Three-Phase Half-Bridge Diode Rectifiers PDF

Three-Phase Half-Bridge Diode Rectifiers Half-bridge diode rectifier 10/10/2022 PECT302_Chpater 1_Lecture 1 1  Objectives For three-phase half-bridge diode rectifier: 1-describe the circuit 2-present some notation 3-identify...

Three-Phase Half-Bridge Diode Rectifiers Half-bridge diode rectifier 10/10/2022 PECT302_Chpater 1_Lecture 1 1  Objectives For three-phase half-bridge diode rectifier: 1-describe the circuit 2-present some notation 3-identify circuit topologies 4- analyze the circuit with R load 5-study the effect of load back EMF 6-evaluate the circuit performance. 10/10/2022 PECT302_Chpater 1_Lecture 1 2  Three-pulse rectifier Supplied from a three-phase, four-wire AC power line. The rectifier consists of three power diodes, DA, DB and DC. The load is connected between the common-cathode point of the diode and the neutral, N, of the supply line. DA DB DC 10/10/2022 PECT302_Chpater 1_Lecture 1 3 At any time only the diode that is supplied with the highest line-to-neutral voltage is conducting the output current, io. For Example when the highest voltage is that of phase B ( vBN > vAN and vBN > vCN) then it is diode DB that is conducting, while diodes DA and DC are reverse biased by line-to-line voltages vBA and vBC, respectively SUPPLY LINE A A B B C C N N iA iB iC io iB io = iB DA DB DC vo LOAD vBA vBC vo = vBN DB 10/10/2022 PECT302_Chpater 1_Lecture 1 4 Each of the three line-to-neutral voltages is higher than the other two for 1/3 of the cycle of input voltage. Consequently, each diode conducts the current for 120◦ interval. Voltage and current waveforms of the three-pulse rectifier with a resistive load (R-load) are shown in the Fig. The waveform pattern of the output voltage is repeated every 120◦. 10/10/2022 PECT302_Chpater 1_Lecture 1 5 Notice that if we consider the start of the conduction of DA at (𝜔𝑡 = 0) it means that the positive zero 𝜋 crossing of vAN occurs at 𝜔𝑡 = − 6 , or 𝜋 𝑣𝐴𝑁 = 𝑉𝐿𝑁,𝑝 sin 𝜔𝑡 + 3 therefore 3 3 𝑉𝑜,𝑑𝑐 3−𝑝𝑢𝑙𝑠𝑒 = 𝑉 2𝜋 𝐿𝑁,𝑝 10/10/2022 PECT302_Chpater 1_Lecture 1 6 The waveform of the phase-A line current, iA, is shown in the figure. Clearly, it is completely different from a sinewave. Such currents drawn by the rectifier would disturb the operation of protection systems. Also, the current has a considerable average, being an ac line current this dc components leads to transformer saturation. (this is the main disadvantage of this rectifier) 10/10/2022 PECT302_Chpater 1_Lecture 1 7 Example 1-1 A three-pulse diode rectifier, fed from a 460-V ac line, supplies a 5-Ω resistive load. Calculate the average output voltage and current of the rectifier. Solution 10/10/2022 PECT302_Chpater 1_Lecture 1 8  Example 1-2 The rectifier given in Example 1-1 charges a 360-V battery pack through a 5Ω resistor. Is the charging current continuous or discontinuous? Sketch the output voltage and current. Solution To determine the current continuity, compare the supply and the battery voltages State 0 DA State 0 DB State 0 0 alpha_1 alpha_2 2pi/3 It is Discontinuous 