Exercice corrigés e p
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This document contains: - Examples of rectifier circuits and calculations of output voltages and currents. - Exercises on modeling inductive and capacitive loads connected to rectifier circuits. - Circuits involving controlled switches for charging batteries from rectified AC sources.
![Électronique de puissance – Exercices © M. ZEGRARI
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Exercice corrigés e p
- 1. Office National Des Aéroports Académie Internationale Mohamed VI de l’Aviation Civile Office National Des Aéroports Académie Internationale Mohamed VI de l’Aviation Civile
- 2. Électronique de puissance – Exercices © M. ZEGRARI
- 3. Électronique de puissance – Exercices © M. ZEGRARI ! " # $ % &'( )* + ) $ $ $ ! " # $ " " % & $ " "& # $ " "" ' " ( " )" $ Charge inductiveicc vs L vcc R is Vm π 3π2π vcc 0 ωt π 2π 3π ωt icc *0 RedresseurSource CA Vm π 3π2π v1 0 ωt Vm π 3π2π v2 0 ωt Vm π 3π2π v3 0 ωt i1 0 T/2 *0 T t +*0 3T/2 i2 0 T/2 *0 T t +*0 3T/2 i3 0 T/2 *0 T t 3T/2td td td
- 4. Électronique de puissance – Exercices © M. ZEGRARI , " # % - * $ ! " "# &, # $ - . / Ω % . 0 1 , /2 / 3 + 4/ 15 % # $ "" &" 6 ( ( 7 7 " $ ! " "# & , # & - . / Ω , /2 / 3 + 4/ 15 %8 $ 9 $ # :&,α .;/< 6 8 6 6 7 7 # $ " " 7 "& # $ T1 220/220 V vp ip Charge R = 20 Ω iT1 T 220 V 50 Hz is icc vcc vT1 vs T2 T3 T4 D1 220/220 V vp ip Charge très inductive L = 2.8 H T 220 V 50 Hz is icc vccvs D2 D3 D4 R = 110 Ω vD1 iD1
- 5. Électronique de puissance – Exercices © M. ZEGRARI . = % . 0 1 & & - • 6 • 7 7 • " $ ! ! "# & & & /2 0/ 3 + 4/ 15 " "# &, # " &, # $ % ./ > 1 - . / Ω %8 $ 9 $ # :&,α . /< 6 7 7 (& $ 9 $ α " 4 = , ! " $ "# & # $ & - . / Ω α .?/< 7 7 - " & "" 7 2 7 α " "& # $ $ α % . / 1 & & - ( α $ 9 $ " $ is iac vac vsvs T1 220 V 50 Hz R = 110 ΩT2 T1vAN ia Charge inductive R = 220 Ωic A 220/380 V 50 Hz ib icc vcc T2 T3 vBN vCN T4 T5 T6 L = 0.7 H B C N T1 220/220 V vp ip Charge très inductive L = 1.8 H iT1 T 220 V 50 Hz is icc vcc vT1 vs T2 T3 T4 R = 20 Ω
- 6. Électronique de puissance – Exercices © M. ZEGRARI / " # ! %0 # - ) & ! $ ! # # & & " & - . Ω 3- . 3 , " : 3 . @0 3 % " # # & & " *AB6 "" & " % &$& , & . C15 "" α 6 " 6 ( % % 6 ( % (& % % /D (& D ! ! # # & " & - " 3- . // 3 " " - . 4// E % # # &" 3 . / 3 % *AB6 "" & " % & # # $ . / C15 "" α 6 % % % % (& % % /D (& / 4D 6 6 ( 6 ( (& & & " Source continue VCC = 120 V iT vL L iL CvC R ChargeiRiD iC D T vR = 300 V PR = 1.5 kW icc Source continue VCC = 48 V iD vL L iT CvC R = 2 Ω ChargeiRiL iC D T vR = 12 V icc vFvD vT
