)= 0.72101, pf (2)= 0.729806 , and pf (3)= 0.723581

c) Circuit with Cd= 1000 F and Ld=Ls

The simulated waveforms are shown in Fig. 5-5.

Regarding Fourier analysis, i s and THD, DPF, and PF for the cases under study we have:FOURIER COMPONENTS OF TRANSIENT RESPONSE I(R_Rs)



1 5.000E+01 1.502E+01 1.000E+00 -1.378E+01 0.000E+002 1.000E+02 9.192E-02 6.120E-03 -9.852E+01 -7.095E+013 1.500E+02 1.054E+01 7.020E-01 1.372E+02 1.785E+02

4 2.000E+02 4.798E-02 3.195E-03 7.338E+01 1.285E+025 2.500E+02 4.681E+00 3.117E-01 -7.951E+01 -1.059E+016 3.000E+02 8.655E-03 5.763E-04 -1.096E+02 -2.689E+017 3.500E+02 1.179E+00 7.850E-02 1.616E+01 1.127E+028 4.000E+02 7.488E-03 4.986E-04 -1.317E+02 -2.143E+019 4.500E+02 1.050E+00 6.988E-02 8.873E+01 2.128E+02

10 5.000E+02 4.126E-03 2.747E-04 4.375E+01 1.816E+02





1 5.000E+01 1.686E+02 1.000E+00 -1.561E+00 0.000E+002 1.000E+02 5.227E-02 3.099E-04 1.719E+02 1.750E+023 1.500E+02 9.957E+00 5.904E-02 4.708E+01 5.176E+01

4 2.000E+02 6.161E-02 3.653E-04 -1.161E+01 -5.365E+005 2.500E+02 7.360E+00 4.364E-02 -1.697E+02 -1.619E+026 3.000E+02 2.274E-02 1.349E-04 1.566E+02 1.660E+027 3.500E+02 2.596E+00 1.539E-02 -7.428E+01 -6.335E+018 4.000E+02 2.010E-02 1.192E-04 1.580E+02 1.705E+029 4.500E+02 2.974E+00 1.763E-02 -1.633E+00 1.242E+01

10 5.000E+02 2.036E-02 1.207E-04 -4.520E+01 -2.959E+01


, and ( ) DF= 0.790302, DPF=0.971217pf (1)= 0.76756, pf (2)= 0.764298 , and pf (3)= 0.764185


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d) Discussion

The table below summarizes the THD, DF, DPF, etc. values for all the simulated cases:

Even though the first circuit (i.e the one without the dc side filter) functions as full-bridgerectifier and produces a v o waveform that is always >=0 (dc), it fails to produce a smoothvoltage waveform. From this point of view the capacitor that was added in parts b and c hasthe beneficial effect of smoothening voltage v o. In fact, from observation of Fig. 5-5 we canclearly see that the larger the value of Cd (capacitor), the smoother the waveform (lessripples).

On the other hand, the usage of the capacitor results in high values of harmonics, both interms of current and voltage. Especially the source current (i s) is severely distorted, and inall cases more than 75% THD. The voltage is less distorted, but still in all cases fail to meetthe relative IEEE standards.

Another figure of merit for this circuit is the power factor. From the table above we observethat the higher the capacitance, the higher the power factor. Therefore, and considering theprevious three arguments, the usage of a large capacitor in this application is advantageous.

Also, from the last graph in Figure 5-5 and the above table we observe that the usage of aninductor further smoothens the electrical features of this circuit, significantly reduces the

THD levels and increases the power factor. Subsequently, the optimized design of thiscircuits may require the appropriate selection of Cd and Ls values.

Figure 5-6: Explanation of the various states during the operation of the circuit at the steadystate.

