B-Institute of Electrical Drives, Power Electronics and Components 3 Four-Quadrant-DC Chopper Figure 3. 1 shows the equivalent circuit diagram of a four- quadrant-d c. chopper LI 14 R D D Figure 3-1: Four-Quadrant-dc chopper The four switches are always cross interlocked Case 1: S1 and S4 closed, S2 and S3 open Vi=Up and V2=0=UL=V1-V2=U Case 2 S2 and S3 closed, S1 and S4 open Vi=0 and V2=UD =UL=V1-V2=-UD Case 1 Case 2 ll Differential U +U equation of the load i,r+L R+L l di r dt R rdi io‖1-e:+i Solution IL 2-io‖1-er+i R
IALB – Institute of Electrical Drives, Power Electronics and Components - 10 - 3 Four-Quadrant-DC Chopper Figure 3.1 shows the equivalent circuit diagram of a four- quadrant-d.c. chopper. D Figure 3-1: Four-Quadrant-dc chopper The four switches are always cross interlocked. Case 1: S1 and S4 closed, S2 and S3 open V1 = UD and V2 = 0 ⇒ uL = V1 – V2 = UD Case 2: S2 and S3 closed, S1 and S4 open V1 = 0 and V2 = UD ⇒ uL = V1 – V2 = – UD Case 1 Case 2 0 1 L0 t L D L L D L L D L L R Inductor L i e i R U i R U dt di R L i u U dt di i R L U U u +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛ = − + = + = = + = − τ Differential equation of the load Solution 0 1 L0 t L D L L D L L D L L R Inductor L i e i R U i R U dt di R L i u U dt di i R L U U u +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ − ⎠ ⎞ ⎜ ⎝ ⎛ = − − + = − + = = − + = − τ
B-Institute of Electrical Drives, Power Electronics and Components Initial Value(V when t→∞ New/V:isU R When t→a R R 2 when t→>∞ NewⅣV 2 The average of voltage load is as following UD≤uL≤UD L=U Figure 3.2 shows the current- and voltage characteristic U Figure 3-2: Current- and voltage characteristic
IALB – Institute of Electrical Drives, Power Electronics and Components - 11 - Initial Value (IV): iL0 = 0 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⇒ = − − τ t D L e R U i 1 New IV: R U i D L0 = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⇒ = − − τ t D L e R U i 2 2 1 when t → ∞ when t → ∞ when t → ∞ New IV: R U i D L0 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⇒ = − − − τ t D L e R U i 2 2 1 New IV: R U i D L0 = The average of voltage load is as following: −UD ≤ u L ≤ UD ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − + − = 2 1 0 1 1 2 1 2 T t U t t t t U L UD D Figure 3.2 shows the current- and voltage characteristic. D D Figure 3-2: Current- and voltage characteristic
IALB-Institute of Electrical Drives, Power Electronics and Components 4 Fourier Series Expansion 2 d(1)=+∑ cK coS kI+∑ ck sin kv a(t)d Time average Φ( t)cos kvt dt (sinko dt Angular frequency of the k-th order harmonic component a a2 2 Bβ2 Figure 4-1: Fourier-Analysis
IALB – Institute of Electrical Drives, Power Electronics and Components - 12 - 4 Fourier Series Expansion ∑ ∑ ∞ = ∞ = Φ = + + 1 1 0 cos sin 2 ( ) k k k k c k t c k t A t ν ν ν π = 2 T A T t dt T 0 0 1 = ∫ Φ( ) Time average c T t k t dt k T = ∫ 2 0 Φ( ) cos ν sk T t k t dt { T = ∫ 2 0 Φ( ) sin ν Angular frequency of the k-th order harmonic component Figure 4-1: Fourier–Analysis