10/10/2022 PECT302_Chpater 1_Lecture 1 9  State 0 DA State 0 DB State 0 0 alpha_1 alpha_2 2pi/3 10/10/2022 PECT302_Chpater 1_Lecture 1 10  Example 1-3 Calculate the average output voltage and current of the rectifier described in Example 1-2 Solution 𝜋 375 sin 𝛼1 + = 360 → 𝛼1 = 43.74° 6 𝛼2 = 76.26° 0 alpha_1 alpha_2 2pi/3 60 1 𝜋 𝑉𝑜,𝑑𝑐 = 𝜋 360 43.74 + න 375 sin 𝜔𝑡 + 30 𝑑𝜔𝑡 180 43.74 3 𝑉𝑜,𝑑𝑐 = 362.7 𝑉𝑜,𝑑𝑐 − 360 𝐼𝑜,𝑑𝑐 = = 0.54𝐴 𝑅 10/10/2022 PECT302_Chpater 1_Lecture 1 11  Performance Evaluation The output current with a large dc component flows through the neutral wire. However, a dc component equal to one-third of the dc-output currents appears also in currents drawn from wires A, B, and C. This property disqualifies the three-pulse rectifier from practical applications. In a power system supplying the rectifier, the dc current would cause saturation of transformer cores, resulting in distortion of voltage waveforms in the system. Another potential trouble stems from the fact that all components of the power system are designed for ac sinusoidal currents of a specific frequency (60 Hz in the USA). The waveform of the phase-A line current, iA. Clearly, it is completely different from a sinewave. Such currents drawn by the rectifier would disturb the operation of protection systems. 10/10/2022 PECT302_Chpater 1_Lecture 1 12  Three-Phase Diode Rectifiers Full-bridge diode rectifier (Analysis of full-bridge diode rectifier with highly inductive and R load) 10/20/2022 PECT302_Chpater 1_Lecture 2 1 Six-pulse Diode Rectifier with Highly Inductive Load A B C The six-pulse (three-phase, full-wave) diode rectifier, shown is iA iB iC the most commonly used ac-to-dc power converter producing a fixed dc voltage. io The circuit of the rectifier consists of six power diodes in a R three-phase bridge configuration. DA DB DC Diodes DA, DB, and DC form the common-cathode group, and vo L diodes DA′, DB′, and DC′ constitute the common-anode group. DA' DB' DC' At any instant, only one diode in the common cathode group E and one in the common-anode group conduct the current. 10/20/2022 PECT302_Chpater 1_Lecture 2 2  Six-pulse diode rectifier with highly inductive load The highly inductive draws a ripple-free current and therefore represented by an ideal dc current supply. At any time, only one diode from the common cathode group (DA, DB, DC): is forward biased which is the diode with the highest voltage at its anode. Also one diode from the common-anode diodes (DA’, DB’, DC’) is forward biased: the diode which has the lowest voltage at its cathode. Or we can say that the conducting diode pair is that supplied with the highest line-to-line voltage, (next slide.) 