参数资料
型号: LT1934IDCB-1#TRPBF
厂商: Linear Technology
文件页数: 9/20页
文件大小: 0K
描述: IC REG BUCK ADJ 60MA 6DFN
标准包装: 2,500
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 1.25 V ~ 28 V
输入电压: 3.2 V ~ 34 V
PWM 型: Burst Mode?
电流 - 输出: 60mA
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 6-WFDFN 裸露焊盘
包装: 带卷 (TR)
供应商设备封装: 6-DFN-EP(2x3)
LT1934/LT1934-1
APPLICATIONS INFORMATION
Table 2. Inductor Vendors
VENDOR
Murata
Sumida
PHONE
(404) 426-1300
(847) 956-0666
URL
www.murata.com
www.sumida.com
PART SERIES
LQH3C
CR43
COMMENTS
Small, Low Cost, 2mm Height
CDRH4D28
CDRH5D28
Coilcraft
(847) 639-6400
www.coilcraft.com
DO1607C
DO1608C
DT1608C
Würth
(866) 362-6673
www.we-online.com
WE-PD1, 2, 3, 4
Electronics
the average inductor current equals the load current, the
maximum load current is:
I OUT(MAX) = I PK – ΔI L /2
where I PK is the peak inductor current and ΔI L is the
peak-to-peak ripple current in the inductor. The ripple
current is determined by the off time, t OFF = 1.8μs, and
the inductor value:
ΔI L = (V OUT + V D ) ? t OFF /L
I PK is nominally equal to I LIM . However, there is a slight
delay in the control circuitry that results in a higher peak
current and a more accurate value is:
I PK = I LIM + 150ns ? (V IN – V OUT )/L
These expressions are combined to give the maximum
load current that the LT1934 will deliver:
I OUT(MAX) = 350mA + 150ns ? (V IN – V OUT )/L – 1.8μs
? (V OUT + V D )/2L (LT1934)
I OUT(MAX) = 90mA + 150ns ? (V IN – V OUT )/L – 1.8μs
? (V OUT + V D )/2L (LT1934-1)
The minimum current limit is used here to be conservative.
The third term is generally larger than the second term,
so that increasing the inductor value results in a higher
output current. This equation can be used to evaluate
a chosen inductor or it can be used to choose L for a
given maximum load current. The simple, single equation
rule given above for choosing L was found by setting
ΔI L = I LIM /2.5. This results in I OUT(MAX) ~0.8I LIM (ignoring
the delay term). Note that this analysis assumes that the
inductor current is continuous, which is true if the ripple
current is less than the peak current or ΔI L < I PK .
The inductor must carry the peak current without satu-
rating excessively. When an inductor carries too much
current, its core material can no longer generate ad-
ditional magnetic ?ux (it saturates) and the inductance
drops, sometimes very rapidly with increasing current.
This condition allows the inductor current to increase
at a very high rate, leading to high ripple current and
decreased overload protection.
Inductor vendors provide current ratings for power induc-
tors. These are based on either the saturation current or
on the RMS current that the inductor can carry without
dissipating too much power. In some cases it is not clear
which of these two determine the current rating. Some data
sheets are more thorough and show two current ratings,
one for saturation and one for dissipation. For LT1934 ap-
plications, the RMS current rating should be higher than
the load current, while the saturation current should be
higher than the peak inductor current calculated above.
Input Capacitor
Step-down regulators draw current from the input sup-
ply in pulses with very fast rise and fall times. The input
capacitor is required to reduce the resulting voltage ripple
at the LT1934 and to force this switching current into
a tight local loop, minimizing EMI. The input capacitor
must have low impedance at the switching frequency to
do this effectively. A 2.2μF ceramic capacitor (1μF for the
LT1934-1) satis?es these requirements.
If the input source impedance is high, a larger value ca-
pacitor may be required to keep input ripple low. In this
case, an electrolytic of 10μF or more in parallel with a 1μF
ceramic is a good combination. Be aware that the input
1934fe
9
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