IALB-Institute of Electrical Drives, Power Electronics and Components Analysing the curve of UL in UL1 and UL2. a line function without sinusoidal function part is developed 20 sine cosa 2COS 2a+ coS 5a U(B)=_2U「 B, sin B sIn B sin 2B 3Bcos3β+ a=B+丌 sInce +B1 2UpIr-a, sin(t-a cos(a-丌)+ 丌 2 Sin3(丌-a1) cos3(a-丌)+ sInc(-cos - sin 2 Sa+ a+131 UL(a)=UL(a)+Ul(a) 2sin 2a 2sin 3a cosa cos 2a+ cos 30 2 since a=2r i and t=2a=a=l1
IALB – Institute of Electrical Drives, Power Electronics and Components - 13 - Analysing the curve of UL in UL1 and UL2, a line function without sinusoidal function part is developed. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = + + + cos3 + ... 3 sin3 cos2 2 sin 2 cos 1 sin 2 2 ( ) 1 1 1 1 1 α α α α α α α π α D L U U ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − + + + cos3 +... 3 sin 3 cos2 2 sin 2 cos 1 sin 2 2 ( ) 1 1 1 1 2 β β β β β β β π β D L U U since, ⎭ ⎬ ⎫ + = = + α β π α β π 1 1 ⎥ ⎦ ⎤ − + − ⎢ ⎣ ⎡ − + − − + ( − + − = − cos3( ) ... 3 sin 3( ) cos 2( ) 2 sin 2( ) cos( ) 1 sin ) 2 2 ( ) 1 1 1 1 2 α π π α α π π α α π π α π α π α D L U U ⎥ ⎦ ⎤ − + ⎢ ⎣ ⎡ + − + − + − = − ( cos3 ) ... 3 sin 3 cos 2 2 sin 2 ( cos ) 1 sin 2 2 ( ) 1 1 1 1 2 α α α α α π α α π α D L U U UU U LL L () () () ααα = + 1 2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + + − = cos3 ... 3 2sin 3 cos2 2 2sin 2 cos 1 2sin 2 2 2 ( ) 1 1 1 1 α α α α α α π α π α D L U U since 2 and 2 0 1 1 1 1 0 T t t T t α = π = α ⇒α = π
cal drives, Power Electronics and Components U0=20n.-n“+40|szn coS2丌 cOS 3·2丌 SIn vi Complex Root-Mean-Square Value(rms value) T Special case: 4,=1=v=13, 5, 7.since sin v=0 when v=2, 4, 6 Generally, Direct component To v∈N Hamornic frequencies are multiples of operating frequency fo: f=v f o R-L-Load 21=R+1joL→|=√R2+(oL)2 Ly R2+(2IvfoL) 2 Lower harmonic current is obtained by higher operating frequency fo. R<< i.wz for high fo and with loRi( Direct component of current
IALB – Institute of Electrical Drives, Power Electronics and Components - 14 - ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − = ⋅ cos 3 2 ... 3 sin 3 + cos 2 2 2 sin 2 sin cos 2 4 2 2 2 ( ) 0 0 0 0 0 0 1 ˆ 0 1 0 T T t t T T t t T t T T U t t U U t D U D L L π π π π π π π π π 1π4 24 4 34 ν νπ π ν ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⋅ 0 1 sin 4 ˆ T t u U D L ~ : Complex Root - Mean -Square Value (r.m.s. value) 0 when 2,4,6,... 2 1,3,5,7,... since sin 2 Special case : 0 1 ⎟ = = ⎠ ⎞ ⎜ ⎝ ⎛ = ⇒ = ν π ν ν T t Generally, ∈ Ν ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − ν 1 2 ˆ Direct component : 0 1 0 T t UL UD 0 0 0 1 Hamornic frequencies are multiples of operating frequency f : f T f =ν ⋅ =ν ⋅ ν R–L–Load: ~ ~ Z R jL Z R L ( ) L L =+ ⇒ = + ω ω 2 2 $ $ ( ) i u R fL L L ν ν πν = +2 0 2 2 Lower harmonic current is obtained by higher operating frequency f0. R T << L ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 0 π $ $ i u f L L L ν ν πν ≈ 2 0 for high f0 and with i U R i L L L 0 0 = = $ (Direct component of current.)