10/20/2022 PECT302_Chpater 1_Lecture 2 3  DA+ DB+ DC+ DA+ DB+ vAn vBn vCn VLN,p wt 0 p/3 2p/3 p 4p/3 5p/3 DB- DC- DA- DB- DC- VP: vAo vBo vCo vAo vBo VN: 10/20/2022 vBo vCo v PECT302_Chpater Ao v 1_Lecture 2 Bo vCnp 4  DA+ DB+ DC+ DA+ DB+ vAn vBn vCn VLN,p wt 0 p/3 2p/3 p 4p/3 5p/3 DB- DC- DA- DB- DC- VP: vAo vBo vCo vAo vBo VN: vBo vCo vAo vBo vCnp vo=vp-vn vAB vAC vBC vBA vCA vCB vAB vAC vBC 10/20/2022 PECT302_Chpater 1_Lecture 2 vo 5  DA+ DB+ DC+ DA+ DB+ vAn vBn vCn VLN,p wt 0 p/3 2p/3 p 4p/3 5p/3 DB- DC- DA- DB- DC- VP: vAo vBo vCo vAo vBo VN: vBo vCo vAo vBo vCnp vo=vp-vn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 6  DA+ DB+ DC+ DA+ DB+ vAn vBn vCn VLN,p wt 0 p/3 2p/3 p 4p/3 5p/3 DB- DC- DA- DB- DC- VP: vAo vBo vCo vAo vBo VN: vBo vCo vAo vBo vCnp vo=vp-vn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo 3VLN,p 10/20/2022 PECT302_Chpater 1_Lecture 2 7  During the first interval …0 ≤ 𝜔𝑡 ≤ 𝜋/3 DA+ DB+ DC+ DA+ DB+ vAn vBn 𝑣𝑜 v= 𝑣𝐴𝐵 Cn VLN,p 𝑖𝐴 = 𝐼𝑜 𝑖𝐵 = −𝐼𝑜 𝑖𝑐 = 0 DB- DC- DA- DB- DC- vAn vBn vCn vAn vAn vBn vCn vAn vBn vCn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo 3VLN,p During the first interval, when diodes DA and DB′ form a path for the output current. The other four diodes are subjected to voltages vAC (diode DC), vCB (diode DC′), and vAB (diodes DA′ and DB) 10/20/2022 PECT302_Chpater 1_Lecture 2 8  During the 2nd interval …π/3 ≤ 𝜔𝑡 ≤ 2𝜋/3 DA+ DB+ DC+ DA+ DB+ vAn vBn vCn 𝑣𝑜 = 𝑣𝐴𝐶 𝑖𝐴 = 𝐼𝑜 𝑖𝐵 = 0 𝑖𝑐 = −𝐼𝑜 DB- DC- DA- DB- DC- vAn vBn vCn vAn vAn vBn vCn vAn vBn vCn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 9  During the 3rd interval …2π/3 ≤ 𝜔𝑡 ≤ 𝜋 DA+ DB+ DC+ DA+ DB+ vAn vBn vCn 𝑣𝑜 = 𝑣𝐵𝐶 𝑖𝐴 = 0 𝑖𝐵 = 𝐼𝑜 𝑖𝑐 = −𝐼𝑜 DB- DC- DA- DB- DC- vAn vBn vCn vAn vAn vBn vCn vAn vBn vCn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 10  During the 4th interval …π ≤ 𝜔𝑡 ≤ 4π/3 DB+ DC+ DA+ DB+ vBn vCn 𝑣𝑜 = 𝑣𝐵𝐴 𝑖𝐴 = −𝐼𝑜 𝑖𝐵 = 𝐼𝑜 𝑖𝑐 = 0 DC- DA- DB- DC- vBn vCn vAn vAn vCn vAn vBn vCn AC vBC vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 11  During the 5th interval …4π/3 ≤ 𝜔𝑡 ≤ 5π/3 DB+ DC+ DA+ DB+ vCn 𝑣𝑜 = 𝑣𝐶𝐴 𝑖𝐴 = −𝐼𝑜 𝑖𝐵 = 0 𝑖𝑐 = 𝐼𝑜 DA- DB- DC- vBn vCn vAn vAn vAn vBn vCn BC vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 12  During the 6th interval …5π/3 ≤ 𝜔𝑡 ≤ 2π + DC+ DA+ DB+ vCn 𝑣𝑜 = 𝑣𝐶𝐵 𝑖𝐴 = 0 𝑖𝐵 = −𝐼𝑜 𝑖𝑐 = 𝐼𝑜 DA- DB- DC- n vCn vAn vAn vAn vBn vCn vBA vCA vCB vAB vAC vBC vo 10/20/2022 PECT302_Chpater 1_Lecture 2 13 Each diode conducts for 120°, and then another diodes in the same group becomes forward-biased. The process of a diode taking over conduction of current from another diode is called natural commutation. 10/20/2022 PECT302_Chpater 1_Lecture 2 14 The line current DA DB DC DA iA Io iA,1 DB ’ DC ’ DA ’ DB ’ DC ’ -A line current 𝑖𝐴 is equal to 𝐼𝑜 when DA is ON and to −𝐼𝑜 when DA’ is ON and to 0 when both DA and DA’ are OFF 10/20/2022 PECT302_Chpater 1_Lecture 2 15 10/20/2022 PECT302_Chpater 1_Lecture 2 16  Derive the expression of the power factor for a highly inductive load 10/20/2022 PECT302_Chpater 1_Lecture 2 17  Full-Bridge 3-phase diode rectifier with R-load VP: vAo vBo vCo vAo vBo VN: vBo vCo vAo vBo vCnp P vo=vp-vn vAB vAC vBC vBA vCA vCB vAB vAC vBC D1 D3 D5 io vo - vAn + A vAB 3VLN,p - vBn + io=vo/R o B R vo vBC - vCn + C D4 D6 D6 N 10/20/2022 PECT302_Chpater 1_Lecture 2 18  VP: vAo vBo vCo vAo vBo The line currents VN: vBo vCo vAo vBo vCnp vo=vp-vn vAB vAC vBC vBA vCA vCB vAB vAC vBC vo (R-Load) 3VLN,p io=vo/R The output current is