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 ≈
−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:
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) = 𝑖𝐿 (𝑡) −
𝑣(𝑡)
𝑅
𝑉
≈ 𝐼𝐿 − 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:
V
DVg - IL.DRon – D’V = <VL> = 0
_
IL
DRon
+
_
DVg
Capacitor node equation:
+
_
𝐷′ 𝐼𝐿 −
𝑉
D’V
= < 𝑖𝐶 > = 0
𝑅
+
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
+
𝑫
𝑽𝒈
𝑫′
𝑹+
𝑽=
𝑹
𝑽
𝑫
𝟏
→
=
𝑫
𝑽𝒈
𝑫′ 𝟏 + 𝑫 𝑹𝒐𝒏
𝑹
(𝑫′)𝟐 𝒐𝒏
(𝑫′)𝟐 𝑹
𝟏
𝑫
𝑹
𝟏+
+ 𝒐𝒏
𝑹
(𝑫′)𝟐
IL =
V
_
And for the efficiency (η):
η=
DRon
𝑽
𝑫′𝑹
Ploss = 𝑰𝟐𝑳 . 𝑫𝑹𝑶𝑵
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:
𝑫𝑻𝒔𝑽𝒈
∆𝒊 =
(∆i = peak ripple in L)
𝟐𝑳
∆𝑽 =
𝑫𝑻𝒔𝑽
(∆V = peak ripple in C)
𝟐𝑹𝑪
𝑽
IL = 𝑫′𝑹
(average inductor current)
Boundary:
𝑰𝑳 > ∆𝑖 𝑓𝑜𝑟 𝐶𝐶𝑀
𝑰𝑳 < ∆𝑖 𝑓𝑜𝑟 𝐷𝐶𝑀
→
𝑽
𝑫′𝑹
𝟐𝑳
𝑫
𝑽=
→
𝑫𝑻𝒔𝑽𝒈
>
𝑫′
𝑫
𝑽𝒈
(𝑫′)𝟐
𝑹
𝟐𝑳
𝑹𝑻𝒔
𝑽𝒈
>
in CCM
𝑫𝑻𝒔𝑽𝒈
𝟐𝑳
> (𝑫′)𝟐
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
 D′ V = -DVg
V
−D

=
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.
−𝑉𝑔 + 𝐿
𝑑𝑖
=0
𝑑𝑡
𝑑𝑖 𝑉𝑔
=
𝑑𝑡
𝐿
Apply KCL to fig.1.
𝐶
𝑑𝑖 𝑉
+ =0
𝑑𝑡 𝑅
𝑑𝑣
𝑉
=−
𝑑𝑡
𝑅𝐶
𝑑 𝑖
[ ]
𝑑𝑡 𝑣
𝑑 𝑖
[ ]
𝑑𝑡 𝑣
1
0
−𝐿
𝑐
− 𝑅𝐶
= [1
1
1
𝑖
] [ ] + [ 𝐿 ] 𝑉𝑔
𝑣
0
𝑖
= 𝐴1 [ ] + 𝐵1 𝑉𝑔
𝑣
→ (1)
→ (2)
By comparing the equation’s (1) and (2) we get values of 𝐴1 and 𝐵1
0
𝐴1 = [
1
𝑐
1
𝐿 ]
1
−
𝑅𝐶
−
1
𝐵1 = [𝐿]
0
D’Ts:
Apply KVL to fig.2.
−𝐿
𝑑𝑖
+𝑉 = 0
𝑑𝑡
𝑑𝑖 𝑉
=
𝑑𝑡 𝐿
Apply KCL to fig.2.
𝑖+𝐶
𝑑𝑣 𝑉
+ =0
𝑑𝑡 𝑅
𝑑𝑣
𝑖
𝑉
=− −
𝑑𝑡
𝐶 𝑅𝐶
𝑑 𝑖
[ ]
𝑑𝑡 𝑣
𝑑 𝑖
[ ]
𝑑𝑡 𝑣
0
=[ 1
−𝐶
1
𝐿
𝑖
1 ][ ] +
− 𝑅𝐶 𝑉
𝑖
= 𝐴1 [ ] + 𝐵1 𝑉𝑔
𝑣
0
[ ] 𝑉𝑔 → (3)
0
→ (4)
By comparing the equation’s (3) and (4) we get values of 𝐴2 and 𝐵2
0
𝐴2 = [ 1
−
𝐶
1
𝐿
0
𝐵2 = [ ]
0
1]
−
𝑅𝐶
𝐴 = 𝐷𝐴1 + 𝐷 ′ 𝐴2
0
𝐴=[
𝐷′
−
𝐶
𝐷′
𝐿 ]
1
−
𝑅𝐶
𝐵 = 𝐷𝐵1 + 𝐷 ′ 𝐵2
𝐷
𝐵 = [𝐿 ]
0
𝑋 = −𝐴‾¹-----𝐵𝑉𝑔
0
𝐼
𝑋 = [ ] = −[
𝐷′
𝑉
−
𝐶
𝐷′
𝐿 ] ‾1
1
−
𝑅𝐶
𝐷𝑉𝑔
′2
𝐼
𝑋=[ ]=[ 𝐷 𝑅 ]
𝐷𝑉𝑔
𝑉
− ′
𝐷
𝐼=
𝐷𝑉𝑔
𝐷 ′2 𝑅
𝑉=−
𝐷𝑉𝑔
𝐷′
𝐷
[ 𝐿 ] 𝑉𝑔
0
Calculating inductor current ripple and capacitor voltage ripple:
We know that the following formula for calculating the ripple values
∆𝑋 = (𝐴1𝑋 + 𝐵1𝑉𝑔)𝐷𝑇𝑠
𝐷𝑉𝑔𝑇𝑠
∆𝐼
𝐿
∆𝑋 = [ ] = [ 2
]
𝐷 𝑉𝑔𝑇𝑠
∆𝑉
𝐷 ′ 𝑅𝐶
Inductor current ripple is
∆𝐼 =
𝐷𝑉𝑔𝑇𝑠
𝐿
Capacitor voltage ripple is
𝐷 2 𝑉𝑔𝑇𝑠
∆𝑉 =
𝐷 ′ 𝑅𝐶