Electronic – Common Collector Amplifier DC Analysis (why not get the VTH and RTH like in Thevenin’s Theorem?)

amplifierbjtcircuit analysisdcemitter-follower

I looked in the book of Electronic Devices and Circuit Theory and in the section of a common collector amplifier configuration with a voltage divider, it shows that:
$$ V_B=\frac{VCCR_2}{R_1+R_2} $$

However on other voltage divider amplifiers with a collector resistor the solution is:
$$ V_{TH} =\frac{VCCR_2}{R_1+R_2} $$

and to get \$ V_B, \$ the solution to be used is \$ V_B = V_{TH}-I_BR_{TH} \$

Obviously, both will yield a different answer. So I was wondering if why the first formula above must be used instead of applying Thevenin's Theorem (VTH, RTH)?

It can't be because of the voltage divider approximation which is \$ \beta R_E \ge 10R_2\$, I calculated it but \$ 10R_2 \$ is still bigger and it still uses the first formula.

I searched for youtube tutorial videos and no one explained why. Here's the schematic and the solution: (There's also on YouTube if that's needed)

Youtube link

Emitter-follower circuit

solution

Best Answer

The schematic is roughly this:

schematic

simulate this circuit – Schematic created using CircuitLab

This is the same as (using a Thevenin divider approach):

schematic

simulate this circuit

From this it is easy to work out (from KVL) that:

$$V_\text{TH}-I_B\cdot R_\text{TH}-V_{\text{BE}_1}-I_E\cdot R_\text{E}=0\:\text{V}$$

But also you know that \$I_E=\left(\beta+1\right)\cdot I_B\$ so it follows that:

$$V_\text{TH}-I_B\cdot R_\text{TH}-V_{\text{BE}_1}-\left(\beta+1\right)\cdot I_B\cdot R_\text{E}=0\:\text{V}$$

And therefore that:

$$I_B=\frac{V_\text{TH}-V_{\text{BE}_1}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}$$

This means that if \$\beta\ge 100\$ then \$I_b\le 63\:\mu\text{A}\$ and that \$V_B=V_\text{TH}-I_B\cdot R_\text{TH}\ge 8.36\:\text{V}\$.

(With \$\beta\ge 300\$ this works out to \$V_B\approx 9.06\:\text{V}\$.)

From there, you can work out the rest.

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