vo/R iA -A line current 𝑖𝐴 , 𝑖𝐵 , 𝑖𝐶 is equal to the DC side current when its line upper diode is ON and to the negative of the DC side iB current when its line lower diode is ON and zero otherwise. iC 10/20/2022 PECT302_Chpater 1_Lecture 2 19 Example 2 A three-phase, full-wave diode bridge is supplied from an ideal three-phase source and is resistively loaded. Show that the rms value of the supply line current is given by 𝐼𝑜 = 0.78 𝑉𝐿𝐿,𝑝 /𝑅 The rms of the diode current = 𝑽𝑳𝑳,𝒑 𝑽𝑳𝑳,𝒑 (… …. ) 𝑹 𝑰𝒐,𝒓𝒎𝒔 = 𝟎. 𝟗𝟓𝟔 𝑹 𝟐 𝑽𝑳𝑳,𝒑 𝑰𝑳 = 𝑰𝒐,𝒓𝒎𝒔 = 𝟎. 𝟕𝟖 𝟑 𝑹 10/20/2022 PECT302_Chpater 1_Lecture 2 20  Example 3 A three-phase full bridge diode rectifier is supplied from a 415V5%, 50Hz line. consider a safety margins for voltage and current of 50%, what is the required voltage and current ratings of the diodes for the following loads: a-Highly inductive load draws a constant current=50A. b-Resistive load of 10 1 + 𝑆𝐼 𝐼𝐷 = = 1.5 ∗ = 25𝐴 3 3 choose …(Irat=30A) as the next available rating b-10Vi) Circuit Diagram  Assumptions The following analysis consider the assumptions: ∙ Constant capacitor voltage: initially we will assume that the capacitor is large enough to allow ripple free capacitor voltage. ∙ Continuous inductor current (iL). ∙ Zero ON-state voltage drop, and zero inductor and capacitor parasitic resistor. Circuit states Switch ON Switch OFF Inductor voltage and output voltage Inductor Current Continuous Conduction Mode Condition Capacitor Current and Voltage Ripple  Example 1 The parameters of a Boost Converter are Vi=20V, d=0.6, R=12.5Ω, C=40μF and the switching frequency =100kHz. (a) Determine the output voltage (b) Determine the minimum L to ensure continuous inductor current. (c) With L=10μ determine the Determine the average, maximum, and minimum inductor currents. Voltage ripple due to Capacitor’s ESR As with the buck converter, equivalent series resistance of the capacitor can contribute significantly to the output voltage ripple. The peak-to-peak variation in capacitor current is the same as the maximum current in the inductor. The voltage ripple due to the ESR is Example 2 A boost converter has an input of 5 V and an output of 25 W at 15 V. The minimum inductor current must be no less than 50 percent of the average. The output voltage ripple must be less than 1 percent. The switching frequency is 300 kHz. Determine the duty ratio, minimum inductor value, minimum capacitor value, and maximum capacitor’s ESR. Buck-Boost Converter Step-down/step-up converter Buck-Boost Converter -19/11/2022 1  Circuit Diagram and assumption iD Assuming that: ii iL io Constant capacitor voltage: the iC capacitor is large enough to allow ripple free capacitor voltage. Vi vL L C vo Continuous inductor current (iL) or (iL>0). Buck-Boost Converter -19/11/2022 2 The circuit states The switch ON The switch OFF 𝑣𝐿 = 𝑉𝑖 𝑣𝐿 = 𝑉𝑜 𝑖𝐶 = 𝑖𝑜 − 𝑖𝐿 𝑖𝐶 = −𝑖𝑜 Buck-Boost Converter -19/11/2022 3  Inductor voltage and output voltage Step Down Mode Step Up Mode The average of inductor voltage is zero, therefore 𝑑𝑇𝑠𝑤 𝑉𝑖 = − 1 − 𝑑 𝑇𝑠𝑤 𝑉𝑜 Gives 𝑉𝑜 𝑑 The negative sign