NPN - linear or saturation region region for GPIO switching

On a sunny day (Wed, 26 Jun 2013 09:52:41 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :

Assuming you use some collector resistor, then for a supply voltage U volt and a resistor R Ohm, you need a minimum collector current of (U - .2) / R (assuming the input is CMOS and draws no current). Look up the beta of that transistor, take its minimum value, the required base current is then ((U - .2) / R) / beta_min.

Now multiply that value by 10 or so, to be on the safe side, and use that as base current.

For higher speeds different rules apply, lower resistors, base emitter charge etc.

Enable pin down.

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 state for several weeks. Are there any potential issues if the transistor is operated in this way.

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.

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.

On Wednesday, June 26, 2013 12:52:41 PM UTC-4, snipped-for-privacy@gmail.com wrote :

's Enable pin down. The base of BC846 is driven a source that can only sour ce 0.2mA of current. Since I only need to turn ON the transistor to pull th e Enable pin down, base drive of 0.2mA is sufficient. Is 0.2mA of base curr ent sufficient to fully turn ON the transistor or it will operate in the li near region? What will be the power dissipation if the transistor is operat ed in linear region compared to saturation? The system may be in this state for several weeks. Are there any potential issues if the transistor is ope rated in this way.

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

Enable pin down.

Thanks. BC846 must sink about 50uA.

C's Enable pin down. The base of BC846 is driven a source that can only sou rce 0.2mA of current. Since I only need to turn ON the transistor to pull t he Enable pin down, base drive of 0.2mA is sufficient. Is 0.2mA of base cur rent sufficient to fully turn ON the transistor or it will operate in the l inear region? What will be the power dissipation if the transistor is opera ted in linear region compared to saturation? The system may be in this stat e for several weeks. Are there any potential issues if the transistor is op erated in this way.

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

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

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 state for several weeks. Are there any potential issues if the transistor is operated in this way.

That's Ic = 50 uA, Ib = 200 uA, a forced beta of 0.25. No problem... that's way more base current than you actually need.

I like to use small mosfets for things like this, 2N7002s or FDV301s. That can save power and often eliminate a base resistor.

pull 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?

The transistor will be on (collector millivolts above ground, saturated) or off (picoamps of collector current, collector voltage +16). In neither case does it dissipate significant power.

an IC's Enable pin down.

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.

?-)