Re: common emitter configuration- voltage divider biasing.

From: Jenny (jennygero1_at_yahoo.com)
Date: 07/26/04 

Date: 26 Jul 2004 13:27:45 -0700

> > V1= R1(R1+R2) Vb= R2(R1+R2) this correct ?
> >
>
> No, it can't : voltages can't be equal to ohms squared.
> This **probably** should be (depending on the schematics ):

ofcourse! sleepy and way too late at night, yes I meant R1 DIVIDED
BY(R1+R2) ANd I forgot the source voltage, sorry about that.

>
> V1 = Vsupply R1/(R1+R2)
> Vb = Vsupply R2/(R1+R2)
>

I get that beta is a function of temperature, doping and many other
things- being intrinsic to every transistor.

> If you "design" the base current a big part of the bias network current,
> then the base voltage will change a lot with temperature, which transistor
> you pick in your stock...

 I am under the impression that one assumes the current that one wants
the Ic to be, and works backwards to get the base current required to
make that happen, Ib=Ic/beta. After that - because we know what Ib is
needed to make the required Ic, we figure out how to bias the base, to
enable such an Ib to flow...

So far so good. this is what I do not get.
             
For this circuit,
                  Vcc +
                  |
                  |_
                  | | Rc with Ic that we want
                  |_|
                  |
                  |
                  |
             /C ------------o Vout
  _____Ib___|
  | \E->
  | .7v | hfe=say 100
  | |
 === Vb Re
  = |
  | |
  ______________________GND___o

In the above circuit, if I assume decide Ic to be X, Ib required will
be Ic/100. Ok. Now how to make Ib what we want it to be ? Attach a
voltage source to it.

Many questions:
What happens if I have a Vb of say 4V and
a] there is no Re, just emitter connected to ground. What happens to
Ib, Ie ?

b] Vb is 4v and Ib and Re is connected between Emitter and ground.

Vb=V drop acess Re + .7, and they say Ib will be small, why ? is it
just a characteristic of the Transistor for base current to be small,
what happens when a large voltage is applied ?

Next : this is the voltage divider circuit, because we can apply a
voltage to the base in anyway either through a new power source like
previously or through using a voltage divider resistors

          Vcc +
    |--------------|
    | |
    | | | Rc with Ic that we want
   | | R1 | |
   | | |
    | |
    | |
    | /C ------------o Vout
    |___Ib___|
    | \E->
    | .7v | hfe=say 100
   | | |
   | | R2 Re
    | |
    | |
    |____________|__________GND___o

Let us ignore the capcitance stuff that is missing for now.
In the above circuit,
IR1 =IR2 +Ib. Ok that is fine. what decides how much current coming
down IR1 is siphoned off into Ib ? Is it related to Re ? How about if
there was no re and a direction connection to the ground.

It does appear that I am seeing it understanding it piecemeal, but not
really GETTING the whole picture.

> A good figure is to make base current 10 times smaller than the bias network
> one.

why this number ?

Thanks

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