Towards understanding the operation and behavior of this family of circuits we refer tofigure 5-6 and to the concept of the charge and discharge of a capacitor that is characterized

THD is(%)

THD vs(%)

DF (is)(deg)

(vs')(deg)

(vs)(deg)

pf(1) pf(2) pf(3) Discr 1-2 Discr 1-3 DPF

no dc side filter 0.00 0.00 1.000 -0.902 -0.902 0.000 0.999 0.999 0.998 0.00% -0.10% 0.999876

200F 77.18 9.21 0.792 28.070 -1.016 0.000 0.699 0.698 0.698 0.11% -0.13% 0.882373500F 105.03 11.89 0.690 15.930 -1.510 0.000 0.663 0.662 0.662 0.15% -0.17% 0.961598

1000F 95.51 10.92 0.723 -5.168 -1.676 0.000 0.720 0.718 0.718 0.24% -0.26% 0.995935

2000F 90.10 9.79 0.743 - 13.950 -1.597 0.000 0.721 0.730 0.724 -1.21% 0.36% 0.970506

2000F+Ls 77.53 7.71 0.790 - 13.780 -1.561 0.000 0.768 0.764 0.764 0.43% -0.44% 0.971217

A B

C

D

E

F


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by the time constant =RC. The higher this constant, the longer it takes for the capacitor tocharge and discharge.

With respect to figure 5-6, we have the following states: AB: The voltage of the source is less than the capacitor voltage. None of the diodes is ON,

and hence the capacitor is being discharged through the resistor Rload. This is why the I loa is>0. Due to the time constant RC, the higher the value the slower the discharge and theless steep is the descent of the voltage

BC: At B the source voltage becomes equal to the capacitor voltage, and thus diodes D1 andD2 are ready to turn ON. However, there are two factor that delay for a short time the flowof current: First, the existence of the inductor Ls that resist changes in the source current i s,and second the existing discharging current of the capacitor that opposes the current flowthat would be required to charge the capacitor.

CD: At C the voltage difference between the source and the capacitor is high enough toovercome the two previous obstacles, the diode is ON and current starts flowing to charge

the capacitor and keep the voltage of the resistor to the levels required by KVL. From B to

D, and since the diodes are considered ideal, the source current is identical to v o.Thisprocedure of charging the capacitor through the current i s is accelerating until the point D,

when the input voltage becomes again equal to the capacitor voltage. Depending on therelative trends of these two voltages the case of Cd=200 uF may be observed (with multiplepeaks of the charging current), or the case of the Cd=500 uF to 2000 uF may occur (withonly one peak of the wave that charges the capacitor).

DE: At D the requirements of discharging the capacitor are met, but the flow of chargingcurrent is a force of inertia that keeps the diodes ON for a short time interval and until E.

EF: From E to F diodes D1 and D2 turn OFF and lock back to v s. At F a cycle similar to theone described here starts, but with the current sign reversed and with the diodes D3 andD4 turning ON and OFF following the same pattern as before D1 and D2.


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The

1/zline1/zline-0

1/zline-1/ZDG2/zline1/zline-

01/zline-1/zline1/ZsysY

0.2435i+0.1017 0.1985i+0.0528 0.1598i+0.0160

0.1985i+0.0528 0.1985i+0.0528 0.1598i+0.0160

0.1598i+0.0160 0.1598i+0.0160 0.1662i+0.0246 Z


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Appendix A

MATLAB CODE


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Problem 2: Matlab Code

% Hart Example 3.2

clear all ; close all ; clc;

% Problem Parameters %R=100; % Ohm %L=0.1; % H %w=377; % rad/sec %Vm=100; % V

f=60; %Hz R=15; % Ohm L=0.08; % H w=2*pi*f; % rad/sec Vm=240*sqrt(2); % V

%========= Calculations ===========================

T=2*pi/w; % sec magZ=sqrt(R^2+(w*L)^2); thetaZ=atan(w*L/R); Tc=L/R; %sec

% beta beta=pi; beta_max=2*pi; tolerance=1e-5; error=1; step=1e-5; while (error>tolerance && beta


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fprintf( '\n PR =%.4f W \n' ,PR); fprintf( '\n d) Power factor:' ); fprintf( '\n pf =%.4f \n' ,pf); fprintf( '\n e) Simulation: Analyzed in the solution \n' );





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Appendix B

Computer Tools


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1) OrCAD Capture 16.5 2) Model Editor 16.5

3)Matlab R2010b 4) PSpice 16.5

hw1 diode bridge rectifiers-rev0

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