Power Electronics Lecture No. 4 PDF

Zagazig University Faculty of Engineering Electrical Power & Machines Department Power Electronics LECTURE NO. 4 By Assoc. Prof.Dr. Mohamed A. Enany Chapter (2) Diode Rectifiers 1-Single phase Half...

Zagazig University Faculty of Engineering Electrical Power & Machines Department Power Electronics LECTURE NO. 4 By Assoc. Prof.Dr. Mohamed A. Enany Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  1-Single-phase half wave rectifier with RL load During diode conduction the circuit is defined by 𝑑𝑖 𝐿 + 𝑅 𝑖 = 𝑉𝑚 sin 𝜔𝑡 𝑑𝑡 Hence 𝑖 𝑡 = 𝑖 𝑓𝑜𝑟𝑐𝑒𝑑 𝑡 + 𝑖 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑡 𝑉𝑚 𝑖 𝑡 = sin 𝜔𝑡 − 𝜑 + 𝐴 𝑒 −(𝑡/𝜏) 𝑍 𝜔𝐿 Where; 𝑍 = 𝑅2 + (𝜔𝐿)2 , 𝜑 = tan−1 and 𝜏 = 𝐿 𝑅 𝑅 Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  1-Single-phase half wave rectifier with RL load And A is constant can be obtained by 𝑖 0 = 0 , so 𝑉𝑚 0= sin 0 − 𝜑 + 𝐴 𝑒 −(0) 𝑍 So 𝑉𝑚 𝐴= sin 𝜑 𝑍 So the Load current will be; 𝑉𝑚 𝑖 𝜔𝑡 = [ sin 𝜔𝑡 − 𝜑 + sin 𝜑 𝑒 −(𝜔𝑡 / tan 𝜑 ) ] 𝑍 Where; (𝑡/𝜏) = (𝜔𝑡/𝜔𝜏) = (𝜔𝑡/(𝜔 𝐿 𝑅 )) = (𝜔𝑡/ tan 𝜑) Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  1-Single-phase half wave rectifier with RL load The value of 𝛽 can be obtained by substituting that 𝑖 𝜔𝑡 = 𝛽 = 0 into previous equation as follow; 𝑉𝑚 𝑖 𝛽 =0= [ sin 𝛽 − 𝜑 + sin 𝜑 𝑒 −(𝛽 / tan 𝜑 ) ] 𝑍 So 0 = sin 𝛽 − 𝜑 + sin 𝜑 𝑒 −(𝛽 / tan 𝜑) The value of 𝛽 can be obtained from the above equation by using numerical methods Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  2-Single-phase half wave rectifier with RL load and FWD  There are two modes of operation: Mode (1) Rectifying Diode (D) Conduction Mode (2) Free Wheeling Diode (Dm)Conduction Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  2-Single-phase half wave rectifier with RL load and FWD 1- During Rectifying Diode conduction the circuit is defined by 𝑑𝑖1 𝐿 + 𝑅 𝑖1 = 𝑉𝑚 sin 𝜔𝑡 𝑑𝑡 Hence 𝑉𝑚 𝑖1 𝑡 = sin 𝜔𝑡 − 𝜑 + 𝐴1 𝑒 −(𝑡/𝜏) 𝑍 And A1 is constant can be obtained by 𝑖1 0 = 𝐼𝑜 , so 𝑉𝑚 𝐼𝑜 = sin 0 − 𝜑 + 𝐴1 𝑒 −(0) 𝑍 So 𝑉𝑚 𝐴1 = 𝐼𝑜 + sin 𝜑 𝑍 So the Load current will be; 𝑉𝑚 𝑉𝑚 𝑖1 𝜔𝑡 = sin 𝜔𝑡 − 𝜑 + (𝐼𝑜 + sin 𝜑)𝑒 −(𝜔𝑡 / tan 𝜑) 𝑍 𝑍 Chapter (2) Diode Rectifiers 1-Single phase Half Wave Diode Rectifier Load Current Expression  2-Single-phase half wave rectifier with RL load and FWD 2- During Free Wheeling Diode conduction the circuit is defined by 𝑑𝑖2 𝐿 + 𝑅 𝑖2 = 0 𝑑𝑡 Hence 𝑖 2 𝑡 = 𝐴2 𝑒 −(𝑡/𝜏) And A2 is constant can be obtained by 𝑖2 𝜔𝑡 = 