FINDING THE CLOSED-LOOP RESPONSE OF THE OUTPUT VOLTAGE FOR THE DIODE RECTIFIER–BOOST CONVERTER A diode rectifier followed by a boost converter is shown in Figure 13.36. The input voltage is vs = 170 sin(120πt). The circuit is operated closed loop, so that the output voltage is Vo = 220 ± 0.2 V, and

Question

FINDING THE CLOSED-LOOP RESPONSE OF THE OUTPUT VOLTAGE FOR THE DIODE RECTIFIER–BOOST CONVERTER

A diode rectifier followed by a boost converter is shown in Figure 13.36. The input voltage is [latex]v_{s}[/latex] = 170 sin(120πt). The circuit is operated closed loop, so that the output voltage is [latex]V_{o}[/latex] = 220 ± 0.2 V, and the input current is sinusoidal with an error of ±0.2 A. Use PSpice to plot the transient response of output voltage [latex]v_{o}[/latex] and the input current [latex]i_{s}[/latex].

The model parameters for the IGBT are:

TAU=287.56E9 KP=50.034 AREA=37.500E-6 AGD=18.750E6
VT=4.1822 KF=.36047 CGS=31.942E9 COXD=53.188E9
VTD=2.6570

The model parameters for the BJT are:

IS=2.33E−27 and BF=13

and those for the diode are:

IS=2.2E−15, BV=1200V, CJO=1PF, and TT=0.

Accepted Answer

The peak input voltage is [latex]V_{m}[/latex] = 170 V. Assuming a diode drop of 1 V, the peak rectified voltage becomes 170 − 2 = 168 V. The input frequency is f = 60 Hz. The block diagram for closed-loop control is shown in Figure 13.37(a). The error of the output voltage after passing through a controller generates the reference current for the current controller, whose output is the carrier signal for the PWM generator. Equating rectified output power to the load power, we get

[latex]V_{in(DC)} I_{in(DC)}=\frac{V^{2}_{o} }{R} [/latex] (13.8)

Because for a rectified sine wave [latex]V_{(average)} = 2V_{(peak)}[/latex] /π, Equation 13.8 becomes

[latex]\frac{2V_{m} }{\pi }\frac{2I_{m}}{\pi } =\frac{V^{2}_{o} }{R} [/latex]

which gives the multiplication constant δ as

[latex]\delta =\frac{I_{m}}{V_{m}} =\frac{\pi ^{2}V^{2}_{o} }{4RV^{2}_{m} } [/latex] (13.9)

[latex]=\frac{\pi ^{2} imes 220^{2} }{4 imes 400 imes 168^{2} } =0.01[/latex]

A value of δ higher than 0.01 would increase the sensitivity and give better transient response. But the PWM generator will be operated in the overmodulation region if it is too high. Let us assume δ = 0.05.

We use a proportional controller with an error band of ±1. The circuit for PSpice simulation of the current controller is shown in Figure 13.37(b). The multiplication is implemented with VALUE. The voltage and current controllers are implemented
by TABLE.

The listing of the subcircuit CONTR for Figure 13.37(b) is as follows:

* Subcircuit definition for CONTR current controller:
.SUBCKT CONTR 2 3 4 5 6
* model desired output rec. input input current carrier
* name voltage voltage voltage signal (voltage) signal
E1 8 0 TABLE{V(2,3)} = ; Voltage controller
+ (−0.2, 2) (0, 1) (0.2, 0)
E2 5 7 VALUE = {0.05*V(4)*V(8)} ; Reference current
E3 6 0 TABLE{V(7)} = ; Current controller
+ (−0.2, 0) (0, 5) (0.2, 10)
.ENDS CONTR

The PWM subcircuit is implemented with an ABM, ABM2 with two inputs. The
listing of the subcircuit PWM is as follows:

* Sub circuit definition PWM:
.SUBCKT PWM 1 2 3
* model reference carrier output
* name voltage signal voltage
E_ABM21 3 0 VALUE {IF(V(2)-V(1)>0, 10, 0)}
.ENDS PWM

Using 20 pulses per half cycle of the input voltage, the switching frequency of the PWM generator is [latex]f_{s}[/latex] = 40 × 60 = 2.4 kHZ, and the switching period is [latex]T_{s}[/latex] = 416.67 μsec.

The PSpice schematic is shown in Figure 13.38 for a unity power factor diode with a boost converter.