indicates that the output voltage is =− 𝑉𝑖 1−𝑑 negative in the polarity shown Buck-Boost Converter -19/11/2022 4  Inductor Current Tsw dTsw To determine the average inductor current, IL,dc: Since: 𝑖𝐿 = 𝑖𝑖 + 𝑖𝐷 , gives 𝐼𝐿,𝑑𝑐 = 𝐼𝑖,𝑑𝑐 + 𝐼𝐷,𝑑𝑐 Vi The diode current: 𝑖𝐷 = −𝑖𝐶 − 𝐼𝑜 Since the capacitor current has zero average: 𝐼𝐷,𝑑𝑐 = −𝐼𝑜 Vo Substitute 𝐼𝐿,𝑑𝑐 = 𝐼𝑖,𝑑𝑐 − 𝐼𝑜 DIL 𝑑 𝐼𝑖,𝑑𝑐 = − 1−𝑑 𝐼𝑜 IL,dc 𝑑 𝑑−1+𝑑 𝟏 𝐼𝐿,𝑑𝑐 = − 𝐼𝑜 − 𝐼𝑜 = − 𝐼𝑜 = − 𝑰 1−𝑑 1−𝑑 𝟏−𝒅 𝒐 Buck-Boost Converter -19/11/2022 5  Inductor Current Tsw dTsw 𝟏 𝐼𝐿,𝑑𝑐 = 𝑰 𝟏−𝒅 𝒐 Vi During ton: ∆𝐼𝐿 Vo 𝑉𝑖 = 𝐿 , 𝑑𝑇𝑠𝑤 Gives: 𝑑𝑇𝑠𝑤 𝑉𝑖 𝑑 DIL ∆𝐼𝐿 = 𝑉𝑖 = 𝐿 𝐿𝑓𝑠𝑤 IL,dc Buck-Boost Converter -19/11/2022 6  Continuous Current Condition ∆𝐼𝐿 < 𝐼𝐿,𝑑𝑐 2 𝑉𝑖 𝑑 1 𝑉𝑜 Tsw < 2𝐿𝑓𝑠𝑤 1 − 𝑑 𝑅 dTsw (1 − 𝑑)𝑑𝑉𝑖 𝑅 𝐿𝑓𝑠𝑤 > 2 𝑉𝑜 By substituting for Vo from (10), the continuous DIL conduction condition is given by: (1 − 𝑑)2 𝑅 (𝟏 − 𝒅)𝟐 𝑽𝒐 𝐿𝑓𝑠𝑤 > = 2 𝟐𝑰𝒐 Buck-Boost Converter -19/11/2022 7 The Capacitor current Tsw dTsw The switch ON 𝑖𝐶 = −𝑖𝑜 = |𝐼𝑜| iD DIL The switch OFF 𝑖𝐶 = 𝑖𝑜 − 𝑖𝑜𝐿 IL,dc iC -Io Buck-Boost Converter -19/11/2022 8  Output Voltage Ripple Tsw dTsw ∆𝑸 𝒅𝑰𝒐 ∆𝑽𝒐 = = iC -Io 𝑪 𝑪𝒇𝒔𝒘 𝒅𝑽𝒐 ∆𝑽𝒐 = 𝑹𝑪𝒇𝒔𝒘 ∆𝑽𝒐 𝒅 = 𝑽𝒐 𝑹𝑪𝒇𝒔𝒘 |Vo| As for Boost converter…. DVo ∆𝑽𝒐, 𝑺𝑬𝑹 = 𝑰𝑳,𝒎𝒂𝒙 𝒓𝒄 Buck-Boost Converter -19/11/2022 9  Example 1 A buck-boost converter has the following parameters Vi=12 V, d=0.6, R =10, L =10H, C =20  with 0.4 ESR, and a switching frequency of 200 kHz. Determine (a) the output voltage, (b) the average, maximum, and minimum inductor currents, (c) the output voltage ripple. (d) Sketch the switching signals and the inductor and capacitor currents. (e) Determine the rms values of the inductor and capacitor currents. Solution: −𝑑 (a) 𝑉𝑜 = 𝑉 = −18𝑉 (1−𝑑) 𝑖 𝑉𝑜 (b) 𝐼𝑜 = 𝑅 = −1.8𝐴 −1 𝑑𝑉𝑖 0.6∗12 𝐼𝐿,𝑑𝑐 = 𝐼 = 4.5𝐴 ∆𝐼𝐿 = = = 3.6𝐴 (1−𝑑) 𝑜 𝑓𝑠𝑤 𝐿 200∗103 ∗10−5 3.6 3.6 𝐼𝐿,𝑚𝑎𝑥 = 4.5 + = 6.3𝐴 𝐼𝐿,𝑚𝑖𝑛 = 4.5 − = 2.7𝐴 2 2 Buck-Boost Converter -19/11/2022 10 (c) ∆𝑉𝑜 𝑑 = 𝑉𝑜 𝑅𝐶𝑓𝑠𝑤 0.6∗18 ∆𝑉𝑜,𝐶 = =0.27V 10∗20∗10−6 ∗200∗103 ∆𝑉𝑜,𝐸𝑆𝑅 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶 =6.3*0.4=2.52V The output voltage ripple (peak-to-peak) is 2.52 V or (2.52/18)=14% of the average output voltage ((RF (vo)5%) Buck-Boost Converter -19/11/2022 11  2 2 ∆𝐼𝐿 𝐼𝐿 = 𝐼𝐿,𝑑𝑐 + = 4,618𝐴 2 3 2 𝐼𝐶 = 1.8 ∗ 0.6 + 1.8292 = 2.3 𝐴 Buck-Boost Converter -19/11/2022 12 Example 2 Design a buck-boost converter to supply a constant power load of 75 W at a voltage ranges between 20- to- 80 V from a 40-V source. The output ripple must be no more than 1 percent. Assume a switching frequency of 50kHz -Specify rang of the duty ratio, -Calculate the inductance so that the minimum inductor current in at least 20% of the inductor average current, -Determine the capacitor required Buck-Boost Converter -19/11/2022 13  Flyback Converter (Isolated Buck-Boost Converter)  Isolated DC supplies A basic disadvantage of non-isolated (AC-DC and DC-DC) converters is the electrical connection between the input and the output. If the input supply is grounded, that same ground will be present on the output and vice-versa. This is undesirable for (user and components) safety reasons. To isolate the output from the input electrically, a transformer is used.  