well, we all know that in transistors beta changes in relationship with Ic.
is there any way to get controll over beta??? for example i know from datasheet that when Vce is 5volts, and Ic is 0.1mA, the beta will be 20. alright, but how about Vce 1.3v and Ic 1.5mA? how much is beta right now???

i have tried to find my transistor's load line, then grabbed my Q point and tried to compare it with hfe graph from datasheet to somehow understand what would be beta in that Q point but i couldn't succeed. any advices?

 

This [variabilty in HFE] is the reason designers go to such trouble to make their transistor circuit Q point as stable as possible in spite of the variability in HFE. The goal is to establish a stable quiescent current notwithstanding the likely variation in HFE.

As you point out this variability can be significant.

Consider the attachments which are snapshots from a data sheet for the 2N3055. The quoted variation in HFE at Ic=4A with VCE=4V is from 20 to 70 even though the graphs of HFE vs Ic and temperature would suggest a much tighter limit at the stated conditions.

Perhaps you might clarify your problem by providing some further details of your methods.

 

Negative feedback is very helpful to reduce the effects of varying HFE and other undesirable non-linearities.

 

The variability in beta is why the typical bias circuit for a transistor AC amplifier uses negative feedback, as wmodavis mentioned. This is provided by an emitter resistor, which generates local negative feedback. Some bias circuits may also use negative feedback with a resistor from collector to base. Either way, the effect of varying beta on the bias point is reduced.

 

http://www9.dw-world.de/rtc/infotheque/semiconamps/semiconductor_amps2.html -A?! did you mean something like this? i liked how they explained the process inside in details. but a person who's trying to learn it must already have good understanding of dc/ac analyses and so called "equivalent circuits" to really brake it down and finally understand it. is there any book which goes through all this step by step? i understand you feel that its just basics but imagine a man who has no idea about stabilizing Q, its not easy at all.
as i get it, the emitter resistor RE (MR. crutschow also said it) tries to reduce voltage to base , whenever Ic goes up, this way RE tries to balance base input in relationship with Ic, and the beta gets fixed. it doesnt changes that much. if it was lets say 200, it will variate but slightly, from 195 to 215.
all this also includes something called negative feedback which i dont know what is and something called input/output impidance Zin Zout which i have no idea what is ))))))))))
aah... and here is good explanation of Z http://www.soundonsound.com/sos/jan03/articles/impedanceworkshop.asp

 

You are not "fixing" the beta or changing its value. The beta of the transistor under a particular set of operating conditions is what it is.

What you are doing is making the circuit largely unaffected by the actual value of beta, provided it at least remains above some value.

Consider an NPN transistor with a 1kΩ resistor between the emitter and ground and a 5kΩ resistor between the collector and a 10V supply.

Now apply 1.6V to the base. Assuming a 0.6V Vbe, that will result in 1V across the 1kΩ resistor and 1mA flowing in it. That includes the base current, so the collector current is going to be slightly less. How much less depends on the beta. If the beta is 199, then there will be 5µA of base current and 995µA of collector current. If beta is 19, there will be 50µA of base current but still 950µA of collector current. So going from a decent beta to a pretty lousy one (for a small signal transistor, anyway) resulted in a change in the collector current of less than 5%.

The beta still changed, it just didn't matter too much.

 

....okay very good; but where the hell a Negative Feedback took place? where is it?

 

The negative feedback comes from the emitter voltage changing in response to a change in the beta.

Note that, in my simple example, I assumed that the base-emitter voltage was a constant 0.6V. Well, that's not exactly true. The base-current is a function of the base-emitter voltage but the relationship is exponential so it stays almost constant with small changes having big impacts. In rought terms, a change of 60mV results in a change of a factor of 10 in base current (and, for constant beta, collector current).

So let's say that we are operating at a beta of 200 and now, for whatever reason (perhaps temperature changing, it doesn't matter) the beta drops to 20. Well, if we had the same base-emitter voltage then the current would drop by a factor of ten. But that would mean that the emitter voltage would drop from 1V to 100mV, which would increase the base-emitter voltage by 900mV to 1.6V. With this kind of voltage across it, the transistor would want to increase the base current by a factor of fifteen orders of magntiude! But long before that happened, a continual process takes place that, as the emitter current drops the emitter voltage drops which increases the base-emitter voltage which increases the base current which results in more collector current (not quite as much as before, but enough to significantly offset the drop due to beta) which inreases the emitter voltage which decreases the base-emitter voltage (not quite to where it was before, but very close). Because we needed an order of magnitude increase in the base current, we can expect the emitter voltage to actually be about 60mV lower than before, so 0.94V instead of 1.00V. This effect will reduce the current by about 6% from the numbers in my simple model, so at a beta of 19 you would have 47uA base current and 894uA of collector current. So a drop of right at 10% instead of the 5% mentioned before.

However, note that this can be reduced by designing the circuit to have more voltage dropped across the emitter resistor. One way to do this would be to connect it to a negative supply and increase the resistance accordingly. If you use a -9V supply and increased the emitter resistor to 10kΩ, you would still expect to have 1mA of current in the emitter resistor with a base voltage of 1.6V, but now when the emitter voltage drops to 0.94V, the decrease in current is by a factor of 60mV/10V instead of 60mV/1V, so the effect would be less than 1%.