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

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)  • Schematic please???? There is a schematic editor available to you within your question editing features here. Help us help you! – jonk May 12 at 8:17
• @jonk I already added it – pxrry May 12 at 8:23
• 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. – jonk May 12 at 8:31
• @jonk Oh, I'm sorry for that. I revised it again. – pxrry May 12 at 8:38
• The first formula neglects the base current - a rather rough approximation because Ib=66µA and the divider current is less than 120µA – LvW May 12 at 8:57

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.

• 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)? – pxrry May 12 at 14:01
• @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. – jonk May 12 at 17:51