The listing of the circuit file for Figure 13.38 is as follows:

Example 13.10 Diode rectifier with PWM current control

.PARAM Vo_ref=220V Freq=2.0kHz
Vr 8 0 PULSE (0 10 0 {1/({2*{Freq}})-1ns} 1ns 1ns {1/{2*{Freq}}}
) ; Reference Signal
Vo_ref 16 0 DC {Vo_ref}SOURCE VS 4 11 SIN (0V 170V 60HZ)
CIRCUIT VZ 4 12 DC 0V ; Measures supply current

D1 12 1 DMOD ; Rectifier diodes
D2 0 11 DMOD
D3 11 1 DMOD
D4 0 12 DMOD
VY 1 2 DC 0V ; Voltage source to measure input current
H1 9 0 VY 0.034
Rg 6 0 100K
RB 7 6 250 ; Transistor base resistance
L 2 3 2.5MH
DM 3 4 DMOD ; Freewheeling diode
VX 4 5 DC 0V ; Voltage source to measure inductor
current
R 5 0 100 ; Load resistance
C 5 0 250UF IC = 0V ; Load filter capacitor
.MODEL DMOD D(IS = 2.2E − 15 BV = 1200V CJO = 1PF
TT = 0) ; Diode model parameters
.MODEL DM D(IS = 2.2E − 15 BV = 1200V CJO = 0 TT = 0) ;
Diode model parameters
Z1 3 6 0 IXGH40N60 ; IGBT switch
* .MODEL IXGH40N60D NPN (IS=2.33E-27 BF=13)
.MODEL IXGH40N60 NIGBT (TAU=287.56E-9 KP=50.034
+ AREA=37.500E-6 + AGD=18.750E-6 VT=4.1822 KF=.36047
+ CGS=31.942E-9 COXD=53.188E-9 VTD=2.6570)
* Subcircuit call for PWM control:
XPW 8 13 10 PWM ;Control voltage for transistor Z1
* Subcircuit call for CONTR current controller:
XCONT 16 15 17 9 13 CONTR
* Subcircuit definition CONTR [latex]\underline{must} [/latex] be inserted.
* Subcircuit definition PWM must be inserted.

ANALYSIS .TRAN 1US 35MS ; Transient analysis

.PROBE ; Graphics post-processor
.FOUR 60HZ I(VZ) ; Fourier analysis
.OPTIONS ABSTOL = 1.00U RELTOL = 0.01 VNTOL = 0.1
ITL5 = 0 ; Convergence

.END

The plots of the output voltage V(5) and input current I(VZ) are shown in Figure 13.39(a).The expanded waveforms are shown in Figure 13.39(b). The current controller forces the input current to be in phase with the input supply voltage and to follow a sinusoidal reference current. This improves the input power factor. A small filter can remove the high-frequency components of the input current. The upper bound of the output voltage is limited by the proportional controller. It should be noted that the load voltage has not yet reached the steady-state condition. The lowerlimit depends on the time constant of the load circuit. It requires careful design to determine the values of L and C, the controller parameters, and the switching frequency. A proportional–integral (PI) controller, coupled with a large value for the load filter capacitor, should give faster response of the output voltage.

The Fourier components of the input current are as follows:

DC COMPONENT = 3.606477E–02

Harmonic No	Frequency (Hz)	Fourier Component	Normalized Component	Phase (Deg)	Normalized Phase (Deg)
1	6.000E+01	8.195E+00	1.000E+00	2.291E+00	0.000E+00
2	1.200E+02	6.478E–02	7.905E–03	5.133E+01	5.591E+01
3	1.800E+02	2.039E+00	2.488E–01	1.678E+02	1.747E+02
4	2.400E+02	5.545E–02	6.766E–03	1.601E+02	1.509E+02
5	3.000E+02	7.729E–01	9.432E–02	3.090E+01	1.944E+01
6	3.600E+02	4.388E–02	5.354E–03	3.395E+01	2.020E+01
7	4.200E+02	2.740E–01	3.344E–02	1.189E+02	1.350E+02
8	4.800E+02	6.214E–02	7.582E–03	1.638E+02	1.455E+02
9	5.400E+02	8.967E–02	1.094E–02	1.693E+02	1.899E+02
10	6.000E+02	8.997E–02	1.098E–02	2.126E+01	4.417E+01
TOTAL HARMONIC DISTORTION = 2.689696E+01 PERCENT

For THD = 26.89% and [latex]φ_{1}[/latex] = −2.291°, Equation 7.3

[latex]PF_{i} =\frac{I_{1(rms)} }{I_{s} } \cos \phi _{1} =\frac{1}{\sqrt{1+(\frac{\%THD}{100} )^{2} } } \cos \phi _{1} [/latex] (7.3)

gives the input power factor of the rectifier as

[latex]PF_{i}=\frac{1}{\sqrt{1+\left\lgroup \frac{\%THD}{100} ight group ^{2} } }\cos \phi _{1} =\frac{1}{\sqrt{1+\left\lgroup\frac{26.89}{100} ight group ^{2} } } \cos (-2.291^{\circ } )[/latex]=0.965 (lagging)

Question 13.10: FINDING THE CLOSED-LOOP RESPONSE OF THE OUTPUT VOLTAGE FOR T...

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