# Polarizated BJT Class A: Conceptual DC analysis

The transistor (figure above) has a following characteristic: $$\V_{BE} = 0.7V\$$ and $$\\beta = 120\$$. It is polarized to class A peform amplifier with $$\V_{CE} = VCC/2\$$. What the value of resistor $$\RB\$$.

My attempts:

Honestly I find 2 answer to same question:

$$\V_{CE} = 6V\quad\therefore\quad V_C =6V\Rightarrow i_C = \frac{V_{CC}-V_{C}}{1k} = 6mA\$$

$$\i_B = \frac{i_C}{\beta} = \frac{6}{120}mA\$$

$$\i_{113k} = i_B + i_{RB}\Rightarrow i_{113k} = \frac{V_{CC}-V_B}{113k} = \frac{6}{120}+i_{RB}\Rightarrow \frac{11.3}{113} = \frac{1}{20}+\frac{0.7}{RB}\$$

$$\\frac{1}{10} - \frac{1}{20} = \frac{0.7}{RB}\Rightarrow \boxed{RB = 14k\Omega}\$$

Second Attempt

If $$\ V_{BE} = 0.7V\Rightarrow V_B = 0.7V\$$, since $$\V_E = 0V\$$

Simple Voltage division:

$$\ V_B = \frac{V_{CC}\cdot RB}{RB+113k}\Rightarrow 0.7 = \frac{12\cdot RB}{113k+RB}\Rightarrow 0.7\cdot 113k = RB(12-0.7)\Rightarrow RB = \frac{0.7\cdot 11.3\cdot 10}{11.3} \Rightarrow \boxed{RB = 7k\Omega}\$$

So, why did I miss/leak the second analysis?

• I think what you mean by "polarized" is usually referred to as "biased" in English; you might want to change that so people can more easily understand. Oct 13, 2019 at 15:12
• But otherwise your English is perfectly clear. Oct 13, 2019 at 15:37
• It's a bad design question, I think. But you can do it in your head. There's 6 mA in the collector resistor, so the base current is 50 microamp. That, times 113k ohm means 5.65 V drop. But you need 11.3 V drop. So, another 5.65 V is needed to get there. That added current, another 50 microamp, must exist in RB. So .7 V divided by 50 microamp is 14k.
– jonk
Oct 13, 2019 at 15:38
• Where do you take base current into account in the second answer? Oct 13, 2019 at 15:38
• I made a mistake and i saw @TimWescott about the negligence the base current into division voltage analysis. I only apply this features only the current of central point is zero (which is not the case). Ty guys! Oct 13, 2019 at 16:13