Thermal Shutdown
(Continued)
heating. When the junction temperature reduces below
155C (typical hysteresis = 20C), the Softstart pin is re-
leased and normal operation resumes.
Applications Information
EXTERNAL COMPONENTS
The procedure for calculating the external components is
illustrated with a design example. The circuit in
Figure 1
is to
be configured for the following specifications:
V
OUT
= 10V
V
IN
= 15V to 75V
F
S
= 625 kHz
Minimum load current = 150 mA
Maximum load current = 1.0A
Softstart time = 5 ms.
R1 and R2:
The ratio of these resistors is calculated from:
R1/R2 = (V
OUT
/2.5V) - 1
R1/R2 calculates to 3.0. The resistors should be chosen
from standard value resistors in the range of 1.0 k
- 10 k
.
Values of 3.0 k
for R1, and 1.0 k
for R2 will be used.
(7)
R
ON
, F
S
:
R
sets the on-time, and can be chosen using
Equation 2 to set a nominal frequency, or from Equation 5 if
the on-time at a particular V
is important. A higher fre-
quency generally means a smaller inductor and capacitors
(value, size and cost), but higher switching losses. A lower
frequency means a higher efficiency, but with larger compo-
nents. If PC board space is tight, a higher frequency is better.
The resulting on-time and frequency have a
±
25% toler-
ance. Re-arranging Equation 2 ,
The next larger standard value (137 k
) is chosen for R
ON
,
yielding a nominal frequency of 618 kHz.
L1:
The inductor value is determined based on the load
current, ripple current, and the minimum and maximum input
voltage (V
IN(min)
, V
IN(max)
). Refer to
Figure 13
.
To keep the circuit in continuous conduction mode, the maxi-
mum allowed ripple current is twice the minimum load cur-
rent, or 300 mAp-p. Using this value of ripple current, the
inductor (L1) is calculated using the following:
(8)
where F
S(min)
is the minimum frequency (F
S
- 25%).
This provides a minimum value for L1 - the next higher
standard value (100 μH) will be used. L1 must be rated for
the peak current (I
PK+
) to prevent saturation. The peak cur-
rent occurs at maximum load current with maximum ripple.
The maximum ripple is calculated by re-arranging Equation
8 using V
, F
, and the minimum inductor value,
based on the manufacturer’s tolerance. Assume, for this
exercise, the inductor’s tolerance is
±
20%.
(9)
I
PK+
= 1.0A + 0.234A / 2 = 1.117A
R
CL
:
Since it is obvious that the lower peak of the inductor
current waveform does not exceed 1.0A at maximum load
current (see
Figure 13
), it is not necessary to increase the
current limit threshold. Therefore R
is not needed for this
exercise. For applications where the lower peak exceeds
1.0A, see the section below on increasing the current limit
threshold.
C2 and R3:
Since the LM5010 requires a minimum of 25
mVp-p of ripple at the FB pin for proper operation, the
required ripple at V
OUT1
is increased by R1 and R2. This
necessary ripple is created by the inductor ripple current
acting on C2’s ESR + R3. First, the minimum ripple current is
determined.
20119922
FIGURE 13. Inductor Current
L
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