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}\$
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?
Best Answer
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.