Re: common emitter configuration- voltage divider biasing.

From: andy (news4_at_earthsong.free-online.co.uk)
Date: 07/27/04 

Date: Tue, 27 Jul 2004 01:02:16 +0100

On Mon, 26 Jul 2004 13:27:45 -0700, Jenny wrote:

>> > 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.

One way to think of it is as a built in negative feedback loop which is
made when you put a transistor in this configuration.

The transistor won't conduct until Vbe > about 0.7V, then Ic increases
exponentially as Vbe goes above that. I.e. big changes in Ic only mean
small changes in Vbe, so Vbe will stay around 0.7V.

Say Re=1kOhm, Vcc=12 v, and R1=R2=10kOhm.
When you switch on, Vbe is 6V, so the transistor is fully on by a big
margin. This makes current flow in Rc and Re, increasing Ve. But as Ve
gets close to Vb, the transistor will start to shut off, and the current
will reduce until it reaches a stable operating point. Because of the
exponential relation between Vbe and Ic, this will be when Vbe is around
0.7-1v. I.e. it stabilises with Ve=6-0.7=5.3V, which gives
Ie=5.3/Re=5.3mA, Ib=Ie/hFE (roughly), and Ic=Ie-Ib. (This isn't quite
right because Hfe is Ic/Ib, i think) This will all happen very quickly.

Then think about what happens if there is a small up or down fluctuation
in Ic, and you'll see how it stabilises itself. 

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