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DTA143ECA

ETL
Part Number DTA143ECA
Manufacturer ETL
Description Digital transistors
Published Dec 20, 2006
Detailed Description www.DataSheet4U.com Digital transistors (built-in resistors) • Features 1) Built-in bias resistors enable the configur...
Datasheet PDF File DTA143ECA PDF File

DTA143ECA
DTA143ECA


Overview
www.
DataSheet4U.
com Digital transistors (built-in resistors) • Features 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 thinfilm 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 device design easy.
• Structure PNP digital transistor (with built-in resistors) •Equivalent circuit (1) DTA143EKA DTA143ECA 2.
9 + 0.
2 1.
9+ 0.
2 0.
95+0.
95 (2) 2.
8+ 0.
2 + 0.
2 - 0.
1 1.
1 - + 0.
2 0.
1 0.
8 + 0.
1 0 ~ 0.
1 (3) 1.
6 IN R2 GND(+) IN OUT GND(+) 0.
4 + 0.
1 - 0.
05 0.
15 + 0.
1 - 0.
06 All terminals have same dimensions EIAJ: SC— 59 2.
9 + 0.
2 1.
9+ 0.
2 0.
95+0.
95 (1) (2) 2.
4+ 0.
2 + 0.
2 - 0.
1 0.
95 + 0.
2 - 0.
1 0.
45 + 0.
1 0 ~ 0.
1 (3) 0.
15 + 0.
1 - 0.
06 1.
3 0.
3 ~ 0.
6 R1 OUT (1) GND (2) IN (3) OUT DTA143EKA 0.
2Min + 0.
1 - 0.
05 0.
4 (1) GND (2) IN (3) OUT All terminals have same dimensions DTA143ECA EIAJ: SOT— 23 • Absolute maximum ratings(Ta=25 °C) Parameter symbol V cc V IN IO I C(Max.
) Pd Tj T stg limits –50 –30~+10 –100 –100 200 150 –55~+150 unit V V mA mW °C °C Supply voltage Input voltage Output current Power dissipation Junction temperature Storage temperature P8–1/3 DTA143EKA DTA143ECA • Elecrical characteristics(T =25°C) a Parameter Input voltage Output Voltage Input current Output current symbol V I(off) Min.
— –3 — — — 20 3.
29 0.
8 — Typ.
— — –0.
1 — — — 4.
7 1 250 Max.
–0.
5 — –0.
3 –1.
8 –0.
5 — 6.
11 1.
2 — Unit V V mA Conditions VCC= – 5V,IO= –100µA VO= – 0.
3V,IO= – 20mA I O/ I I = –10mA / –0.
5mA V I= – 5V V CC V I(on) V O(on) II I O(off) µA — KΩ — MHz = – 50V,V I = 0 V DC current gain GI Input resistance R1 Resistance ratio R 2/ R 1 Transition frequency fT *Transition frequency of the device V O= – 5V,I O= – 10mA — — V CE= – 10V,I E = 5 mA,f=100MH...



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