Isolated DC supply with line frequency transformer If the converter has a first stage that rectifies an ac power source to dc, a transformer could be used on the ac side. This design has two disadvantages: -First, not all applications require ac to dc conversion (some are dc/dc). -Second, a transformer operating at a low frequency (50/60 Hz) requires a large magnetic core and is therefore large, heavy, and expensive. The use of line frequency transformer is an inferior design. Isolated DC supplies A more efficient method of providing electrical isolation between input and output of a dc-dc converter is to use a transformer in the switching system. The switching frequency is much higher than the ac power-source frequency, enabling the transformer to be small. Besides isolation, the transformer turns ratio provides practical method to match voltage levels. Also with the use of multiple transformer windings, switching converters can be designed to provide multiple output voltages. Transformer and model Ideal Transformer L1 L2 Model used for most power 𝐿1 𝑁1 2 electronics circuits. = 𝐿2 𝑁2 Transformer versus inductor L S1 S2 i 1 2 𝑊= 𝐿𝑖 2 i1 A V1/L1 i2 i1 i1 i2 i2 S1 1:1 S2 L1 L2 V1 V2 A -V2/L2 1 𝑊= 𝐿1 𝑖12 + 𝐿2 𝑖22 2  i1 More general case.. V1/L1 A 𝐴𝑁1 = 𝐵𝑁2 i2 i1 N1:N2 i2 B S1 S2 -V2/L2 V1 V2 V1 2 𝐿1 𝑁1 = 𝐿2 𝑁2 V2(N1/N2) Equal Area (zero energy change)  D N1:N2 i2 To Flyback converter … S1 i1 Vo (isolated Buck-Boost) V1 C R i1 N1:N2 i2 S1 S2 i1 N1:N2 i2 V1 V2 S1 D Vo V1 C R D i1 N1:N2 i2 i1 N1:N2 i2 S1 D Vo V1 C R V1 V2 R  Continuous Conduction mode In continuous mode, at any time (either i1 or i2 >0) Additional assumptions for the analysis are made: 1. The output capacitor is very large, resulting in a constant output voltage Vo. 2. The circuit is operating in the steady state, implying that all voltages and currents are periodic, beginning and ending at the same points over one switching period. 3. The duty ratio of the switch is D, being closed for time DT and open for (1 D)T. 4. The switch and diode are ideal Flyback converter analysis: conversion ratio Tsw dTsw (1-d)Tsw 𝑁1 S 𝑑𝑇𝑠𝑤 𝑉𝑠 = (1 − 𝑑)𝑇𝑠𝑤 𝑉𝑜 𝑁2 𝑉𝑜 𝑁2 𝑑 = 𝑉𝑠 𝑁1 1−𝑑 Vs 𝑉𝑠 𝑑𝑇𝑠𝑤 ∆𝐼𝐿𝑀 (𝑐𝑙𝑜𝑠𝑒𝑑) = -Vo(N1/N2) 𝐿𝑚 −𝑉𝑜 (1 − 𝑑)𝑇𝑠𝑤 𝑁1 ∆𝐼𝐿𝑀 (𝑜𝑝𝑒𝑛) = 𝐿𝑚 𝑁2 DILM Flyback converter analysis DILM Continuous Conduction Condition ∆𝐼𝐿𝑚 < 𝐼𝐿𝑚,𝑑𝑐 2 iS (N1/N1)iD 𝐼𝑜 𝑁2 𝐼𝐿𝑚,𝑑𝑐 = (1 − 𝑑) 𝑁1 𝑉𝑜 𝑁2 𝑉𝑠 𝑑𝑇 > iD 𝑅(1 − 𝑑) 𝑁1 𝐿𝑚 2 (1 − 𝑑)2 𝑅 𝑁1 𝐿𝑚 𝑓𝑠𝑤 > 2 𝑁2 ic ∆𝑉𝑜 𝑑 = Io 𝑉𝑜 𝑅𝐶𝑓𝑠𝑤 Flyback Converter Topologies Switch ON Flyback Converter Topologies Switch OFF Example A flyback converter of has the following circuit parameters: Determine (a) the required duty ratio d; (b) (b) the average, maximum, and minimum values for the current in Lm; and (c) the output voltage ripple. Assume that all components are ideal.  Drill Problem The flyback converter has parameters Vs =4.5 V, d=0.6, N1/N2 =0.4, R =15 Ohm, Lm =10 H, and C =10 F, and the switching frequency is 250 kHz. Determine (a) the output voltage; (b) the average, maximum, and minimum inductor currents; and (c) the output voltage ripple.  