𝜋 = 𝐼𝜋 , so 𝐼𝜋 = 𝐴2 𝑒 −(𝜋 /𝜔𝜏) So 𝐴2 = 𝐼𝜋 𝑒 (𝜋/𝜔𝜏) So the Load current will be; 𝑖2 𝜔𝑡 = 𝐼𝜋 𝑒 (𝜋/𝜔𝜏) 𝑒 −(𝜔𝑡 /𝜔𝜏) = 𝐼𝜋 𝑒 −(𝜔𝑡 −𝜋 )/𝜔𝜏 As at 𝑖2 𝜔𝑡 = 2𝜋 = 𝐼2𝜋 = 𝐼𝑜. So the Load current will be; 𝐼2𝜋 = 𝐼𝑜 = 𝐼𝜋 𝑒 −(2𝜋 −𝜋 )/𝜔𝜏 = 𝐼𝜋 𝑒 −𝜋/𝜔𝜏 So 𝐼𝜋 = 𝐼𝑜 𝑒 𝜋/𝜔𝜏 = 𝐼2𝜋 𝑒 𝜋/𝜔𝜏 Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  There are two types of Single phase Full Wave Diode Rectifier circuits 1- Full-wave bridge rectifier 2-Full-wave rectifier with center-tapped transformer Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-1-Single-phase Full-wave bridge rectifier with R load Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-1-Single-phase Full-wave bridge rectifier with R load PIV Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-1-Single-phase Full-wave bridge rectifier with R load Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-1-Single-phase Full-wave bridge rectifier with R load TUF 81 % 1.11 0.482 0.81  Note 1/ TUF = 1/ 0.81 = 1.235 signifies that the transformer must be 1.235 times larger than that when it is being used to deliver power from a pure ac voltage.  this rectifier has ripple factor, 48 %; a medium efficiency 81 % and a TUF , 0.81. Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-2-Single-phase Full-wave rectifier with center-tapped transformer with R load This circuit is called Bi-phase Half wave rectifier Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-2-Single-phase Full-wave rectifier with center-tapped transformer with R load Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-2-Single-phase Full-wave rectifier with center-tapped transformer with R load PIV Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-2-Single-phase Full-wave rectifier with center-tapped transformer with R load Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-2-Single-phase Full-wave rectifier with center-tapped transformer with R load TUF 81 % 1.11 0.482 0.573  Note 1/ TUF = 1/ 0.573 = 1.745 signifies that the transformer must be 1.745 times larger than that when it is being used to deliver power from a pure ac voltage.  this rectifier has ripple factor, 48 %; a medium efficiency 81 % and a TUF , 0.573. Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  Different design parameters of basic single-phase full wave rectifier circuits VA of transformer PIV full-wave rectifier with center-tapped 0.573 transformer full-wave bridge rectifier 0.81  The advantage full-wave bridge rectifier are summarized below 1- The need for center-tapped transformer is eliminated, 2- The output is twice that of the center tapped circuit for the same secondary voltage. 