Power Electronic System Design I Winter 2010
Steven Trigno, Satya Nimmala, Romeen Rao
Buck-Boost Converter Analysis
PWMVg
LC
VcR
iL
+
-
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
Vg
LC
VcR
iL
+
-
iC
Transistor ON, Diode OFF
VL
+
_
V
ig
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 ≈ −𝑉
𝑅
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:
Vg
LC
VcR
iL
+
-
iC
Transistor OFF, Diode ON
VL
+
_
V
ig
Figure 3 – Schematic Buck-boost converter when the switch is OFF
VL(t) = -v(t) ≈ -V
Ic(t) = 𝑖𝐿(𝑡) −𝑣(𝑡)
𝑅 ≈ 𝐼𝐿 −
𝑉
𝑅
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: DVg - IL.DRon – D’V = <VL> = 0
+
_ +
_DVg D’V
DRon
IL
Capacitor node equation: 𝐷′ 𝐼𝐿 − 𝑉
𝑅= < 𝑖𝐶 > = 0
D’IL
R
+
_
V
Input current (node) equation: Ig = D.IL
D’IL
+
_
Vg
Ig
Then we draw the circuit models together as shown below:
DIL
+
_
Vg
Ig
DVg D’V D’IL
+
+
_
_
V
IL
+_
1:D transformer
reversed polarity marksD’:1 transformer
DRon
Model including ideal dc transformers:
+
_
Vg
Ig +
_
V
ILDRon
1:D D’:1
R
𝑽 = 𝑫
𝑫′𝑽𝒈
𝑹
𝑹 + 𝑫
(𝑫′)𝟐𝑹𝒐𝒏
→ 𝑽
𝑽𝒈=
𝑫
𝑫′
𝟏
𝟏 +𝑫
(𝑫′)𝟐𝑹𝒐𝒏𝑹
And for the efficiency (η):
η = 𝟏
𝟏+𝑫
(𝑫′ )𝟐+
𝑹𝒐𝒏𝑹
IL = 𝑽
𝑫′𝑹
Ploss = 𝑰𝑳𝟐. 𝑫𝑹𝑶𝑵
Discontinuous Conduction Mode in Buck-Boost Converter
PWM
Vg
L
+
_
C
VR
iL
+
-
Figure 4-Buck-boost converter
During D1Ts:
Vg
L
+
_
C
VR
iL
+
-
Figure 5-Transistor ON, Diode OFF
VL = Vg
During D2Ts:
Vg
L
+
_
C
VR
iL
+
-
VL
Figure 6 - Transistor OFF, Diode ON
VL = -V
During D3Ts:
Vg
L
+
_
C
VR
iL
+
-
VL
Figure 7 - Transistor OFF, Diode OFF
VL = 0
Boundary between modes:
CCM:
∆𝒊 = 𝑫𝑻𝒔𝑽𝒈
𝟐𝑳 (∆i = peak ripple in L)
∆𝑽 =𝑫𝑻𝒔𝑽
𝟐𝑹𝑪 (∆V = peak ripple in C)
IL = 𝑽
𝑫′𝑹 (average inductor current)
Boundary:
𝑰𝑳 > ∆𝑖 𝑓𝑜𝑟 𝐶𝐶𝑀
𝑰𝑳 < ∆𝑖 𝑓𝑜𝑟 𝐷𝐶𝑀
→ 𝑽
𝑫′𝑹 >
𝑫𝑻𝒔𝑽𝒈
𝟐𝑳
𝑽 = 𝑫
𝑫′𝑽𝒈 in CCM
→ 𝑫
(𝑫′)𝟐𝑽𝒈
𝑹 >
𝑫𝑻𝒔𝑽𝒈
𝟐𝑳
𝟐𝑳
𝑹𝑻𝒔 > (𝑫′)𝟐
Buck Boost Convertor
DTS D’TS
KVL
-Vg + VL = 0 -VL + V = 0
VL = Vg VL = V
KCL
ic =
−V
VR ic = iL −
VVR
Inductor volt sec balance:
< VL > = DVg + D′V = 0
D′V = -DVg
V
Vg =
−D
1−D
Capacitor Charge Balance:
< ic > = D −VVR
+ D′ i −V
VR
= −DV
VR+ D′iL − D′ V
VR
= D′ iL − DV
VR−D′ V
VR
= D′ iL −V
VR(D + D′)
= D′ iL − V
VR = 0
iL = 1
1−D
V
R
State Space Analysis
DTs:
Apply KVL to fig.1.