Flyback Converter (Isolated Buck Converter) The forward converter, is a magnetically coupled dc-dc converter. Steady-state operation is assumed for the analysis of the circuit, and the current in inductance Lx is assumed to be continuous. The transformer has three windings: windings 1 and 2 transfer energy from the source to the load when the switch is closed; winding 3 is used to provide a path for the magnetizing current when the switch is open and to reduce the magnetizing current to zero before the start of each switching period In forward Converter: The magnetic core is charged during Ton (𝑖𝐿𝑀 increases during Ton) at the same time the power is transferred to the secondary side through the diode. During Toff the more must be demagnetized completely (by returning 𝑖𝐿𝑀 to zero) The demagnetization is done through the trinary coil: the magnetization energy returned to the supply. In continuous conduction mode, the current through Ls remains >0. When the switch is on.. Assuming Vc constant and iLS >0 When the switch is OFF..  Demagnetization condition The magnetic core must be demagnetized during each switching cycle, otherwise the magnetic core will saturate (𝑖𝐿𝑀 will grow without limit) During ∆𝑇𝑥 … → 1 𝐷𝑚𝑎𝑥 = 𝑁3 1+ 𝑁1 Currents and voltages Example: Chapter 4 Inverters Introduction -Inverter is a DC-to-AC power converter. -An inverter is supplied from a dc source while the ac-output voltage has a fundamental component with adjustable frequency and amplitude. - The Inverters are required in many applications such as: PV systems (photovoltaic system) Variable speed AC machines drives, Uninterruptable power supplies (UPS). -There are two types of inverters according to the dc supply circuit: The voltage source inverter (VSI) and current source inverter (CSI).  Voltage Source Inverter (VSI) Lf Ideal Inverter DC Inverter AC line 3 Cf AC output AC output circuit circuit Modeled source n-pulse as: rectifier VSI has a dc supply circuit with constant voltage. With large shunt capacitor filter the input voltage is almost constant and modeled as a pure DC voltage source. Most of the practical Inverters are VSIs.  Current Source Inverter (CSI) CSIs have a dc supply circuit with constant current Using large series inductor filter the input current is almost constant and modeled by a pure DC current source. Practically CSI are not so common, and this inverter is not studied in this chapter!  Single-phase inverter circuit –Full Bridge Switches in one branch of the inverter cannot be turned on at the same time, as this would short-circuit the supply. a b SA DA SB DB a A vo B Vi io Load a a b Practically semiconductor switches, have nonzero on-off time. SA DA SB DB Therefore, in practice, to avoid short circuit, a switch is turned off shortly before the turn-on of the other switch in the same leg. dead time In the subsequent discussion, the dead time a will be assumed zero. a Switching variables Two switching variables a and b have been assigned as the control