Experiment 1 Singlephase Rectifiers PDF

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Department of Electrical and Electronics Engineering EE4201 Power Electronics Experiment 1 Single Phase Diode Rectifier Circuit Oğuzhan Gaser Abstract Many applications have been found for diodes in electronics and electronical engineering circuits. Power diodes play a significant role in power electronics circuits for conversion of electiric power. A diode acts as a switch to perform various functions, such as rectifiers, freewheeling in switching regulators, charge reversal of capacitor and energy transfer between companents,voltage isolation, energy feedback from the load to the power source, and trapped energy recovery. In this experiment we used a diode connected in series with a switch and diode bridge rectifier to exibit the characteristics of a power device and analyze switching circuits consisting of R and L. The diode allows a unidirectional current flow and the switch performs the on and off functions. Diodes are extensively used in rectifiers. A rectifier is a circuit that converts an ac signal into a unidirectional signal. A rectifier is a type of ac-dc converter. Rectifiers varies depending their input types, in this experiment we examine single phase rectifiers. Also this single phase rectifiers can have two types of output half wave and full wave. Keywords Diode, Full wave rectifier, Half wave rectifier, Single Phase 1. Introduction Diodes are widely used in rectifiers. A rectifier is a circuit that converts an ac signal into a unidirectional signal. In such diode rectifiers, the power flow can onlt be from the utility ac side to the dc side. A majority of the power electronics applications such as switching dc power supplies, ac motor drives, dc servo drives, and so on, us such uncontrolled rectifiers. The dc output voltage of rectifier should be as ripple free as possible. In real life applications output of a rectifier contains harmonics and ripples. Also at the same time , rectifier should maintain input current as sinusoidal as possible and in phase with input voltage so that the power factor is near unity. The power processing quality of a rectifier requires the determination of harmonic contents of the input current, the output voltage and the output current. We can use Fourier series expansion to find harmonic contents.

In analysis i used some shortenings and symbols showed in the table below; Table 1 Symbols and shortenings Symbols Meaning ID(av), ID(rms) Average and rms diode currents IO(av), IO(rms) Average and rms output currents IP,IS Rms primary and currents of transfoermer Pdc,Pac Dc and ac output powers FF,RF,TUF,PF Output Form Factor, Ripple Ractor,Transformer Utilication Ractor and Power Factor vD(t),iD(t) Instantaneous diode voltage and current Vs(t), VO(t) Instantaneous input supply and output voltages Vm,VO(av), Peak, average and rms output VO(rms) voltage Vp, Vs rms primary and secondary voltages

2. Rectifier model 2.1 Single phase halfwave rectifier Halfwave diode rectifier is shown below in figure 1. It’s output should be look like the figure 2 in ideal conditions. While arranging output i used

1

diode’s forvard voltage as 10V to observe it clearly but it is normally 0.7 for silicon diodes. The waveforms shows that diode will conduct like a closed switch when it’s anode voltage is positive with respect to it’s cathode, and behaves like open switch when it’s cathode voltage higher than it’s anode voltage level. This rectifier is the simpliest rectifier but in real life it reraly used in applications.

voltage, D2 conducts and diode D1 is in a blocking condition. The negative portion of the input voltage appears across the load as a positive voltage. In this application peak inverse voltage of diodes are 2Vm. The output voltage exists whole period of cycle as shown in figure 4.

Fig. 1 Halfwave rectifier circuit

Fig. 4 Fullwave rectifier with a center tapped transformer’s input-output-diode-transformer voltages

2.2.b Diode bridge rectifier Instead of using center tapped transformer we can use four diodes as shown in figure 5. Using this circuit we can directly connect to the grid without using transformer.

Fig. 2 Half wave rectifier input- output voltage

2.2 Single phase fullwave rectifier There is two application of this circuit one is fullwave rectifier with a center tapped transformer and the other is diode bridge rectifier. 2.2.a Fullwave rectifier with center tapped transformer It consists of two diode and center tapped transformer, it’s circuit is shown below in figure 3.

Fig. 5 Diode bridge rectifier

During the positive half cycle diodes D1 and D2 conducts, at the negative half cycle current flows through D3 and D4 diodes. The peak inverse voltage of diode is only Vm.

