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LT1962EMS8-1.5 参数 Datasheet PDF下载

LT1962EMS8-1.5图片预览
型号: LT1962EMS8-1.5
PDF下载: 下载PDF文件 查看货源
内容描述: 300mA,低噪声,微功率LDO稳压器 [300mA, Low Noise, Micropower LDO Regulators]
分类和应用: 稳压器
文件页数/大小: 16 页 / 292 K
品牌: Linear [ Linear ]
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LT1962 Series  
U
U
U
PI FU CTIO S  
R
P
8
1
SHDN pin can be driven either by 5V logic or open-  
collector logic with a pull-up resistor. The pull-up resistor  
is required to supply the pull-up current of the open-  
collector gate, normally several microamperes, and the  
SHDN pin current, typically 1µA. If unused, the SHDN pin  
must be connected to VIN. The device will not function if  
the SHDN pin is not connected.  
IN  
OUT  
LT1962  
5
2
+
+
SHDN SENSE  
LOAD  
V
IN  
GND  
4
R
P
1962 F01  
NC (Pins 6, 7): No Connect. These pins are not internally  
connected. For improved power handling capabilities,  
these pins can be connected to the PC board.  
Figure 1. Kelvin Sense Connection  
pin.TheADJpinvoltageis1.22Vreferencedtogroundand  
the output voltage range is 1.22V to 20V.  
IN (Pin 8): Input. Power is supplied to the device through  
the IN pin. A bypass capacitor is required on this pin if the  
device is more than six inches away from the main input  
filter capacitor. In general, the output impedance of a  
battery rises with frequency, so it is advisable to include a  
bypass capacitor in battery-powered circuits. A bypass  
capacitor in the range of 1µF to 10µF is sufficient. The  
LT1962 regulators are designed to withstand reverse  
voltages on the IN pin with respect to ground and the OUT  
pin. In the case of a reverse input, which can happen if a  
battery is plugged in backwards, the device will act as if  
there is a diode in series with its input. There will be no  
reverse current flow into the regulator and no reverse  
voltage will appear at the load. The device will protect both  
itself and the load.  
BYP (Pin 3): Bypass. The BYP pin is used to bypass the  
reference of the LT1962 to achieve low noise performance  
from the regulator. The BYP pin is clamped internally to  
±0.6V (one VBE). A small capacitor from the output to this  
pin will bypass the reference to lower the output voltage  
noise. A maximum value of 0.01µF can be used for  
reducing output voltage noise to a typical 20µVRMS over a  
10Hz to 100kHz bandwidth. If not used, this pin must be  
left unconnected.  
GND (Pin 4): Ground.  
SHDN (Pin 5): Shutdown. The SHDN pin is used to put the  
LT1962 regulators into a low power shutdown state. The  
output will be off when the SHDN pin is pulled low. The  
W U U  
U
APPLICATIO S I FOR ATIO  
TheLT1962seriesare300mAlowdropoutregulatorswith  
micropowerquiescentcurrentandshutdown.Thedevices  
are capable of supplying 300mA at a dropout voltage of  
300mV. Output voltage noise can be lowered to 20µVRMS  
over a 10Hz to 100kHz bandwidth with the addition of a  
0.01µFreferencebypasscapacitor. Additionally, therefer-  
ence bypass capacitor will improve transient response of  
the regulator, lowering the settling time for transient load  
conditions. The low operating quiescent current (30µA)  
drops to less than 1µA in shutdown. In addition to the low  
quiescentcurrent, theLT1962regulatorsincorporatesev-  
eral protection features which make them ideal for use in  
battery-powered systems. The devices are protected  
against both reverse input and reverse output voltages. In  
battery backup applications where the output can be held  
up by a backup battery when the input is pulled to ground,  
the LT1962-X acts like it has a diode in series with its  
output and prevents reverse current flow. Additionally, in  
dual supply applications where the regulator load is re-  
turnedtoanegativesupply,theoutputcanbepulledbelow  
groundbyasmuchas20Vandstillallowthedevicetostart  
and operate.  
Adjustable Operation  
The adjustable version of the LT1962 has an output  
voltage range of 1.22V to 20V. The output voltage is set by  
theratiooftwoexternalresistorsasshowninFigure2.The  
device servos the output to maintain the ADJ pin voltage  
at 1.22V referenced to ground. The current in R1 is then  
equalto1.22V/R1andthecurrentinR2isthecurrentinR1  
11