sert said:
I was reading an analysis of this CE amplifier:
http://xs126.xs.to/xs126/08174/complete168.png
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:
http://xs126.xs.to/xs126/08174/dc_equiv421.png
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?
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.
Jon