Fig. 3 Fullwave rectifier with a center tapped transformer

During positive half cycle of the input voltage, diode D1 conducts and diode D2 is in a blocking region. The input voltage appears across the load. During negative half cycle of the input

Fig.6 Diode bridge rectifier’s output-input-diode voltages

2

3. Performance Parameters The average value of the output voltage,Vdc ; 1 𝑇 𝑉𝑑𝑐 = 𝑇 ∫0 𝑉𝑚 sin(𝑤𝑡)𝑑𝑡 The average value of output current,Idc ; 𝑉 𝐼𝑑𝑐 = 𝑑𝑐 𝑅 The output dc power,Pdc; 𝑃𝑑𝑐 = 𝑉𝑑𝑐 𝐼𝑑𝑐 The rms value of the output voltage, Vrms ; 1 1 𝑇 𝑉𝑟𝑚𝑠 = [ 𝑇 ∫0 (𝑉𝑚 sin(𝑤𝑡))2 𝑑𝑡 ] ⁄2 The rms value of the output current, Irms ; 𝑉 𝐼𝑟𝑚𝑠 = 𝑟𝑚𝑠 𝑅 The output ac power,Pac ; 𝑃𝑎𝑐 = 𝑉𝑟𝑚𝑠𝐼𝑟𝑚𝑠 The efficiency (retrification ratio) ; 𝑃 ŋ = 𝑃𝑑𝑐 𝑎𝑐

The effective(rms) value of the ac component of output voltage;

𝑅𝐹 = √(1.57)2 − 1 = 1.21𝑜𝑟121% 0.5𝑉𝑚 𝑉𝑠 = 0.707𝑉𝑚 ,𝐼𝑠 = 𝑇𝑈𝐹 = 𝐶𝐹 =

𝑉𝑚 ⁄ 𝑅 0.5𝑉𝑚

𝑉𝑑𝑐 =

𝑇 𝑉𝑚 𝛱

𝑃𝑎𝑐 =

ŋ=

𝐼𝑠

𝛱

𝑇

1

1⁄ 2

𝑉𝑟𝑚𝑠 = [ 𝑇 ∫0 (𝑉𝑚 sin(𝑤𝑡))2 𝑑𝑡 ]

𝑉𝑟𝑚𝑠 = 𝑃𝑑𝑐 =

𝑃𝑎𝑐 = ŋ=

𝑉𝑚

= 0.5𝑉𝑚 , 𝐼𝑟𝑚𝑠 =

2

(0.318)2 𝑉𝑚 𝑅

(0.5)2 𝑉𝑚 𝑅

(0.318)2𝑉𝑚 𝑅 (0.5)2𝑉𝑚 𝑅 0.5𝑉𝑚

𝐹𝐹 =

0.318𝑉𝑚

0.5𝑉𝑚 𝑅

1⁄ 2

𝑉𝑚

= 0.707𝑉𝑚 , 𝐼𝑟𝑚𝑠 =

2

(0.6366)2 𝑉𝑚

𝑅 (0.707)2 𝑉𝑚 𝑅

(0.6366)2 𝑉𝑚 𝑅 (0.707)2 𝑉𝑚 𝑅 0.707𝑉𝑚

;

;

0.707𝑉𝑚 𝑅

;

;

;

=81%

𝑚

3.1 The performance expectation of halfwave rectifier Using the circuit in figure 1 and assuming input signal as 𝑉𝑠 = 𝑉𝑚 sin(𝑤𝑡) we can calculate the theoretical performance values. 1 𝑇 𝑉𝑑𝑐 = ∫0 𝑉𝑚 sin(𝑤𝑡) 𝑑𝑡 ; 𝑉𝑚 𝑅

𝑇

𝑉𝑚 𝑅

𝐹𝐹 = 0.6366𝑉 = 1.11𝑜𝑟111%

𝑉𝑠 𝐼𝑠

Crest factor, CF ; 𝐼 𝐶𝐹 = 𝑠(𝑝𝑒𝑎𝑘)

= 0.318𝑉𝑚 , 𝐼𝑑𝑐 = 0.318

=2

= 0.6366𝑉𝑚 , 𝐼𝑑𝑐 = 0.6366

1

𝑃𝑑𝑐 =

𝑇 𝑉𝑚

= 0.286

𝑉𝑟𝑚𝑠 = [𝑇 ∫0 (𝑉𝑚 sin(𝑤𝑡))2 𝑑𝑡 ]

