Pre-biased transistor DDTC144EUAQ-13-F turn on voltage

[ku637]

hello,

im planning to use a transistor with built in resistors say 47kOhm PN: DDTC144EUAQ-13-F for enabling an LDO output.

The idea is to add a pull up resistor (47k) to the collector pin to 3.8V and connnect the collector pin to LDO EN pin.Then control the base pin from MCU logic working in 1.8V.

The turn ON voltage of the transistor is specified as typical 1.4V and maximum as 3V. I can understand 1.4V because the resistor divider divides that by 2 and supply 0.7V to base of transistor to turn it on. But 3V maximum turn on specification im unable to understand. If i calculate the worst case resistor tolerance then also Im getting only 1.575V at the input to base pin to generate 0.7V at the Base to emitter voltage.

Or is it like the device will always turn on at 1.4V only the collector current will be different.

 

[00kam]

Hello
from data sheet "input voltage" pin 1 to pin 2 is( -10-->+40 volt) .then max. out put. current "Io" is 100mA.
"Vioff" voltage input off is 1.1 volt . and" Vion" voltage input on 1.4 typical or 3 volt max. under test condition vcc= 5 volt. (pin 2 to pin 3) only that is all about DDTC144EUAQ-13-F // Vioff make DDTC144EUAQ-13-F " cut off) whlie Vion makes DDTC144EUAQ-13-F (saturation = ON )

kamal

 

[ku637]

So does it mean i cant reliably drive this transistor to ON from a 1.8V digital I/O right?

 

[betwixt]

Correct! It would make more sense to use a conventional transistor with a base current limiting resistor rather than one with a potential divider.

Brian.

 

[00kam]

So does it mean i cant reliably drive this transistor to ON from a 1.8V digital I/O right?

why you can't as in the data sheet you can turn the DDTC144EUAQ-13-F from cutoff to ON state because
Vion" voltage input on 1.4 typical or 3 volt max. and 1.8 volt within the rang (1.4 volt-3 volt)
and you will get (Vo ON= pin 1 to pin 3=0.3 volt) and 2mA drive current . try to ignore minima value" 1.9 volt" .
because maximum "Ii" input current is 180uA if we applay 1.8 volt the current passes through
B-E junction will be (1.8 volt/47k+47k=19.2uA) THE major currnet passes through B-E
junction
kamal