3- The peak inverse voltage is one half of the center-tap circuit.  The disadvantage of full-wave bridge rectifier are summarized below 1- It requires four diodes instead of two, in full wave circuit, and, 2- There are always two diodes in series are conducting. Therefore, total voltage drop in internal resistance of the diodes and losses are increased. Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier  2-3-Single-phase Full-wave rectifier with center-tapped transformer with RL load Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier Load Current Expression  2-3-Single-phase Full-wave rectifier with center-tapped transformer with RL load During diode conduction the circuit is defined by 𝑑𝑖 𝐿 + 𝑅 𝑖 = 𝑉𝑚 sin 𝜔𝑡 𝑑𝑡 Hence 𝑖 𝑡 = 𝑖 𝑓𝑜𝑟𝑐𝑒𝑑 𝑡 + 𝑖 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑡 𝑉𝑚 𝑖 𝑡 = sin 𝜔𝑡 − 𝜑 + 𝐴 𝑒 −(𝑡/𝜏) 𝑍 𝜔𝐿 Where; 𝑍 = 𝑅2 + (𝜔𝐿)2 , 𝜑 = tan−1 and 𝜏 = 𝐿 𝑅 𝑅 Chapter (2) Diode Rectifiers 2-Single phase Full Wave Diode Rectifier Load Current Expression  2-3-Single-phase Full-wave rectifier with center-tapped transformer with RL load And A is constant can be obtained by 𝑖 𝜔𝑡 = 0 = 𝑖 𝜔𝑡 = 𝜋 = 𝑖 𝜔𝑡 = 2𝜋 , so 𝑉𝑚 𝑉𝑚 sin 0 − 𝜑 + 𝐴 𝑒 −(0) = sin 𝜋 − 𝜑 + 𝐴 𝑒 −(𝜋 /𝜔𝜏) 𝑍 𝑍 𝑉𝑚 𝑉𝑚 − sin 𝜑 − sin 𝜑 = 𝐴 𝑒 −(𝜋 /𝜔𝜏) − 𝐴 𝑒 −(0) 𝑍 𝑍 So 2𝑉𝑚 1 𝐴= sin 𝜑 𝑍 (1 − 𝑒 −(𝜋 /𝜔𝜏) ) So the Load current will be; 𝑉𝑚 2 sin 𝜑 𝑖 𝜔𝑡 = [ sin 𝜔𝑡 − 𝜑 + −(𝜋 /𝜔𝜏) 𝑒 −(𝜔𝑡 / tan 𝜑 ) ] 𝑍 (1 − 𝑒 ) Where; (𝑡/𝜏) = (𝜔𝑡/𝜔𝜏) = (𝜔𝑡/(𝜔 𝐿 𝑅 )) = (𝜔𝑡/ tan 𝜑) Chapter (2) Diode Rectifiers Single Phase Diode Rectifier 1-phase Full Wave Diode 1- phase Half Wave 1-phase Full Wave Rectifier With center tap Diode Rectifier Bridge Rectifier transformer 40.5 % 81 % 81 % 1.57 1.11 1.11 1.21 0.482 0.482 0.286 0.573 0.81 VA of transformer PIV Chapter (2) Diode Rectifiers Single Phase Diode Rectifier Load current equation of Single phase Half Wave Diode Rectifier with RL load 𝑉𝑚 𝑖 𝜔𝑡 = [ sin 𝜔𝑡 − 𝜑 + sin 𝜑 𝑒 −(𝜔𝑡 / tan 𝜑 ) ] 𝑍 Load current equation of Single phase Half Wave Diode Rectifier with RL load+ FWD 𝑉𝑚 𝑉𝑚 𝑖1 𝜔𝑡 = sin 𝜔𝑡 − 𝜑 + (𝐼𝑜 + sin 𝜑)𝑒 −(𝜔𝑡 / tan 𝜑) 𝑍 𝑍 𝑖2 𝜔𝑡 = 𝐼𝑜 𝑒 −(𝜔𝑡 −2𝜋)/𝜔𝜏 Load current equation of Single phase Full Wave Diode Rectifier with RL load 𝑉𝑚 2 sin 𝜑 −(𝜔𝑡 / tan 𝜑 ) 𝑖 𝜔𝑡 = [ sin 𝜔𝑡 − 𝜑 + 𝑒 ] 𝑍 (1 − 𝑒 −(𝜋/𝜔𝜏 ) )