ECE 320:
Energy Systems I
Session 39; Page 1/5
Fall 2003
ECE 320: Lecture 39
Notes
Single Switch DC-DC Converter Configurations:
1. Buck Converter (or down converter). The output voltage is less than or equal to the input volta
S
+
L
D
V
C
Vd
o
-
Vo
Vd
= D
Io
Id
=
1
D
2. Boost Converter (up converter). The output voltage larger than the input voltage. The inductor
current is on the input, and it is pumped up with the switch.
+
L
D
C
S
Vd
V
-
Ideal, steady-state continuous conduction:
Vo
1
=
Vd
1−D
Io
Id
= 1−D
o
ECE 320:
Energy Systems I
Session 39; Page 2/5
Fall 2003
3. Buck-Boost Converter (up/down converter). The output voltage can be smaller or larger than
the input voltage. The inductor current is now an intermediate stage. Note output voltage polarity
S
-
D
L
V
C
Vd
o
+
Ideal, steady-state continuous conduction:
Vo
Vd
=
D
1−D
Io
Id
=
( 1 − D)
D
Polarity of the voltage and the current is reversed from the other two topologies.
Boundary of Discontinuous Conduction Through The Inductor:
•
Boost Converter
When switch is closed, the slope of the inductor current is: :
∆iL
t
=
Vd
L
When the switch is open, the slope is:
∆iL
t
=
Vd − Vo
L
Since the current follows these slopes while the switch position is maintained, in continuous
counduction the steady-state Current Ripple in inductor is:
∆iL =
Vd
( Vd − Vo)
⋅ D ⋅ Ts =
⋅ ( 1 − D ) ⋅ Ts
L
L
The average input current is now the average current through the inductor
ECE 320:
Energy Systems I
Session 39; Page 3/5
Fall 2003
Power Balance:
Po =
Vo
R
2
Vo
and
Vd
=
1
1−D
So:
Vd
Po =
2
2
( 1 − D) ⋅ R
But:
Po = Pi = Vd⋅ ILave
Vd
Therefore:
2
2
( 1 − D) ⋅ R
= Vd⋅ ILave
or
ILave =
Vd
2
( 1 − D) ⋅ R
Adding on the current ripple, the minimum current through the inductor is:
ILmin = ILave −
∆iL
2
Adding on the current ripple, the maximum current through the inductor is:
ILmax = ILave +
∆iL
2
At the boundary of discontinuous conduction we see:
ILmin = 0 =
Vd
2
( 1 − D) ⋅ R
Vd 
⋅ D ⋅ Ts
 2⋅ L 
− 
So, we can determine Lmin by solving this equation for L:
2
Lmin =
D ⋅ ( 1 − D ) ⋅ R ⋅ Ts
2
ECE 320:
Energy Systems I
Session 39; Page 4/5
Fall 2003
We can also state this in terms of voltages and currents using:
Vo
Io
R=
IoB = ( 1 − D) ⋅ Id = ( 1 − D) ⋅ ILB
2
Lmin =
D⋅ ( 1 − D) ⋅ Vo⋅ Ts
2⋅ IoB
=
D⋅ ( 1 − D) ⋅ Vo⋅ Ts
2⋅ ILB
or in terms of the output current at the boundary:
2
D⋅ ( 1 − D) ⋅ Vo⋅ Ts
IoB =
•
2⋅ Lmin
Buck-Boost Converter
When switch is closed, the slope of the inductor current is: :
∆iL
t
=
Vd
L
When the switch is open, the slope is:
∆iL
t
=
Vo
L
Since the current follows these slopes while the switch position is maintained, in continuous
counduction the steady-state Current Ripple in inductor is:
∆il =
Vd
( −Vo)
⋅ D ⋅ Ts =
⋅ ( 1 − D ) ⋅ Ts
L
L
Now the inductor current isn't the input current or the output current. It is only an intermediate step
•
At the boundary between continuous and discontinuous conduction, we will have:
ILB =
∆iL  1 
1
⋅ ILpeak =
= 
⋅ D⋅ Ts⋅ Vd
2
2
 2⋅ L 
ECE 320:
Energy Systems I
Session 39; Page 5/5
Fall 2003
or we could also write this in terms of Vo as:
1 
1 − D
1 
ILB = 
⋅ D⋅ Ts⋅ 
⋅ Vo = 
⋅ Ts⋅ ( 1 − D) ⋅ Vo
 2⋅ L 
 D 
 2⋅ L 
or
Lmin = 
1 
⋅ Ts⋅ ( 1 − D) ⋅ Vo
2
⋅
I
LB


However we usually want this in terms of the output current. If we write a node equation at the
terminal of the inductor
Io = IL − Id
or
D 
1 − D + D
Io
= Io 
=
1 − D
 1−D  1−D
IL = Io + Id = Io + Io⋅ 
1 

ILB( 1 − D) = 
⋅ Ts⋅ ( 1 − D) ⋅ Vo ⋅ ( 1 − D)
 2⋅ L 

Vo 
2
IoB = 
⋅ Ts⋅ ( 1 − D)
 2⋅ L 
Uncontrolled Rectifier Circuits (Chapter 5)
•
•
•
•
•
We have discussed the use of DC-DC converters to regulate a dc voltage to level different tha
the input dc voltage.
But we haven't discussed where this dc voltage comes from.
In most applications we start with the 120Vrms from a wall outlet
Now we want to convert that dc with a rectifier circuit
Next week we will combine the rectifier with a transformer and a dc-dc converter in one circu