Buck-Boost (group 4)
advertisement
Power Electronic System Design I Winter 2010 Steven Trigno, Satya Nimmala, Romeen Rao Buck-Boost Converter Analysis iL - R Vc + Vg C L PWM iC Figure 1-Buck-boost circuit schematic implemented with practical switch When the transistor is turned ON, the diode is reverse-biased; therefore, not conducting (turned OFF) and the circuit schematic looks like as follows: 0 < t < DTs Transistor ON, Diode OFF + + Vg R Vc VL C L iC ig V _ iL Figure 2-Schematic of buck-boost converter when the switch is ON VL(t) = Vg – iL.RON ≈ Vg – iL.RON iC(t) = -v(t) / R ≈ −π π ig(t) = iL(t) ≈ IL When the transistor is OFF, the Diode is turned ON (DTs < t < Ts). The circuit is shown in fig 3: Transistor OFF, Diode ON + + Vg R Vc VL C L iC iL ig Figure 3 – Schematic Buck-boost converter when the switch is OFF VL(t) = -v(t) ≈ -V Ic(t) = ππΏ (π‘) − π£(π‘) π π ≈ πΌπΏ − π Ig(t) = 0 volt.second balance: <VL(t)> = 0 = D(Vg - IL.RON) + D’(-V) charge balance: <ic(t)> = 0 = D(-V/R) + D’(IL – V/R) Average input current: <ig> = Ig = D(IL) + D’(0) Next, we construct the equivalent circuit for each loop equation: Inductor loop equation: V DVg - IL.DRon – D’V = <VL> = 0 _ IL DRon + _ DVg Capacitor node equation: + _ π·′ πΌπΏ − π D’V = < ππΆ > = 0 π + R V D’IL _ Input current (node) equation: Ig = D.IL _ D’IL Vg + Ig Then we draw the circuit models together as shown below: DRon IL _ Vg Ig + DIL + _ DVg D’V + _ + D’IL V _ 1:D transformer reversed polarity marks Model including ideal dc transformers: D’:1 transformer IL 1:D _ D’:1 Ig + R Vg + π« π½π π«′ πΉ+ π½= πΉ π½ π« π → = π« π½π π«′ π + π« πΉππ πΉ (π«′)π ππ (π«′)π πΉ π π« πΉ π+ + ππ πΉ (π«′)π IL = V _ And for the efficiency (η): η= DRon π½ π«′πΉ Ploss = π°ππ³ . π«πΉπΆπ΅ Discontinuous Conduction Mode in Buck-Boost Converter iL _ + Vg PWM R V L C + - Figure 4-Buck-boost converter During D1Ts: iL _ + Vg L R V C + - Figure 5-Transistor ON, Diode OFF VL = Vg During D2Ts: iL _ + Vg VL L R V + - C Figure 6 - Transistor OFF, Diode ON VL = -V During D3Ts: iL _ + Vg VL L R V + - C Figure 7 - Transistor OFF, Diode OFF VL = 0 Boundary between modes: CCM: π«π»ππ½π βπ = (βi = peak ripple in L) ππ³ βπ½ = π«π»ππ½ (βV = peak ripple in C) ππΉπͺ π½ IL = π«′πΉ (average inductor current) Boundary: π°π³ > βπ πππ πΆπΆπ π°π³ < βπ πππ π·πΆπ → π½ π«′πΉ ππ³ π« π½= → π«π»ππ½π > π«′ π« π½π (π«′)π πΉ ππ³ πΉπ»π π½π > in CCM π«π»ππ½π ππ³ > (π«′)π Buck Boost Convertor DTS D’TS KVL -Vg + VL = 0 -VL + V = 0 VL = Vg VL = V KCL ic = −V ic = iL − VV VR R Inductor volt sec balance: < VL > = DVg + D′ V = 0 ο° D′ V = -DVg V −D ο° = Vg 1−D Capacitor Charge Balance: < ic > = D( −V V ′ VR ) + D (i − VR ) = −D V ′ ′ V VR + D i L − D VR = D′ iL − D V ′ V VR −D VR = D′ iL − V (D VR + D′) = D′ iL − V VR = 0 iL = 1 1−D V ( ) R State Space Analysis DTs: Apply KVL to fig.1. −ππ + πΏ ππ =0 ππ‘ ππ ππ = ππ‘ πΏ Apply KCL to fig.1. πΆ ππ π + =0 ππ‘ π ππ£ π =− ππ‘ π πΆ π π [ ] ππ‘ π£ π π [ ] ππ‘ π£ 1 0 −πΏ π − π πΆ = [1 1 1 π ] [ ] + [ πΏ ] ππ π£ 0 π = π΄1 [ ] + π΅1 ππ π£ → (1) → (2) By comparing the equation’s (1) and (2) we get values of π΄1 and π΅1 0 π΄1 = [ 1 π 1 πΏ ] 1 − π πΆ − 1 π΅1 = [πΏ] 0 D’Ts: Apply KVL to fig.2. −πΏ ππ +π = 0 ππ‘ ππ π = ππ‘ πΏ Apply KCL to fig.2. π+πΆ ππ£ π + =0 ππ‘ π ππ£ π π =− − ππ‘ πΆ π πΆ π π [ ] ππ‘ π£ π π [ ] ππ‘ π£ 0 =[ 1 −πΆ 1 πΏ π 1 ][ ] + − π πΆ π π = π΄1 [ ] + π΅1 ππ π£ 0 [ ] ππ → (3) 0 → (4) By comparing the equation’s (3) and (4) we get values of π΄2 and π΅2 0 π΄2 = [ 1 − πΆ 1 πΏ 0 π΅2 = [ ] 0 1] − π πΆ π΄ = π·π΄1 + π· ′ π΄2 0 π΄=[ π·′ − πΆ π·′ πΏ ] 1 − π πΆ π΅ = π·π΅1 + π· ′ π΅2 π· π΅ = [πΏ ] 0 π = −π΄‾¹-----π΅ππ 0 πΌ π = [ ] = −[ π·′ π − πΆ π·′ πΏ ] ‾1 1 − π πΆ π·ππ ′2 πΌ π=[ ]=[ π· π ] π·ππ π − ′ π· πΌ= π·ππ π· ′2 π π=− π·ππ π·′ π· [ πΏ ] ππ 0 Calculating inductor current ripple and capacitor voltage ripple: We know that the following formula for calculating the ripple values βπ = (π΄1π + π΅1ππ)π·ππ π·ππππ βπΌ πΏ βπ = [ ] = [ 2 ] π· ππππ βπ π· ′ π πΆ Inductor current ripple is βπΌ = π·ππππ πΏ Capacitor voltage ripple is π· 2 ππππ βπ = π· ′ π πΆ