- 7. Électronique de puissance – Exercices © M. ZEGRARI 1 ! ! % , : & $ % " F F & "& 6 & + " / G G6 2 F & F + " 6 2 G G6 F F & 6 # # " # $ (& " "& & - ! , ! "# & " &" " % . @ 1 , " 3 .@0 3 % 9 $ " & &" & & . C15 ( :" "& 6 6 7 & ," & & & (& " & $ " 7 .@/ 3 : % Vcc = 48 V icc T1 T3 T4 T2 L = 2.4 mH iac vac D1 D3 D4 D2 K2 K1 i2 i1 ich vch R = 50 Ω E1 = 40 V E2 = 40 V
- 8. Électronique de puissance – Exercices © M. ZEGRARI 2 3' & # $ &'( )* + ) $ $ 3 • 6 " $ "# & " " &$ "" & ( ) [ ] π =θ− π =θθ π = π π ( ) ( )==θ θ− π =θθ π = ππ • 6 " $ "# & " &$ & ( ) [ ] π =θ− π =θθ π = π π ( ) ( )==θ θ− π =θθ π = ππ • 6 " $ "# & " H " I" J ( ) [ ] π =θ− π =θθ π = π π π π ( ) π π π π θ θ− π =θθ π = Vm π 3π2π v1 0 ωt Vm π 3π2π v2 0 ωt Vm π 3π2π v3 0 ωt π/6 5π/6
- 9. Électronique de puissance – Exercices © M. ZEGRARI 4 3 • " $ , & & # $ & " % "& ( ) =−+= ( ) ( )=== • " $ , & & & # $ & " % "& ( ) =−+= + ( ) ( ) −=−== • " $ , & & # $ " % "& ( ) =+−−+−= ( ) ( )==−= $ ! " # $ "& ( ) ( )[ ] ( ) π =×=×= "" ( ) ( )=×= ) " $ π = π == Vm π 3π2π vcc 0 ωt *0 π 2π 3π ωt icc i1 0 T/2 *0 T t +*0 3T/2 i2 0 T/2 *0 T t +*0 3T/2 i3 0 T/2 *0 T t 3T/2td td td +*0
- 10. Électronique de puissance – Exercices © M. ZEGRARI $5 3' & # - * $ ! " "# &, # $ - . / Ω % . 0 1 - " & 6 ( ) = π ×× = π = ( ) ( ) === " "& =×= π = "" ( ) ( ) =×=×= ) " === ! " "# &, # & - . / Ω - " & 6 ( ) ( ) ( ) =°+ π =α+ π = ( ) ( ) !=== "& & - ( ) ( ) ( ) != π + π − π π× × = α + α − π π = % $ " % . 0 1 & - • ) • 6 ( ) =° π × =α π = • ( ) ( ) === • ) " =°× π =α π =
- 11. Électronique de puissance – Exercices © M. ZEGRARI $$ ! ! " "# & ( # &" "# & /2 0/ 3 - " & 6 ( ) =° π × =α π = ( ) ( ) === & : , 4 = °=α= × ××π = × ×π =α=α π " , ! " $ "# & # $ & - . / Ω 6 ( ) ( ) # == π α+α−π = ( ) ( )# # === 3 " "" , α 6 ( ) ( ) ( ) =π=α =π=α ==α π α+α−π == !# # $ ( )# === % ./ 1 & & - = $ 9 $ ( ) °= π× = ω ==α !#$ % & #$ %'#$%
- 12. Électronique de puissance – Exercices © M. ZEGRARI $ 3' & # ! 0 # - ) & ! $ ! # # & &" & - . Ω K 3- . 3 3 .@0 3 "" ===α & # # $ ( ) (!)(( =α−==α== - " & 3 % ( ) ( )& ==== ( ) ( )&& =×=×=∆ ( ) ( )& & =−= ∆ −= ( ) ( )& &#* =+= ∆ += * $ +!& & =×× − = ∆ − = − " & $ ( ) ( ) ( ) ( ) (, !& - = ××× ××− = ∆ α− = − − ! ! # # & - . 4// E " 3- . // 3 % # # &, 3 . / 3 ) & # # $ . / C15 "" =−=−=α & # # $ ( ) ()(( =α−==α=== - " & 3 % ( )& === ( ) ( )&& =×=×=∆ ( ) ( )& & =−= ∆ −= ( ) ( ) !& &#* =+= ∆ += * $ +& & =××= ∆ = −
- 13. Électronique de puissance – Exercices © M. ZEGRARI $! - α = ∆ -& # $ Ω=== " & $ ( ) (,- = ×× ×× = ∆ α = − − & " 367 : . // 3 K 3( 7 : .+ // 3 ! ! % , : & $ ) 3 # $ ( ) ../ === ( ) ( )/ / === "& ( )// =×== ! , ! "# & " 3 .@0 3 K % . @ 1 K . C15 L:" # $ - "" =α K & (( ===== ) 7 % $ % $ 7 , : 3 : & = ×× ×× == − − 3 ( )# === & & & " 7 .@/ 3 ( ) ( ) (!# # =− × =−=−= − vch 0 t ich 0.8 A T/2 -40 V 40 V -0.8 A T/2