The form factor, FF ; 𝑉 𝐹𝐹 = 𝑟𝑚𝑠

𝑉𝑑𝑐 =

⁄𝑅

𝑅 (0.318)2 𝑉𝑚 ⁄ 𝑅 0.707×0.5𝑉𝑚 ⁄ 𝑅

3.2 The performance expectation of fullwave rectifier Using the circuit in figure 5 and assuming input signal as 𝑉𝑠 = 𝑉𝑚 sin(𝑤𝑡) we can calculate the theoretical performance values. 1 𝑇 𝑉𝑑𝑐 = ∫0 𝑉𝑚 sin(𝑤𝑡) 𝑑𝑡 ;

𝑉𝑟𝑚𝑠 =

The ripple factor, RF; 𝑅𝐹 = √𝐹𝐹 2 − 1 Transformer utilization factor, TUF ; 𝑃 𝑇𝑈𝐹 = 𝑑𝑐

=

𝑃𝐼𝑉 = 𝑉𝑚

2 − 𝑉2 𝑉𝑎𝑐 = √𝑉𝑟𝑚𝑠 𝑑𝑐

𝑉𝑑𝑐

𝑃𝑑𝑐 𝑉𝑠 𝐼𝑠

;

𝑅𝐹 = √(1.11)2 − 1 = 0.482𝑜𝑟48.2% 0.5𝑉𝑚 𝑉𝑠 = 0.707𝑉𝑚 ,𝐼𝑠 = 𝑃

𝑇𝑈𝐹 = 𝑉 𝑑𝑐𝐼 = 𝑠 𝑠

𝑃𝐼𝑉 = 𝑉𝑚 𝐶𝐹 =

𝑉𝑚 ⁄ 𝑅 0.5𝑉𝑚 ⁄ 𝑅

𝑅 (0.318)2 𝑉𝑚 ⁄ 𝑅 (0.707)2 𝑉𝑚 ⁄𝑅

= 81.1%

= √2

4. Matlab analysis of experiment 1-Halfwave rectifier with R load

; ;

;

; =40.5% = 1.57𝑜𝑟157%

3

3-Halfwave rectifier with RL load

Fig.6 Halfwave rectifier voltage and current output waveforms with only resistive load(192Ω)

n = 0.4188 FF = 1.5452 RF = 1.1780 TUF = 0.3040 2- Halfwave rectifier with RL load

Fig.8 Halfwave rectifier voltage and current output waveforms with series connected resistive(192Ω) and inductive load(50.8mH)(current is amplified 50 times for better observation)

N= 0.4405 FF = 1.9016 RF = 1.6175 TUF = 0.1928 4-Fullwave rectifier with R load

Fig.7 Halfwave rectifier voltage and current output waveforms with series connected (192Ω) and inductive load(0.61H)(current is amplified 50 times for better observation)

n =0.4012 FF =1.5911 RF =1.2376 TUF = 0.2793

Fig.9 Fullwave rectifier voltage and current output waveforms with only resistive load(192Ω)

n =0.8029 4

FF = 1.1160 RF = 0.4954 TUF =0.8337 5-Fullwave rectifier with RL load

n =0.8120 FF =1.1156 RF = 0.4945 TUF=0.9220 5. The experimental outputs 1-Halfwave rectifier with R load

2-Halfwave rectifier with RL load Fig.10 Halfwave rectifier voltage and current output waveforms with series connected (192Ω) and inductive load(0.61H)(current is amplified 50 times for better observation)

n = 0.9413 FF =1.1186 RF =0.5013 TUF =0.8483 6-Fullwave rectifier with RL load

3-Halfwave rectifier with RL load

Fig.11 Halfwave rectifier voltage and current output waveforms with series connected (192Ω) and inductive load(50.8mH)(current is amplified 50 times for better observation)