−𝑉𝑔 + 𝐿𝑑𝑖
𝑑𝑡= 0
𝑑𝑖
𝑑𝑡=
𝑉𝑔
𝐿
Apply KCL to fig.1.
𝐶𝑑𝑖
𝑑𝑡+
𝑉
𝑅= 0
𝑑𝑣
𝑑𝑡= −
𝑉
𝑅𝐶
𝑑
𝑑𝑡 𝑖𝑣 =
0 −1
𝐿1
𝑐−
1
𝑅𝐶
𝑖𝑣 +
1
𝐿
0 𝑉𝑔 → (1)
𝑑
𝑑𝑡 𝑖𝑣 = 𝐴1
𝑖𝑣 + 𝐵1 𝑉𝑔 → (2)
By comparing the equation’s (1) and (2) we get values of 𝐴1 and 𝐵1
𝐴1 = 0 −
1
𝐿1
𝑐−
1
𝑅𝐶
𝐵1 = 1
𝐿0
D’Ts:
Apply KVL to fig.2.
−𝐿𝑑𝑖
𝑑𝑡+ 𝑉 = 0
𝑑𝑖
𝑑𝑡=
𝑉
𝐿
Apply KCL to fig.2.
𝑖 + 𝐶𝑑𝑣
𝑑𝑡+
𝑉
𝑅= 0
𝑑𝑣
𝑑𝑡= −
𝑖
𝐶−
𝑉
𝑅𝐶
𝑑
𝑑𝑡 𝑖𝑣 =
01
𝐿
−1
𝐶−
1
𝑅𝐶
𝑖𝑉 +
00 𝑉𝑔 → (3)
𝑑
𝑑𝑡 𝑖𝑣 = 𝐴1
𝑖𝑣 + 𝐵1 𝑉𝑔 → (4)
By comparing the equation’s (3) and (4) we get values of 𝐴2 and 𝐵2
𝐴2 = 0
1
𝐿
−1
𝐶−
1
𝑅𝐶
𝐵2 = 00
𝐴 = 𝐷𝐴1 + 𝐷′𝐴2
𝐴 = 0
𝐷′
𝐿
−𝐷′
𝐶−
1
𝑅𝐶
𝐵 = 𝐷𝐵1 + 𝐷′𝐵2
𝐵 = 𝐷
𝐿0
𝑋 = −𝐴‾¹-----𝐵𝑉𝑔
𝑋 = 𝐼𝑉 = −
0𝐷′
𝐿
−𝐷′
𝐶−
1
𝑅𝐶
‾1 𝐷
𝐿0 𝑉𝑔
𝑋 = 𝐼𝑉 =
𝐷𝑉𝑔
𝐷′2𝑅
−𝐷𝑉𝑔
𝐷′
𝐼 =𝐷𝑉𝑔
𝐷′2𝑅
𝑉 = −𝐷𝑉𝑔
𝐷′
Calculating inductor current ripple and capacitor voltage ripple:
We know that the following formula for calculating the ripple values
∆𝑋 = (𝐴1𝑋 + 𝐵1𝑉𝑔)𝐷𝑇𝑠
∆𝑋 = ∆𝐼∆𝑉
=
𝐷𝑉𝑔𝑇𝑠
𝐿𝐷2𝑉𝑔𝑇𝑠
𝐷′𝑅𝐶
Inductor current ripple is
∆𝐼 =𝐷𝑉𝑔𝑇𝑠
𝐿
Capacitor voltage ripple is
∆𝑉 =𝐷2𝑉𝑔𝑇𝑠
𝐷′𝑅𝐶
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