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参数资料
| 型号： | LTC3632IMS8E#TRPBF |
| 厂商： | LINEAR INTEGRATED SYSTEMS INC |
| 元件分类： | 稳压器 |
| 英文描述： | 0.06 A SWITCHING REGULATOR, PDSO8 |
| 封装： | LEAD FREE, PLASTIC, MSOP-8 |
| 文件页数： | 7/20页 |
| 文件大小： | 247K |
| 代理商： | LTC3632IMS8E#TRPBF |
LTC3632
15
3632f
APPLICATIONS INFORMATION
Other losses, including CIN and COUT ESR dissipative
losses and inductor core losses, generally account for
less than 2% of the total power loss.
Thermal Considerations
The LTC3632 does not dissipate much heat due to its high
efciency 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 specications: VIN = 24V,
VOUT = 3.3V, IOUT = 20mA, f = 250kHz. Furthermore, as-
sume for this example that switching should start when
VIN is greater than 12V and should stop when VIN is less
than 8V.
First, calculate the inductor value that gives the required
switching frequency:
L
V
kHz
mA
V
=
33
250
50
1
33
24
220
.
–
.
μ
μH
Next, verify that this value meets the LMIN requirement.
For this input voltage and peak current, the minimum
inductor value is:
L
Vns
mA
μH
MIN =
24
100
50
48
Therefore, the minimum inductor requirement is satised,
and the 220μH inductor value may be used.
Next, CIN and COUT are selected. For this design, CIN should
be size for a current rating of at least:
ImA
V
mA
RMS
=
20
33
24
33
17
.
.
–
Due to the low peak current of the LTC3632, decoupling
the VIN supply with a 1μF capacitor is adequate for most
applications.
COUT will be selected based on the ESR that is required to
satisfy the output voltage ripple requirement. For a 50mV
output ripple, the value of the output capacitor ESR can
be calculated from:
ΔVOUT = 50mV ≤ 50mA ESR
A capacitor with a 1Ω ESR satises this requirement. A 10μF
ceramic capacitor has signicantly 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:
R
V
Rk
OUT
1
08
1
2 750
=
=
.
–
The undervoltage lockout requirement on VIN can be
satised with a resistive divider from VIN to the RUN and
HYST pins. Choose R1 = 2M and calculate R2 and R3 as
follows:
R
V
VV
Rk
R
IN RISING
2
121
1 224
3
1
=
=
.
–.
()
..
–.
–
.
()
1
11
12 90 8
V
VV
RR
k
IN FALLING
=
Choose standard values for R2 = 226k and R3 = 91k.
The ISET pin should be left open in this example to select
maximum peak current (50mA). Figure 9 shows a complete
schematic for this design example.
VIN
LTC3632
RUN
2M
1μF
226k
91k
HYST
3632 F09
SW
VIN
24V
VOUT
3.3V
20mA
ISET
SS
VFB
GND
750k
10μF
220μH
240k
Figure 9. 24V to 3.3V, 20mA Regulator at 250kHz
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