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LDTB114EWT1G

产品描述Small Signal Bipolar Transistor, 0.5A I(C), 1-Element, PNP, Silicon,
产品类别分立半导体    晶体管   
文件大小290KB,共3页
制造商LRC
官网地址http://www.lrc.cn
标准
下载文档 详细参数 全文预览

LDTB114EWT1G概述

Small Signal Bipolar Transistor, 0.5A I(C), 1-Element, PNP, Silicon,

LDTB114EWT1G规格参数

参数名称属性值
是否Rohs认证符合
厂商名称LRC
包装说明,
Reach Compliance Codeunknown
最大集电极电流 (IC)0.5 A
最小直流电流增益 (hFE)56
元件数量1
极性/信道类型PNP
最大功率耗散 (Abs)0.2 W
表面贴装YES
晶体管元件材料SILICON
Base Number Matches1

文档预览

下载PDF文档
LESHAN RADIO COMPANY, LTD.
Bias Resistor Transistor
PNP Silicon Surface Mount Transistor
with Monolithic Bias Resistor Network
Applications
Inverter, Interface, Driver
3
LDTB114EWT1G
Features
1
2
1) Built-in bias resistors enable the configuration of an
inverter circuit without connecting external input
resistors (see equivalent circuit).
2) The bias resistors consist of thin-film resistors with
complete isolation to allow positive biasing of the input.
They also have the advantage of almost completely
eliminating parasitic effects.
3) Only the on / off conditions need to be set for operation,
making the device design easy.
We declare that the material of product compliance with
RoHS requirements.
Absolute maximum ratings
(Ta=25°C)
Parameter
Symbol
Limits
Unit
SOT–323 (SC–70)
1
BASE
R1
R2
3
COLLECTOR
2
EMITTER
Supply voltage
Input voltage
Output current
Power dissipation
Junction temperature
Storage temperature
V
CC
V
IN
I
C
P
D
Tj
Tstg
−50
−40
to
+10
−500
200
150
−55
to
+150
V
V
mA
mW
C
C
DEVICE MARKING AND RESISTOR VALUES
Device
LDTB114EWT1G
LDTB114EWT3G
Marking
Q6
Q6
R1 (K)
10
10
R2 (K)
10
10
Shipping
3000/Tape & Reel
10000/Tape & Reel
Electrical characteristics
(Ta=25°C)
Parameter
Input voltage
Output voltage
Input current
Output current
DC current gain
Input resistance
Resistance ratio
Transition frequency
Characteristics of built-in transistor
Symbol
V
I(off)
V
I(on)
V
O(on)
I
I
I
O(off)
G
I
R
1
R
2
/R
1
f
T
Min.
−3
56
7
0.8
Typ.
−0.1
10
1
200
Max.
−0.5
−0.3
−0.88
−0.5
13
1.2
Unit
V
V
mA
µA
kΩ
MHz
Conditions
V
CC
=
−5V,
I
O
=
−100µA
V
O
=
−0.3V,
I
O
=
−10mA
I
O
/I
I
=
−50mA/−2.5mA
V
I
=
−5V
V
CC
=
−50V,
V
I
=0V
V
O
=
−5V,
I
O
=
−50mA
V
CE
=
−10V,
I
E
=50mA, f=100MHz
1/3

 
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