Datasheet

Application information B5973D
34/44 DocID14117 Rev 4
8.6 Positive buck-boost regulator
The device can be used to implement a step-up/down converter with a positive output
voltage.
The output voltage is given by:
Equation 32
where the ideal duty cycle D for the buck boost converter is:
Equation 33
However, due to power losses in the passive elements, the real duty cycle is always higher
than this. The real value (that can be measured in the application) should be used in the
following formulas.
The peak current flowing in the embedded switch is:
Equation 34
while its average current is equal to:
Equation 35
This is due to the fact that the current flowing through the internal power switch is delivered
to the output only during the OFF phase.
The switch peak current must be lower than the minimum current limit of the overcurrent
protection (see Table 4 on page 6 for details) while the average current must be lower than
the rated DC current of the device.
As a consequence, the maximum output current is:
Equation 36
where I
SW MAX
represents the rated current of the device.
The current capability is reduced by the term (1-D) and so, for example, with a duty cycle of
0.5, and considering an average current through the switch of 2 A, the maximum output
current deliverable to the load is 1 A.
Figure 22 shows the schematic circuit of this topology for a 12 V output voltage and
5 V input.
V
OUT
V
IN
D
1D–
-------------=
D
V
OUT
V
IN
V
OUT
+
------------------------------=
I
SW
I
LOAD
1D–
---------------
I
RIPPLE
2
--------------------+
I
LOAD
1D–
---------------
V
IN
2L
-----------
D
f
SW
---------+==
I
SW
I
LOAD
1D–
---------------=
I
OUT MAX
I
SW MAX
1D–