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Zeta converter DCM - Exam desgin of Z converter - M(D,K) ratio
Typology: Exams
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(^1) Institución Universitaria ITM
I. Análisis del convertidor en CCM
Fig. 1. Convertidor Z
Fig. 2. Subcircuito del convertidor Z - switch ON
Fig. 3. Subcircuito del convertidor Z - switch OFF
(Switch ON)
− V S + V L 1 = 0 ; V L 1 = V S
− V L 1 − V C 1 + V L 2 + V o = 0 ; V L 2 = V S + V C 1 − V o
− i C 1 − i L 2 = 0 ; i C 1 = − i L 2
i L 2 − i C 2 − V o R = 0 ; i C 2 = i L 2 − V o R
(Switch OFF)
− V L 1 − V C 1 = 0 ; V L 1 = − V C 1
V L 2 + V o = 0 ; V L 2 = − V o
i C 1 − i L 1 = 0 ; i C 1 = i L 1
i L 2 − i C 2 − V o R = 0 ; i C 2 = i L 2 − V o R
II. Ecuaciones de balance
(L1)
< V L 1 > = DT s ( V S ) + D ′ T s (− V C 1 ) = 0
V C 1 =
D′^ V S
(L2)
< V L 2 > = DT s ( V S + V C 1 − V o) + D ′ T s (− V o ) = 0
DV S + DV C 1 − V o = 0 ; V o = D ( V S + V C 1 )
V o = D
( V S +
V S
) = D
([ V S
( 1 +
)])
V o = D
([ V S
( D ′ +D D′
)]) = D
([ V S
( (1−D)+D D′
)])
V o = V C 2 =
V S
(C2)
< i C 2 > = DT s
( i L 2 − V o R
)
T s
( i L 2 − V o R
) = 0
i L 2 = V o R =
·
(C1)
< i C 1 > = DT s (− i L 2 ) + D
T s ( i L 1 ) = 0
D ′ i L 1 = Di L 2
i L 1 = D D′^ i L 2 =
( D D′
) 2 V S R
III. Rizados de corriente - K, Kcritico y
Rcritico
(Rizado de corriente L1) V s L 1 =^
2∆i 1 DT s
∆ i 1 =
(V s )(DT s ) 2 L 1
(Rizado de corriente L2) V s +V C 1 −V o L 2 = 2∆i 2 DT s
Accionamiento y electronica de potencia | HKχ | July 19, 2021 | 1–
Como: V C 1 =
V s y V C 2 = V o =
V s
V s + D D ′^ V s − D D ′^ V s L 2 =^
2∆i 2 DT s
∆ i 2 =
(V s )(DT s ) 2 L 2
(K,Kcritico y Rcritico)
I 1 + I 2 < ∆ i 1 + ∆ i 2
( D D′
) 2 V S R
<
(V s )(DT s ) 2 L 1
(V s )(DT s ) 2 L 2
V S R
(( D D′
) 2
) < D
( (V s )(T s ) 2 L 1 +^
(V s )(T s ) 2 L 2
)
( D^2 +DD′ D′^2
) < D
( 2 V s T s (L 1 +L 2 ) 4(L 1 )(L 2 )
)
(R)(T s )(L 1 +L 2 ) < D
Por tanto:
K =
(R)(T s )(L 1 +L 2 )
K Critico ( D ) = D ′^2 y R Critico ( D ) =
(D′^2 )(T s )(L 1 +L 2 )
IV. Análisis en DCM
(Switch ON/OFF)
Fig. 4. Subcircuito del convertidor Z - switch ON/OFF
(L1)
D 1 T s ( V s) + D 2 T s (− V C 1 ) + D 3 T s (0) = 0
V C 1 =
V s
D 1 T s ( V s + V C 1 − V O) + D 2 T s (− V o) + D 3 T s (0) = 0
D 1 ( V s + V C 1 ) − V o ( D 1 + D 2 ) = 0
V o =
( V s + D 1 D 2 V s
)
(D 1 +D 2 ) =^
D 1 V s
( 1+ D 1 D 2
)
(D 1 +D 2 ) =^
D 1 V s
( (^) D 1 + D 2 D 2
)
V o =
D 1 V s (D 1 +D 2 ) D 2 (D 1 +D 2 ) =^
D 2 V s
V. Gráficas DCM
Fig. 5. Gráfico - Voltaje en los inductores
Fig. 6. Gráfico - Corriente de inductores y corriente en el diodo
VI. Ecuaciones en el nodo del diodo
Fig. 7. Convertidor Z - Corrientes en el nodo del diodo
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