My problem with the analysis was in the DC section. The author removes the circuit that lies beyond the capacitors and he then claims that he replaces the R1, R2 voltage divider with their Thevenin equivalent Vbb, Rb and reaches the following circuit for the DC signals:
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With:
Rb = R1//R2
Vbb = Vcc*R2/(R1+R2)
Now, I was sure that the Thevenin theorem states that you can replace a part of a circuit between two nodes. Are the nodes in question here the Vcc and Ground? But the whole DC circuit lies between Vcc and Ground. Also, we have the base current leaving between R1 and R2. We could ignore that current since it's very small but he doesn't ignore in in the DC equivalent.
I'm lost here, can someone provide a helping hand? How do we reach the second circuit starting from the first?
The effect of base current drawn will be modelled by Rb in the Thevenin equivalent version.
I think your main problem may be in your understanding of the Thevenon equivalent. Its whole purpose to is to simplify by reducing the number of components. I think you need to explain more of *what* it is you don't exactly understand.
. . . . . Vcc VB . | current thru R1= I = IB + -- . | 1 R2 . | . I2 | [R1] IB | . v | -> |/ Vcc= I1 x R1 + VB . +---------| . VB | + |> VB . -- | [R2] | = (IB + -- ) x R1 + VB . R2 v | VB R2 . | - . --- Vcc- IB x R1 . making VB= ----------- . R1 . 1 + -- . R2 . . Vcc IB x R1 . Then rewriting VB= ------- - ------- . R1 R1 . 1 + -- 1 + -- . R2 R2 . . . . R2 R1 x R2 . making VB = Vcc x ------- - IB x ------- . R1 + R2 R1 + R2 . . . . same result as : . . . IB | R2 . -> |/ where VBO=Vcc x ------- . ---[Rb]-----| R1 + R2 . | + |>
. |+ | R1 x R2 . --- VBO VB and RB= ------- . - - R1 + R2 . |- . | . --- . . . . but VBO and RB are Thevenin equivalent of VCC and R1,R2 . . . so the results are the same using either the TE or solving . . . from scratch.... . . .
. . . . . Vcc VB . | current thru R1= I = IB + -- . | 1 R2 . | . I1 | [R1] IB | . v | -> |/ Vcc= I1 x R1 + VB . +---------| . VB | + |> VB . -- | [R2] | = (IB + -- ) x R1 + VB . R2 v | VB R2 . | - . --- Vcc- IB x R1 . making VB= ----------- . R1 . 1 + -- . R2 . . Vcc IB x R1 . Then rewriting VB= ------- - ------- . R1 R1 . 1 + -- 1 + -- . R2 R2 . . . . R2 R1 x R2 . making VB = Vcc x ------- - IB x ------- . R1 + R2 R1 + R2 . . . . same result as : . . . IB | R2 . -> |/ where VBO=Vcc x ------- . ---[Rb]-----| R1 + R2 . | + |>
. |+ | R1 x R2 . --- VBO VB and RB= ------- . - - R1 + R2 . |- . | . --- . . . . but VBO and RB are Thevenin equivalent of VCC and R1,R2 . . . so the results are the same using either the TE or solving . . . from scratch.... . . .
You cannot ignore the base current! The whole point of a transistor is that the base current is "Amplified". If you are ignoring it then what are you amplifying?
Look, you need to learn about biasing a transitor. The voltage divider is just a good method to bias the transistor. You can also bias it with just a resistor. There are pro's and cons of each.
The difference between the first and second circuit is that he simply changed the bias method to an equivalent one.
In the first, the voltage divider network looking out of the base supplies a certain amount of current into the base. It only depends on R1 and R2 and not anything after C1(since we are talking about DC). Since R1 and R2 form a voltage divider we are supplying a voltage to the base(technically its a current but obviously ohms law makes them proportional).
In the second case, we supply the same amount of current the voltage divider is but directly.
Point being, if X amps are flowing into the base then it doesn't matter how we supply it. Through a voltage divider, a single resistor, or any other combination. If X amps are flowing in then it is DC equivalent to any other method. (although again, some are better than others).
The way to analyze this circuit is to note that
VB = VE + 0.7 IE ~= IC IB = IC/Beta IE = IB + IC
Those are your 4 equations and for the most part all you need to know for basic analysis.
If you know VB you know VE. If you know VE you know IE. If you know IE then you know IB and IC and then VC.
BUT YOU DO KNOW VB!! At least in the first case with the voltage divider you do which is why it is used. Its an easy way to bias the circuit and get a very good approximation to VB. (its not exact cause the base does draw some current but if its small compared to the total then its ok)
i.e., VB = Vcc*R2/(R1 + R2)
All you assume is that there is no transistor connected... i.e., no current flowing into base! (which kinda contradicts what I said before but this is only an approximation because we assume it won't have a huge effect on the base current).
suupose R1 = R2 = 1k, VCC = 10V then VB = 5V
This means VE = 4.3V and IE = 4.3/RE ~= IC.
Its only an approximation cause IB ~= 0 like we assumed to get the voltage. But if we choose R1 and R2 then IB will have a neglible effect on the voltage divider.
In any case we can supply the same current into the base with just a single resistor and some voltage:
We need IE = 4.3/RE, this gives IE. We can find IB if we know Beta using the formulas I gave. Knowing IB we can then find the necessary voltage and and resistor.
e.g., suppose RE = 4.3k and Beta = 100, then IE = 1mA.
IB ~= 10uA So we we have
IB = V/R
We can choose V and R any way so that there ratio is IB. Of course this isn't completely true cause a transistor has its limits.
But also notice that this method depended on Beta. If Beta changed then are ratio changed... hence biasing like this will depend heavily on Beta.
The voltage divider method did not depend on Beta(well, it did but in a way that didn't matter much... Beta/(1 + Beta) ~= 1 if Beta is large)
So the whole point has to do with biasing. In the first he uses a voltage divider which is the standard method cause it is "independent" of Beta. In the second case he uses an equivalent biasing method(i.e., idealy supplies the same amount of current) but it does depend heavily on beta. Idealy both are the same, in practice they are not and the voltage divider is prefered. But sometimes its easier to analyze the second and the first.
In any case, those 4 equations(technically 3 if Beta is large) I gave are enough to analyze almost all basic transistor circuits.
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