NPN - linear or saturation region region for GPIO switching

[sridhar09.cherukuri@gmail.com]

I am using a general purpose BC846 NPN transistor collector to pull an IC'sEnable pin down. The base of BC846 is driven a source that can only source0.2mA of current. Since I only need to turn ON the transistor to pull the Enable pin down, base drive of 0.2mA is sufficient. Is 0.2mA of base current sufficient to fully turn ON the transistor or it will operate in the linear region? What will be the power dissipation if the transistor is operatedin linear region compared to saturation? The system may be in this state for several weeks. Are there any potential issues if the transistor is operated in this way.

thanks
srcherukuri

 

[bloggs.fredbloggs.fred@gmail.com]

I am using a general purpose BC846 NPN transistor collector to pull an IC's Enable pin down. The base of BC846 is driven a source that can only source 0.2mA of current. Since I only need to turn ON the transistor to pull the Enable pin down, base drive of 0.2mA is sufficient. Is 0.2mA of base current sufficient to fully turn ON the transistor or it will operate in the linear region? What will be the power dissipation if the transistor is operated in linear region compared to saturation? The system may be in this statefor several weeks. Are there any potential issues if the transistor is operated in this way.

What textbook are you using? Anyone who would think enough about those various issues would not be so ignorant of the answer.

 

[sridhar09.cherukuri@gmail.com]

Thanks. BC846 must sink about 50uA.

 

[sridhar09.cherukuri@gmail.com]

What's the pullup resistor value, to what voltage?



Shouldn't be a problem as long as the max collector current is a mA or

two. It will saturate to a tenth of a volt or less.









--



John Larkin Highland Technology, Inc



jlarkin at highlandtechnology dot com

http://www.highlandtechnology.com



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Thanks for the response. Pull up is 330K to 16V.

 

[sridhar09.cherukuri@gmail.com]

This depends on how much current the IC pin might source. I know of no

modern IC's that would source enough to cause problems, but if you're

using something oddball you should check.








That depends on what's hanging off of the collector, and on the

transistor. If you load presented by your pull-up resistor and the IC

pin sources less than 2mA when the collector voltage is 0.2V, then it

will almost certainly be saturated. It'll probably be saturated with

higher collector currents, but that depends on the transistor in question

-- check the data sheet.









Well, transistors, like all other physical objects, follow the laws of

physics. The power dissipation of the transistor will be equal to the

collector-emitter voltage drop times the collector current, plus the base-

emitter voltage drop times the base current. So the dissipation will

depend on the CE voltage drop and the collector current, and you haven't

given enough information for anyone to predict those values.



Check the allowable dissipation of the transistor, and compare that

dissipation with your calculated dissipation. If the calculated value is

less than the data sheet value by a healthy margin, you're OK.



--



Tim Wescott

Wescott Design Services

http://www.wescottdesign.com

The collector current is always limited by few tens of uA due to high valuepull up resistor and the IC pin internal limitation. Base drive is max 0.2mA. With this limitation of base current and collector current will the transistor ever reach saturation? If it doesn't, will operating it in leaner mode continuously for long periods cause any issues due to high Vce?

 

[josephkk]

Thanks. BC846 must sink about 50uA.

You are extremely unlikely to need any more base drive than collector
current. You could probably cut it the 1/10 of required collector
current.

?-)