Electronic – Thevenin equivalent of voltage divider bias

(1) It MAY be of assistance to you to note that equation 4.29 in the above cited text COULD be called "the voltage divider rule" as it refers to R1, R2 and Vcc. ie changing what it says just slightly without changing the meaning:

• Vout = Vin x R2 / (R1 + R2)

ie R1 & R2 form a voltage divider and the above equation defines a "rule" of the result.

BUT

(2) There IS NO "voltage divider rule" as such.

Even if somebody uses that term there is still no such rule.
BECAUSE the terminology is much too general.
That's even more general than saying eg "The Ohm's law rule"
where you at least have some guide.

If you have a specific question you should explain it clearly in words and not use general terms or few words or the real requirement is liable to be missed.

Added:

Re question:

• Can you tell me how this 'rule' is derived? I'm a beginner so I didn't understand how he got the relation Vr2= (R2)(Vcc)/(R1+R2)

(1) Short answer.
Voltage across each resistor is proportional to current in it (Ohm's law).

As current in both resistors
= battery current
= the same
THEN the voltages across each resistor are proportional to their resistance value.
THIS IS THE KEY FACTOR THAT MAKES THIS WORK

Vout = Vr2 = ib x R2
Vcc = Vr1+Vr2 = ib x R1 + ib x R2 = ib x (R1 + R2)

So Vout / Vcc
= Vr2 / (Vr1 + VR2)
= ib x R2 / (ib x (R1 + R2) ) Cancel ib's Vout/Vcc= R2/(R1 + R2)
Multiply both sides by Vcc.
Vout = Vcc x R2 / (R1 + R2)
QED.

(2) Longer answer.

You MUST know Ohms law. If you don't know Ohms law and it's various re arrangements, stop reading this now, drop all lse and learn it. Wikipedia and Google know all about it N time over

... time lapse ... or no time at all as the case may be ...
So we know you know Ohm's law.

So - one version of Ohm's law says, as you know

• V = i x R

ie the voltage drop across a resistor is equal to the value of the resistor multiplied by the current flowing in it.

Now look at fig 4-29

Take this circuit in isolation.
The current from the battery flows from B+ at the top left of R1, via R1, then via R2 and back to B- and the bottom left. Look at the diagram and be SURE that you agree with the above.
Now, lets call the battery current Ib.

Call the current in R1 I_R1. It can be seen "by inspection that I_R1 = Ib.

Call the current in R2 I_R2. It can be seen "by inspection that I_R2 = Ib.

So I_R1 = IR2 = Ib.
ie the current is the same in each resistor and out of and into the battery.

Now, the voltage across R1 = VR1 is, based on Ohm's law = I_R1 x R1.
And, the voltage across R2 = VR2 is, based on Ohm's law = I_R2 x R2.

BUT I_R1 = Ib and IR2 = Ib.

So VR1 = I_R1 x R1 = Ib x R1
And VR2 = I_R2 x R2 = Ib x R2

The ratio of VR2 / VR1 = Ib x R2 / Ib x R1 = R2/R1
ie the voltages across the two resistors are proportional to their resistance values.

Look at the diagram.
Vbattery = Vcc
Vcc = the voltage across R1 + the Voltage across R2
Vcc = VR1 + VR2
Vcc = ib x R1 + ib x R2
Vcc = ib (R1 + R2)

So
To determine the ratio Vout / Vcc:

Vout / Vcc = V_R2 / Vcc
= ib x R2 / ib (R1 + R2)
but the ib's cancel so
Vout/ Vcc = R2 / (R1 + R2)
and rearranging

Vout = Vcc x R2 / (R1 + R2)

So the voltage across R2 compared to battery voltage = Vr2= (R2)(Vcc)/(R1+R2)

I think Thevenin (or Norton) equivalent circuits do not consider variable sources. The same refers to non-linear resistors (and other elements in AC scope). But I understand what you mean: you would like to have something like these.

In your case you should first select all the elements that are not dependent on other and do not alter other elements, and simplify them. The next step is to find all independent voltage/current sources.

Now combine non-linear static elements, like resistors. The combination of a linear object and a non-linear object is also non-linear object (but there is a theoretical possibility that two non-linear functions make a linear one).

At this moment you get: combined resistances that are (generally speaking) non-linear and do not alter anything and independent and dependent sources, and the elements that alter sources. If possible, combine independent sources.

That's the hardest task now: to combine independent sources with dependent. The Kirchoff's laws might be necessary here.

UPDATE

According to your circuit, this is not that difficult as it seems on the first sight. Please forgive me there are no exact calculations as I did them last time almost 20 years ago...

First of all, take a look at the non-ideal current source I1. Because it has R1 in parallel you can convert it to a non-ideal voltage source, which has resistance in series. This voltage source would have internal resistance 1 Ohm too and voltage R1 * 4Ix that is 4*Ix volts as R1 = 1 Ohm. I will name this new source as V2.

At the moment on the left side of the circuit you have non-ideal voltage source V2 (equivalent to I1 current source), its internal resistance (equivalent to R1), than voltage source V1. The R1 resistance is gone as it became internal load of voltage source. More reading about source transformation.

Because in the same branch there are two voltage sources you can combine them. So it is E = V1 + V2 which leads to (4 Ix - 10) V (- because V1 is in opposition to V2).

Now we have the first part of our task, the source. Now we're going to find equivalent resistance, and, moreover, we need to drive out Ix from source equation, because after combining resistances to one there will be no Ix.

As we know from Mr. Kirchoff, the load current (the one in R3), say I, divides in two: Ix and IL (IL flows through R3). The Ix is U2 / R2 and IL is U2 / (R3 + RL). You can write down proper equations yourself :).

Now you can find relation between Ix and IL (you need IL in equation of voltage source) and make E function of IL. If this source is no more function of Ix, you can combine other resistances to one equivalent. Do not forget source E internal resistance (the one driven from R1).

Please note that this method will lead you to have voltage source that is a function of load current (so in fact load resistance RL). This is normal as U2 depends on this load (that's why I've written at the very beginning it is not true Thevenin method).