ML4895
V
C
t
ON
V
SR
Q (ONE SHOT)
INDUCTOR CURRENT (I )
L
Figure 4. One Shot and Synchronous Rectifier Timing
DESIGN CONSIDERATIONS
TON × V − VOUT
b
g
= F ×IOUT(MAX)
A typical design can be implemented by using the
following design procedure. Note that this procedure is
not intended to give final values, but to give a good
starting point, and provide the relationships necessary to
make trade-off decisions. Some experimentation will be
necessary to optimize values and to verify that the design
operates over worst case conditions.
IN
∆IL =
(4)
(5)
IRC
L
where FIRC = ratio of inductor ripple current to the
maximum output current, or:
TON × V − VOUT
b
g
IN
L =
0.5× IOUT(MAX)
DESIGN SPECIFICATIONS
Calculate the inductance using the volt-seconds value
It is important to start with a clear definition of the design
specifications. Make sure the specifications reflect worst
case conditions. Key specifications include the minimum
and maximum input voltage and the output voltage and
load current.
given in Figure 3 at the maximum input voltage. Choose
the nearest standard value, realizing the trade-offs
mentioned before. Then, using the inductance value
chosen, determine the actual inductor ripple current at
the maximum and minimum input voltage using Equation
4 and Figure 3.
INDUCTOR AND SENSE RESISTOR SELECTION
The sense resistor value can be determined using the
inductor ripple current value calculated above and
Equation 3 rearranged as follows:
Figure 5 shows the inductor current of the buck regulator.
The inductor current is made up of two components: the
DC current level set by the transconductance amplifier,
VSENSE(MIN)
RSENSE
=
I
, and the inductor ripple current, ∆I . The figure also
SENSE
shows that I
L
1
2
(6)
is the summation of I
and ½∆I .
IOUT(MAX)
−
∆IL(MIN)
OUT
SENSE L
TON × V − VOUT
b
g
(3)
VSENSE
1
IN
Having determined the values for the inductor and sense
resistor, we can now specify the inductor peak current
rating. This value is calculated at current limit and at the
maximum input voltage, and is given by:
IOUT =ISENSE
+
I =
2 L RSENSE
+
2× L
Therefore, the selection of the inductance value
determines how much of the output current is made up of
the ripple current. Higher inductor ripple current allows
smaller inductor values, but results in higher peak
currents, lower efficiency, and higher output voltage
ripple.
IL(PEAK(MAX)) = ISENSE(MAX) + ∆IL(MAX)
(7)
V
SENSE MAX
a f
ILPEAKMAX
+∆ILMAX
a f =
a f
c
h
R
SENSE
0.1V
Inductor ripple currents in the range of 30% to 70% of the
maximum output current are typical. As a good starting
point, set the inductor ripple current to 50% of the
maximum output current:
IL PEAK MAX
=
+ ∆IL MAX
a f
a f
c
h
RSENSE
6
MICRO-LINEAR [ MICRO LINEAR CORPORATION ]
