1
\$\begingroup\$

enter image description here

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?

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – Hearth Oct 13 '19 at 15:12
  • \$\begingroup\$ But otherwise your English is perfectly clear. \$\endgroup\$ – TimWescott Oct 13 '19 at 15:37
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – jonk Oct 13 '19 at 15:38
  • \$\begingroup\$ Where do you take base current into account in the second answer? \$\endgroup\$ – TimWescott Oct 13 '19 at 15:38
  • \$\begingroup\$ 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! \$\endgroup\$ – miguel747 Oct 13 '19 at 16:13
1
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.