LTC3632 LINER [Linear Technology], LTC3632 Datasheet - Page 15

• Manufacturer Part Number: LTC3632
• Description: High Effi ciency, High Voltage 20mA Synchronous Step-Down Converter
• Manufacturer: LINER [Linear Technology]

APPLICATIONS INFORMATION
2. I
Thermal Considerations
The LTC3632 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 LTC3632 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:
OUT
IN
internal switches, R
switching, the average output current fl owing through
the inductor is “chopped” between the high side PMOS
switch and the low side NMOS switch. Thus, the series
resistance looking back into the switch pin is a function
of the top and bottom switch R
duty cycle (DC = V
The R
be obtained from the Typical Performance Characteris-
tics curves. Thus, to obtain the I
R
average output current:
Other losses, including C
losses and inductor core losses, generally account for
less than 2% of the total power loss.
L =
2
is greater than 12V and should stop when V
SW
R losses are calculated from the resistances of the
R
I
= 3.3V, I
2
SW
R Loss = I
to R
250kHz • 50mA
DS(ON)
= (R
L
and multiply the result by the square of the
3.3V
OUT
DS(ON)TOP
for both the top and bottom MOSFETs can
O
= 20mA, f = 250kHz. Furthermore, as-
2
(R
OUT
SW
SW
)DC + (R
, and external inductor R
/V
+ R
• 1–
IN
IN
) as follows:
L
)
and C
3.3V
24V
DS(ON)BOT
DS(ON)
2
R losses, simply add
OUT
220μH
ESR dissipative
values and the
)(1 – DC)
IN
IN
L
. When
= 24V,
is less
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 220μH inductor value may be used.
Next, C
be size for a current rating of at least:
Due to the low peak current of the LTC3632, decoupling
the V
applications.
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 1Ω ESR satisfi es this requirement. A 10μF
ceramic capacitor has signifi cantly less ESR than 1Ω.
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:
OUT
ΔV
I
L
R1=
R2 =
R3 =
RMS
MIN
IN
will be selected based on the ESR that is required to
OUT
IN
supply with a 1μF capacitor is adequate for most
=
=
and C
0.8V
V
V
V
= 50mV ≤ 50mA • ESR
20
OUT
24
IN(RISING)
IN(FALLING)
mA
V
50
OUT
– 1 • R2 = 750k
•
1.21V
1.1V
mA
100
•
are selected. For this design, C
3 3
24
.
– 1.21V
ns
V
– 1.1V
V
≅
•
48
3 3
24
μH
.
• R1– R2 = 90.8k
• R1= 224k
V
V
–
1 7
≅
IN
LTC3632
MIN
mA
to the RUN and
requirement.
RMS
IN
IN
15
can be
should
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