variables to the inverter legs and defined as follows: a b a b SA DA SB DB SA DA SB DB A vo B A vo B Vi Vi io Load io Load a b a b SA DA SB DB SA DA SB DB a=1 a=0  Switching state a SA DA A vo b B SB DB a SA DA A vo b B SB DB Vi Vi io Load io Load An inverter state is designated as ab2. If, a b a b SA DA SB DB SA DA SB DB for instance, a = 1 and b = 1, the inverter State 0 State 1 is said to be in state 3 as 112 = 3. Clearly, (00) (01) four states are possible, from state 0 to state 3. a a b SA DA b SB DB SA DA SB DB vo A vo B A B Vi Vi io Load io Load a b a b SA DA SB DB SA DA SB DB State 2 State 3 (10) (11)  The inverter output voltage It can be shown that … a b SA DA SB DB 𝑣𝐴 = 𝑎𝑉𝑖 and 𝑣𝐵 = 𝑏𝑉𝑖 A vo B Vi The output voltage, vo, of the inverter can be expressed io Load as: a b vA vB SA DA SB DB 𝑣𝑜 = 𝑣𝐴 − 𝑣𝐵 𝑣𝑜 = 𝑉𝑖 (𝑎 − 𝑏) gives the output voltages for various states State Vo 0 (00) 0 1 (01) -Vi 2 (10) Vi 3 (11) 0  Input current The input current is defined as a a a function of the output current SA DA b SB DB SA DA vo=Vi b SB DB and the switching state A vo B A vo B Vi Vi io Load io Load vo=0 a b a b SA DA SB DB SA DA SB DB State 2 state 0 State Vo Input current (ii) 0 (00) 0 0 a b a b 1 (01) -Vi -io SA DA vo=-Vi SB DB SA DA vo=0 SB DB vo vo 2 (10) Vi io Vi A B Vi A B io Load io Load 3 (11) 0 0 a b a b SA DA SB DB SA DA SB DB state 1 State 3  Half-Bridge Inverter A simpler inverter circuit a Vi/2 the circuit is controller by one binary io switching variable (a) Vi Vo a Vo 0 -Vi/2 a Vi/2 1 Vi/2  Square-wave mode 1 𝑓𝑜𝑟 0 < 𝜔𝑡 < 𝜋 0 𝑓𝑜𝑟 0 < 𝜔𝑡 < 𝜋 𝑎=ቊ 𝑎𝑛𝑑 𝑏 = ቊ 0 𝑓𝑜𝑟 𝜋 < 𝜔𝑡 < 2𝜋 1 𝑓𝑜𝑟 𝜋 < 𝜔𝑡 < 2𝜋 Vo,1 = 0.9Vi and Vo,h, =0.435Vi Gives. The total harmonic distortion, THD, is 0.483. 2 2𝑉𝑖 The amplitude of the kth harmonics of vo, is 𝑉𝑜,𝑘 = ; 𝑘𝜋 k=1,3,5,... H.W. -Derive Vo,1as a function Vi and the THD of Vo -Draw the shape of vo for the half-bridge inverter, mark the amplitudes  Example The single-phase full-bridge inverter is operated in square mode to supply an RL load. The switching frequency is 60 Hz. The dc supply is ( Vi=) 100 V, and the RL load has R=10 Ω and L = 25 mH. Determine (a) the expression for load current (𝑖𝑜 (𝑡)) , (b) the power absorbed by the load, (c) the average current in the dc source. (d) Over one operation cycle draw the waveforms of the load voltage and current and identify the conducting switch and the operating quadrant. Solution 𝑉𝑖 𝑉𝑖 𝜔𝑡 − 𝑖𝑜 = 𝑖1 = − + 𝐼𝑝 𝑒 tan ∅ 𝑓𝑜𝑟 0 < 𝜔𝑡 < 𝜋 Ip 𝑅 𝑅 i1 𝑉𝑖 𝑉𝑖 𝜔𝑡−𝜋 − 𝑖𝑜 = 𝑖2 = − + + 𝐼𝑝 𝑒 tan ∅ 𝑓𝑜𝑟 𝜋 < 𝜔𝑡 < 2𝜋 𝑅 𝑅 i2 Applying the boundary condition: - Ip 𝑖1 𝜔𝑡 = 𝜋 = 𝑖2 (𝜔𝑡 = 𝜋) Gives: 𝜋 − 𝑉𝑖 1 − 𝑒 𝑡𝑎𝑛∅ 𝝎𝒕 𝐼𝑝 = 𝜋 𝟏𝟎 − 𝟏𝟗. 𝟑𝟏𝒆−𝟎.𝟗𝟒𝟐 , 𝒇𝒐𝒓 𝟎 < 𝝎𝒕 < 𝝅 𝑅 − 𝒊𝒐 = ൞ 1+𝑒 𝑡𝑎𝑛∅ 𝝎𝒕−𝝅 −𝟎.𝟗𝟒𝟐 −𝟏𝟎 + 𝟏𝟗. 𝟑𝟏𝒆 , 𝒇𝒐𝒓 𝝅 < 𝝎𝒕 < 𝟐𝝅 Accordingly: 𝜔𝐿 120𝜋 ∗ 0.025 tan ∅ = = = 0.942 𝑅 10 𝐼𝑝 = 9.31𝐴 (b) To determine the power absorbed by the load, find Io 2 1 𝜋 − 𝜔𝑡 𝐼𝑜 = න 10 − 19.31𝑒 0.942 𝑑𝜔𝑡 𝜋 0 Solved numerically using calculator, Io=6.644A The load power is the power dissipated in the resistor, 𝑷 = 𝑰𝟐𝒐 𝑹 = 𝟒𝟒𝟏. 𝟔𝑾 (c) The dc source current is equivalent to the load current for 0