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


  • \$\begingroup\$ Schematic please???? There is a schematic editor available to you within your question editing features here. Help us help you! \$\endgroup\$
    – jonk
    Commented May 12, 2021 at 8:17
  • \$\begingroup\$ @jonk I already added it \$\endgroup\$
    – pxrry
    Commented May 12, 2021 at 8:23
  • \$\begingroup\$ Do you think it might help in interpreting your equation to label your components??? I mean, I might guess at \$R_1\$. But I'd rather not. That said, I think I know what to say. But I think it would still improve your question to label things. \$\endgroup\$
    – jonk
    Commented May 12, 2021 at 8:31
  • 1
    \$\begingroup\$ @jonk Oh, I'm sorry for that. I revised it again. \$\endgroup\$
    – pxrry
    Commented May 12, 2021 at 8:38
  • \$\begingroup\$ The first formula neglects the base current - a rather rough approximation because Ib=66µA and the divider current is less than 120µA \$\endgroup\$
    – LvW
    Commented May 12, 2021 at 8:57

1 Answer 1


The schematic is roughly this:


simulate this circuit – Schematic created using CircuitLab

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


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.

  • \$\begingroup\$ I clearly understand this but why on websites and videos that I search for and also the answer that I provided from my post (it is from a solution manual), ignored the base current? So it's just VB=VCC*RC/(R1+R2)? \$\endgroup\$
    – pxrry
    Commented May 12, 2021 at 14:01
  • \$\begingroup\$ @pxrry It would be really bad to ignore the base current if the resistor divider at the base has a high impedance, because of the voltage drop then caused. But most designs will set the divider to a relatively low impedance. Doing so means that the error in using that over-simple formula (the unaccounted voltage drop) is relatively small. Given that anything within 10% of where you think it is can be considered "good enough" and given the complexity of trying to teach someone a harder-t--grasp formula, the simpler idea is often used instead. \$\endgroup\$
    – jonk
    Commented May 12, 2021 at 17:51

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