Tentin tekstisisältö

Tampere University of Technology —DEE-33116 Power Electronics Converters — 18.12.2018
Electrical Energy Engineering
Jenni Rekola Programmable calculator allowed

Answers in English or in Finnish allowed

5 guestions/ ä 6 p, answer 4 guestions out of 5. Maximum 24 p/exam.

1. Rectifier

The grid phase voltage, DC voltage and grid current are shown in Fig. 1

a)
b)
c)
d)

e)
n

What is the used rectifier topology according to voltage and current waveforms shown in Fig. 1?
Draw the used rectifier topology

Calculate the average output voltage value

What should be the maximum peak repetitive reverse voltage rating of the power semiconductor
switching components used in the rectifier if 1.5 safety margin is used?

What is the lowest freguency of the produced grid current harmonic component?

What kind of problems the non-sinusoidal grid-current may cause for the grid?

 

600 = T

400

200 i

-200

Voltage(V) and current(A)
o

-400 G

 

 

 

-600 i
0.02 0.03 0.04 0.05 0.06 0.07 0.08

time(s)
Fig. 1. Grid phase voltage and grid current

2. Power supply of LED lights

The power supply of LED lights is shown in Fig. 2.

a)
b)
e)
d)
e)

n

What is the average output voltage of diode rectifier when LED lights are connected to Finnish
utility grid?

Why PFC-boost converter is included in the circuit?

What is the duty cycle of the buck-converter is the voltage of the LED lights is 2,2V?

The energy efficiency of the power supply is supposed to be 97%. The power of LED lights is
8,5W/each. The power level of regulating resistor is SW. How large is the reguired grid current?

How large inductor is reguired in the buck-converter if the switching freguency is 20 kHz. The
allowed inductor current ripple is 30% of the average current value.

Why transformer-isolated DC/DC converter topologies are often used in LED light applications?

1(5)
Tampere University of Technology —DEE-33116 Power Electronics Converters — 18.12.2018
Electrical Energy Engineering
Jenni Rekola Programmable calculator allowed

Answers in English or in Finnish allowed

5 guestions/ a 6 p, answer 4 guestions out of 5. Maximum 24 p/exam.

PFC Boost Converter Buck Regulator Stage

1 1

 

 

 

[ i
Buck

FFC eH
|Controlter | | Controller

 

 

 

 

 

 

 

 

 

 

|
|
|
[—==i—

Fig 2. Schematic diagram of a power supply of LED lights
3. Single-phase inverter

The single-phase inverter is used in the solar power system shown in Fig. 3a and the output voltage
waveform is shown in Fig. 3b.

a) What is the inverter bridge topology used in the solar power system according to the output
voltage waveform? Why this converter topology is often used?

b) Draw the inverter bridge.

c) Whatis the switching freguency of the transistors?

d) Whatis the modulation index of the inverter? Why this modulation method is often used?

e) What is the maximum output voltage rms value in linear modulation region?

f) What should be the modulation index if the load is designed to be grid connected (230 Vrms)
and the DC voltage is the same as in Fig. 3b?

 

265)
Tampere University of Technology —DEE-33116 Power Electronics Converters — 18.12.2018
Electrical Energy Engineering
Jenni Rekola Programmable calculator allowed

Answers in English or in Finnish allowed

5 guestions/ ä 6 p, answer 4 guestions out of 5. Maximum 24 p/exam.

1

 

 

 

 

 

 

 

 

400 r T T T T
200
=
5 o
o
5
-200
-400
o 0.01 0.02 0.03 0.04 0.05 0.06
Time (s)
1)

Fig. 3. a) Solar power system and b) inverter bridge voltage Viridse

4. Single-phase inverter

Continue to analyze the single-phase inverter in the solar power system as shown in Fig. 3a. The
output voltage waveform is shown in Fig. 3b.

a) Whatis the lowest freguency of the produced output current harmonic component?

b) Why LC-filter is connected in the converter output? How the inductor and capacitor values are
chosen?

c) Why DC/DC converter is often connected between the solar power panel and the inverter?

d) What is the maximum reverse voltage capability of the IGBTs in the inverter?

€) Why capacitor is connected between the panel and the inverter?

£) Why three-phase inverters are used in high-power solar power plants instead of single-phase
inverters?

3(5)
Tampere University of Technology —DEE-33116 Power Electronics Converters — 18.12.2018
Electrical Energy Engineering
Jenni Rekola Programmable calculator allowed

Answers in English or in Finnish allowed

5 guestions/ ä 6 p, answer 4 guestions out of 5. Maximum 24 p/exam.
5. Space-vector modulation

Three-phase inverter is shown in Fig. 4a and the corresponding space vector sectors (i.e., hexagon) in Fig.
4b. Assume that DC voltage eguals 560 V and switching freguency is 10 kHz.

a) Present the switching seguence of the inverter bridge when a conventional space-vector

pulse-width modulation (SV-PWM) algorithm is used and the reference voltage eguals
7

v" =200 V-e'* where 200V is the peak value of the phase voltage (2 p)

b) What is the maximum output voltage with the analyzed inverter in the linear modulation
region when SV-PWM modulation method is used?
c) Calculate the switching times of each switching vector to realize the reference vector

ja
v =200 V-e'* (3p)

The complex space vector of three-phase variable is defined as
Zn

2 N jän
2=3la tan +ax). where = a=e

The reference vector is

 

- ; — lact1 act?
Vafmix — Vrdf-a Y IVrg-p > y +

 

 

The alpha and beta components of the reference vector are
Vapa = Jv cos(wr)

yj = IV sin(or)
The vectors are shown in Table 1.

sv3 -+-

—++
svd

 

 

 

 

Fig. 4. a) Three-phase inverter b) vector diagram

45)
Tampere University of Technology —DEE-33116 Power Electronics Converters — 18.12.2018
Electrical Energy Engineering

Jenni Rekola Programmable calculator allowed
Answers in English or in Finnish allowed
5 guestions/ä 6 p, answer 4 guestions out of 5. Maximum 24 p/exam.

Table 1. Vectors

 

 

 

 

 

 

 

 

 

Vector SW Van + VhW VeN Va vp v
1 100 Vdc 0 0 2 0 2
3 Vac 3 Vac
2 110 Vdc Ve 0 1, BB 2 =
3 & = e 5 Ve
3 010 0 Vdc 0 1 NJ 2 ji
3 | 5 5Yue ?
4 011 0 Vdc Vdc 2 0 2 Je 2
"3 Vac 3 VuoE = 3 Vac
5 001 0 0 Vuc N 4 2 a
- 3 * 3 Vic 3 Vue
6 101 Vac 0 Vdc 1 N 43 2 ja
3 a Ve 3 vy
7 000 0 0 0 0 0 0
8 111 Vdc Vdc Vdc 0 0 0

 

 

 

 

 

 

 

 

 

 

5(5)