13.12
SWITCH-MODE POWER SUPPLIES 2.3.3 Discontinuous current in the coil In this case the energy in the coil is completely discharged before the transistor switch recloses. Once again there is a dead time td (discontinuous current). Fig. 13-10b shows the associated waveforms. Closing the switch S results in a voltage across the coil, whereby I1 ∆i __ ___ Vi = vL = L . ∆t = L . δ.T or: L . I1 = δ . T . Vi
(13-17)
After the switch opens during the time interval (1 − δ ) . T − td almost all the energy in the coil is gone. The voltage across the coil is now V i and the current changes fromI1 to zero, so that: (0 − I1 ) _______________ Vi = vL = L . (1 − δ ) . T − →→ L . I1 = − Vo . [ (1 − δ ) . T − t ] d td ] [
(13-18)
From (13-17) and (13-18) it follows that: δ . T . Vi = − Vo . [ (1 − δ ) . T − td ] from which:
δ _________ Vo = − Vi . 1 − δ − t /T
(13-19)
d
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2.3.4 Remarks 1. 2. 3. 4.
From (13-16) and (13-19) it follows that the buck-boost converter results in voltage inversion. Depending on the value of δ we have | V o | ≤ or ≥ | V i | . This is therefore a step-up /stepdown converter. The maximum voltage across the (transistor) switch is V i + V o as can be seen in fig. 13-10 at the bottom . The maximum reverse voltage across the diode is V i + V o .
2.3.5 Numeric example 13-3: For a buck boost converter with continuous current mode we are given: input voltage: Vi = 3 to 15V max. output current: Io = 3A output voltage: Vo = 9V ± 0.1% chopper frequency: f = 100kHz
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Required: determine the values of L and C Solution: δmin Vo 9 _________ _________ From (13-16) →→ V o = Vi max . (1 − →→ δmin = V + = ______ = 0.375 . V δmin ) 15 + 9 o i max
Assumption ∆iL max = 0.2 x Io = 0.6A − 5 (1 − δ min ) . T __________________ 9 . (1 − 0.375) . 10 ____________ From (13-15) →→ L max = V o . = = 93.75µH ∆iL 0.6 ∆iL 0.6 ___________ ______________ From (13-7) →→ C min = 8 . f . ∆ = = 83.4µF vC max 8 . 10 5 . 9 . 10 − 3 ∆vo max _______ 0.1% . 9 _______ and: E SR max = ∆i = 0.6 = 0.015 Ω L