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DaMoos

Transistor biasing help

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Say I have a 2n3904 transistor. I bias it using the voltage divider method described in virtually every book I have.

The math all makes sense to me.

Here's where I'm lost. A specific Beta is used in all the calculations in all the examples I read, but usually the same examples say that the current gain of even different transistors of the same part number can be quite different.

So is the voltage divider method  intended to force a specific current gain?  Do you even really use the characteristics curves when biasing a transistor?

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The datasheet for a 2N3904 transistor says that its current gain can be anywhere from 100 to 300 when its callector current is 10mA and its temperature is 25 degrees C.
Many 2N3904 transistors have different current gains and it changes with collector current changes and with temperature changes.

A voltage divider can be used to bias the base only if an emitter resistor is also used to provide DC negative feedback which cancels many differences in transistors and cancels most of the effect of temperature and collector current.

The curves on a datasheet show a "typical" transistor that you cannot buy. Most transistors are typical but many of them are not. 

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So the characteristics graph is really only academic, and useless for selecting specific resistors? Does this mean that ypu would never truley draw an actual loadline when designing a bjt amplifier?


So basically, the combination of the voltage divider on the base, and the emmiter resistor form a "cap" on current gain.  Say the actual transistor I have has a current gain of 250, but I want to make sure it realizes a current gain of only 100. is that the point of this form of biasing?

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I drew load lines 47 years ago when I was in university. Never again.

Pick an operating voltage for the collector so that the output will be symmetrical at full output, use an emitter resistor value that is 1/10th to 1/20th the collector resistor and use a divider current that is 10 times the transistor's base current when its gain is halfway from minimum to maximum. Then calculate the divider resistor values.
Then a transistor with a current gain of 100 or a gain of 300 will work almost the same. It also will be almost the same when the Vbe and temperature are different for each transistor.

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I hope I'm not being aggrivatingly dense here, but I just want to fully understand this stuff.

Does this form of biasing force a SPECIFIC current gain? I.E., i want a current gain of exactly 100, and I know the transistor can give me at least that. Does this form of biasing do this?

What is the reason for making the emmiter resistor 1/10 to 1/20 the size of the collector ressistor? I always read this, but the reason is never really explained. This seems like the actual value is almost arbitrary.

Same thing with making the current through the voltage divider being 10 times the current through the base.  Why is this?

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A transistor is rarely used for current gain. It is almost always used for voltage gain.
Without an emitter resistor a common-emitter transistor is a thermometer that is saturated when hot, when it has a high current gain and when its Vbe is low.
It is cutoff when the conditions are the opposite.

The divider with a current of 1/15th the transistor's base current sets a base voltage that barely changes when the gain of the transistor is different.
The emitter resistor reduces the effect of different Vbe (each transistor has a different voltage and temperature affects the voltage) and different current gains.

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Lower beta transistors used for power applications have low value emitter resistors. Since it is difficult to use low value base biasing resistors, because of loading, the base current may be 1/3 the current of the biasing resistors. So the voltage gain is lower than the collector resistor divided by the emitter resistor.

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