Polarizated BJT Class A: Conceptual DC analysis

Question

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:

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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}\$

Correct Answer

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}\$

Wrong Answer

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

Answer 1

In your second analysis, you're ignoring the current flowing into the base of the transistor. You're assuming that all of the current flowing through the 113 k resistor is also flowing through RB.

Answer 2

It's not simple voltage divisor because there is connected transistor with it's resistance in parallel to voltage divisor, and there is current flowing into it.