5

4-Fullwave rectifier with R load

5-Fullwave rectifier with RL load

6-Fullwave rectifier with RL load

6. Examining Matlab results My first condition on matlab simulation is halfwave rectifier with purely resistive load rectifier analysis. I calculated it’s theoretical values in section 3.1 in this report. As we see our matlab results are very close theoretical values. Because both diode and resistive are almost linear equipments and we did this experiment in their linear regions. Thus our experimental values are similiar with theoretical values. Both second and third part of experiment we used RL load halfwave rectifier. As we see in the matlab outputs when when supply signal’s negative part supplied to circuit inductive load transfers it’s sttored energy and creates like a opposite battery to signal. So that we observe ringing when diode stops conducting or signal is in negative cycle. This condition slightly increased efficiency but decreased transformer utilization factor. Our second condition is fullwave rectifier with a purely resistive load. Again it’s matlab results are similiar to theoretical results which calculated in section 3.2. Second and third part of fullwave rectifier experiment increasing inductive companent’s value increased efficiency without effecting transformer utilization factor much. 7. Matlab codes 1-For purely resistive load clc; clear all; close all; hwr=xlsread('1.xlsx'); time=hwr(:,1); voltage=hwr(:,2); current=hwr(:,3); R=192; Vdc=mean(voltage); Vrms=rms(voltage); Vs=max(voltage)/sqrt(2); Idc=abs(Vdc)/R; Irms=abs(Vrms)/R; Is=Vrms/R; Pdc=Vdc*Idc; Pac=Vrms*Irms; Pin=Vs*Is; n=Pdc/Pac FF=Vrms/Vdc RF=sqrt((FF^2)-1) TUF=Pdc/Pin subplot(2,1,1) plot(time,voltage); xlabel('Time'); ylabel('Output Voltage'); title('Output Voltage Waveform'); grid on;

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subplot(2,1,2) plot(time,current); xlabel('Time'); ylabel('Output Current'); title('Output Current Waveform'); grid on; hold on plot(time,voltage) plot(time,50*current) xlabel('Time'); ylabel('Output Voltage'); title('Output Voltage Waveform'); grid on;

2-For RL load clc; clear all; close all; hwr=xlsread('8.xlsx'); time=hwr(:,1); voltage=hwr(:,2); current=hwr(:,3); R=192; L=0.0508; Z=sqrt((R^2)+(L^2)); Vdc=mean(voltage); Vrms=rms(voltage); Vs=max(voltage)/sqrt(2); Idc=mean(current); Irms=rms(current); Is=Vs/Z; Pdc=(Idc^2)*R; Pac=(Irms^2)*R; Pin=Vs*Is; n=Pdc/Pac FF=Vrms/Vdc RF=sqrt((FF^2)-1) TUF=Pdc/Pin subplot(2,1,1) plot(time,voltage); xlabel('Time'); ylabel('Output Voltage'); title('Output Voltage Waveform'); grid on; subplot(2,1,2) plot(time,current); xlabel('Time'); ylabel('Output Current'); title('Output Current Waveform'); grid on; hold on plot(time,voltage) plot(time,50*current)

8. Conclusion In this experiment we observed halfwave rectifier and fullwave rectifier. Before experimental results we calculated both halfwave and fullwave rectifier’s theoretical results, then analyzing experimental results we saw that rectifiers has consistent working conditions. Our experimental results showed us we can analyze these circuits with theoretical ideal conditions. As we expected halfwave rectifier’s efficiency is very low and it is not useful in real life. Also transformer utilization factors are very low and adding inductive load decreases it too.(0.3040.1928-0.2793) Using these TUS values we can basically calculate how much bigger transformer has to we use to use all power that supplied to circuit.Using 1⁄𝑇𝑈𝑆 equation (3.289-5.18-3.58) times bigger transformer we need. We used inductive load because in real life a great majority of load are inductive. As we see in experiment increasing inductive load increases the fullwave rectifier’s efficiency so that this type of rectifier usefull in real life. Also it’s TUS values are (0.8337-0.9220-0.8483) and applying equation we get(1.199-1.084-1.178). As we see we almost use all the power that we supply in fullwave rectifiers. References Muhammad H. Rashid-Power electronics_devices,circuits,and applicationsPearson(2014)-pg 104 chapter Diode rectifiers Cyril W. Lander-Power Electronics- third edition-Mcgraw-Hill publishing company-pg42 chapter Rectifying circuits

xlabel('Time'); ylabel('Output Voltage'); title('Output Voltage Waveform'); grid on;

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