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# Using β when the BJT is saturated.

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#### alphacat

##### New Member
Hey,
I see here and also in the every day life that even when the BJT is saturated then when given an IC, and you want to determine the required IB in order to set RB, then people say:
"
Okay IC is 200mA.
β is 200 minimum by datasheet.
Therefore IB = 1mA, and RB equals (3.3V - VBE) / 1mA (for a 3.3V port).
"

But when the transistor is saturated, then there's no current amplification, and the equation IC = β*IB doesnt hold anymore since you can't neglect IE anymore (the Base-Collector junction is forward biased).

So what is the right way to determine the required IB (in order to set RB), when IC is given and the transistor is saturated (acting as a switch)?

Thank you.

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Often a saturated beta is given in the data sheet.

If it's not given, I'll use 10 to 20 for big BJTs or 200 to 1000 for big Darlingtons. I sometimes use the HFE/20 for small BJTs.

note: Sometimes saturation doesn't matter, and in fact you might not want the speed penalty. It's seen by some engineers as the Holy Grail of a closed switch when all you really wanted was enough power to the load and a switch that isn't too hot.

Last edited:
Hey, thanks for helping out.

In saturation, according to ebers moll equations,
IC = βF * IB - Is/αR * e(VBC/Vth).

so if βF is 200 minumum, and IC is 200mA, and you set RB so IB will be 1mA, then you can still receive IC which is lower than 200mA since you didnt take into account the factor - Is/αR * e(VBC/Vth)

Isnt it a problem?

Sorry, I expected this was about practical design. Maybe you'll have better luck in the Theory forum.

In theory, you probably have a valid point. In practice, all I ever want is enough power to the load and a switch that isn't too hot.

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