LTC3631 Linear Technology Corporation, LTC3631 Datasheet - Page 15

• Manufacturer Part Number: LTC3631
• Description: High Voltage 100mA Synchronous Step-Down Converter
• Manufacturer: Linear Technology Corporation

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APPLICATIONS INFORMATION
Thermal Considerations
The LTC3631 does not dissipate much heat due to its high
effi ciency and low peak current level. Even in worst-case
conditions (high ambient temperature, maximum peak
current and high duty cycle), the junction temperature will
exceed ambient temperature by only a few degrees.
Design Example
As a design example, consider using the LTC3631 in an
application with the following specifi cations: V
V
sume for this example that switching should start when
V
than 8V.
First, calculate the inductor value that gives the required
switching frequency:
Next, verify that this value meets the L
For this input voltage and peak current, the minimum
inductor value is:
Therefore, the minimum inductor requirement is satisfi ed,
and the 47μH inductor value may be used.
Next, C
be sized for a current rating of at least:
Due to the low peak current of the LTC3631, decoupling
the V
applications.
OUT
IN
Other losses, including C
losses and inductor core losses, generally account for
less than 2% of the total power loss.
I
L =
L
RMS
is greater than 12V and should stop when V
MIN
IN
= 3.3V, I
IN
supply with a 1μF capacitor is adequate for most
250kHz • 225mA
=
=
and C
100
24
OUT
225
V
3.3V
mA
OUT
•
100
= 100mA, f = 250kHz. Furthermore, as-
mA
are selected. For this design, C
•
3 3
24
ns
.
V
V
≅
•
10
• 1–
IN
μH
3 3
24
.
and C
V
V
3.3V
24V
–
OUT
1 35
≅
MIN
ESR dissipative
47μH
mA
requirement.
RMS
IN
IN
IN
should
= 24V,
is less
C
satisfy the output voltage ripple requirement. For a 50mV
output ripple, the value of the output capacitor ESR can
be calculated from:
A capacitor with a 200mΩ ESR satisfi es this requirement.
A 10μF ceramic capacitor has signifi cantly less ESR than
200mΩ.
The output voltage can now be programmed by choosing
the values of R1 and R2. Choose R2 = 240k and calculate
R1 as:
The undervoltage lockout requirement on V
satisfi ed with a resistive divider from V
HYST pins. Choose R1 = 2M and calculate R2 and R3 as
follows:
Choose standard values for R2 = 226k and R3 = 91k. The
I
mum peak current (225mA). Figure 9 shows a complete
schematic for this design example.
SET
OUT
ΔV
R1=
R2 =
R3 =
pin should be left open in this example to select maxi-
24V
V
will be selected based on the ESR that is required to
OUT
IN
Figure 9. 24V to 3.3V, 100mA Regulator at 250kHz
0.8V
V
V
V
= 50mV ≤ 225mA • ESR
OUT
IN(RISING)
IN(FALLING)
1μF
– 1 • R2 = 750k
1.21V
1.1V
2M
226k
91k
– 1.21V
V
RUN
HYST
– 1.1V
IN
LTC3631
GND
I
V
SW
SET
SS
FB
• R1– R2 = 90.8k
• R1= 224k
47μH
IN
750k
240k
LTC3631
to the RUN and
3631 F09
10μF
IN
V
3.3V
100mA
OUT
15
can be
3631fa