AC to DC Rectifier Analysis PDF

Analysis of Single-Phase Full- Bridge Controlled Rectifier with RL Load Known also as: 2-pulse converter and phase-controlled bridge rectifier  Continuous-current mode The converter current will be Continuous if the output current is always positive: 𝒊𝒐 𝝎𝒕 > 𝟎 f...

Analysis of Single-Phase Full- Bridge Controlled Rectifier with RL Load Known also as: 2-pulse converter and phase-controlled bridge rectifier  Continuous-current mode The converter current will be Continuous if the output current is always positive: 𝒊𝒐 𝝎𝒕 > 𝟎 for all t T1 and T2 conduct for 𝛼𝑓 < 𝜔𝑡 < 𝛼𝑓 + 𝜋 T3 and T4 are conduct for 𝛼𝑓 + 𝜋 < 𝜔𝑡 < 𝛼𝑓 + 2𝜋  RL-Load Current (𝑖𝑜,𝐶 ) Vp The load current satisfies the equation: 0 To Determine the current equation for one period (𝛼𝑓 < 𝜔𝑡 < 𝛼𝑓 + 𝜋) 𝑣𝑜 = 𝑣𝑠 = 𝑉𝑝 sin(𝜔𝑡) The current is found by solving the DE: 𝑑𝑖𝑜 𝑉𝑝 sin(𝜔𝑡) = R𝑖𝑜 + 𝐿 𝑑𝑡 The complete solution of this equation has the form: 𝑖𝑜 = 𝑖𝑜,𝑁 + 𝑖𝑜,𝐹 a a+p a+2p where 𝑡−𝑡𝑜 − 𝑍 = 𝑅2 + 𝜔𝐿 2 𝑖𝑜,𝑁 = 𝐴𝑒 𝐿/𝑅 and and 𝜔𝐿 𝑉𝑝 ∅= tan−1 𝑖𝑜,𝐹 = sin(𝜔𝑡 − ∅) 𝑅 𝑍  RL-Load Current (𝑖𝑜,𝐶 ): the complete expression 𝑡−𝑡𝑜 𝑉𝑝 (𝜔𝑡−𝛼𝑓 ) 𝑉𝑝 − − 𝑖𝑜 = 𝑖𝑜,𝑁 + 𝑖𝑜,𝐹 = 𝐴𝑒 𝐿/𝑅 + sin 𝜔𝑡 − ∅ → 𝑖𝑜 𝜔𝑡 = 𝐴𝑒 𝑡𝑎𝑛∅ + sin 𝜔𝑡 − ∅ 𝑍 𝑍 To determine A, consider the boundary condition: 𝑖𝑜 𝜔𝑡 = 𝛼𝑓 = 𝑖𝑜 𝜔𝑡 = 𝛼𝑓 + 𝜋 2𝑉𝑝 sin(𝛼𝑓 −∅) Gives 𝐴 = 𝜋 − 𝑍 𝑒 𝑡𝑎𝑛∅ −1 𝑉𝑝 2sin(𝛼𝑓 − ∅) −(𝜔𝑡−𝛼𝑓 ) 𝑖𝑜,𝐶 𝜔𝑡 = sin 𝜔𝑡 − ∅ + 𝜋 𝑒 𝑡𝑎𝑛∅ 𝑍 − 𝑒 𝑡𝑎𝑛∅ − 1 for (𝛼𝑓 < 𝜔𝑡 < 𝛼𝑓 + 𝜋)  Example 1 A single-phase full-bridge rectifier is supplied from a 220V, 50Hz AC source and operates in continuous mode supplying an RL load with 50mH inductor and 10Ω resistor. The firing angle is set to 45°. Determine Simulink verification  Continuous current condition Example 2: Show that the rectifier is in Example 1 is operating in continuous conduction mode. ? Sol. 𝛼𝑓 ∅ → 45 < 57.5  < ∴ 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 Discontinuous-current mode In this case, the output current reduces to zero before 𝜔𝑡 = 𝛼𝑓 + 𝜋 and the convertor goes to state zero (ALL 4 SCRs OFF) until the next triggering af ae p+af In this case, the output current is determined by applying the initial condition (𝑖𝑜 𝜔𝑡 = 𝛼𝑓 = 0), gives 𝑉𝑝 (𝜔𝑡−𝛼𝑓 ) − 𝑖𝑜 (𝜔𝑡) = sin 𝜔𝑡 − ∅ − sin 𝛼𝑓 − ∅ 𝑒 𝑡𝑎𝑛∅ 𝑍 𝛼𝑓 < 𝜔𝑡 < 𝛼𝑒 1 𝛼 and 𝑉𝑜,𝑑𝑐 = 𝜋 ‫ 𝑝𝑉 𝑒 𝛼׬‬sin 𝜔𝑡 𝑑𝜔𝑡 𝑓 =0 𝛼𝑒 < 𝜔𝑡 < 𝛼𝑓 + 𝜋 𝑉𝑝 𝑉𝑜,𝑑𝑐 = cos 𝛼𝑓 − cos 𝛼𝑒 𝜋  Example 3 A single-phase full-bridge rectifier is supplied from a 220V, 50Hz AC source and operates in continuous mode supplying an RL load with 50mH inductor and 10Ω resistor. The firing angle is set to 90°. Determine the conduction mode and find Solution: To check the conduction mode, check Now find 𝛼𝑒 by solving the condition 𝑖𝑜 (𝛼𝑒 ) = 0 numerically 𝛼𝑓 < ∅ ∅ = 57.5° t i(t) 90 0 ? 𝛼𝑓 ∅ → 90 > 57.5 Discontinuous Mode < 120 8.4 150 12.1 220 2 (𝜔𝑡−90) − 89.94 180 10.8 𝑖𝑜 (𝜔𝑡) = sin 𝜔𝑡 − 57.5 − sin 90 − 57.5 𝑒 18.6 210 5.36 (𝜔𝑡−90) − 89.94 𝑖𝑜 (𝜔𝑡) = 16.73 sin 𝜔𝑡 − 57.5 − 0.537𝑒 𝛼𝑒 231 0  𝑉𝑝 220 2 𝑉𝑜,𝑑𝑐 = cos 𝛼𝑓 − cos 𝛼𝑒 = (0 − cos 231)=63V 𝜋 𝜋 𝑉𝑜,𝑑𝑐 𝐼𝑜,𝑑𝑐 = = 6.3𝐴 𝑅 2 4 𝜋 (𝜔𝑡− ) 𝐼𝑜,𝑎𝑐 7.072 − 6.32 1 − 𝜋/2 2 𝑅𝐹𝑖𝑜 = = 𝐼𝑜 = න 16.73 sin 𝜔𝑡 − 1.004 − 0.537𝑒 𝑑𝜔𝑡 = 7.07𝐴 𝐼𝑜,𝑑𝑐 6.3 𝜋 𝜋/2 = 0.51 1 4 2 2 4 1−cos 2𝜔𝑡 𝑉𝑜 = ‫׬‬ 𝜋 𝜋/2 220 2 sin 𝜔𝑡 𝑑𝜔𝑡 = 220 𝜋 ‫𝜋׬‬/2 2 𝑑𝜔𝑡 220 𝜋 cos 8 = [4 − 2 − ] =196.3V 𝜋 2 𝑉𝑜,𝑎𝑐 185.6 𝑅𝐹𝑣𝑜 = = = 2.94𝑉 𝑉𝑜,𝑑𝑐 63 𝑉𝑜,𝑎𝑐 = 𝑉𝑜2 − 𝑉𝑜,𝑑𝑐 2 = 185.6