The task is to find the operating point of this transistor:

Given information:

\$E_{C}=12V\$, \$V_{BE}=0.6V\$, \$V_{IN}=5V\$, \$\beta=200\$, \$R_{B}=440k\Omega\$, \$R_{C}=5k\Omega\$, \$R_{E}=3.3k\Omega\$

The answer to this excercise is \$I_{C}=0.8mA\$ and \$V_{CE}=5.36V\$.

My problem is that I cannot get the right \$I_{C}\$. I start with calculating \$I_{B}\$ which is: $$ I_{B}=\frac{V_{R_{B}}}{R_{B}}=\frac{V_{IN}-V_{BE}}{R_{B}}=\frac{5-0.6}{440000}[A]=0.01mA $$ Then, using the formula \$I_{C}=\beta\times I_{B}\$ I get \$2mA\$ which is clearly wrong.

If I use the \$I_{C}\$ straight from the answer, I can finish this exercise, using this equation: $$V_{CE}=E_{C}-I_{C}R_{C}-I_{C}R_{E}=$$$$=12-0.8\times10^{-3}\times5\times10^{3}-0.8\times10^{-3}\times3.3\times10^{3}=$$$$=5.36[V]$$

So the question is: "How to calculate the base- and the collector current?".

  • \$\begingroup\$ Vrb is not (Vin - Vbe). You haven't taken Vre into account. \$\endgroup\$ – brhans Jan 5 '16 at 16:04

Applying Kirchoof's voltage law in the base to emitter loop we get

\$V_{in} = (I_b \cdot R_b) + V_{be} + (I_e \cdot R_e)\$

Now you can write the emitter current \$I_e\$ as \$I_e=(1+b) \cdot I_b\$


\$b = \dfrac{I_c}{I_b}\$

\$1 + b = \dfrac{I_c + I_b}{I_b} = \dfrac{I_e}{I_b}\$

Since we know that \$I_c+I_b=I_e\$, we can write this as \$\dfrac{I_e}{I_b}\$.

Therefore \$\dfrac{I_e}{I_b} = (1+b)\$, when transistor in forward active mode.

Then your answer for \$I_b = \dfrac{V_{in}-V_{be}}{R_b + (1+b) \cdot R_c}\$

\$I_b=\dfrac{5-0.6}{440k+(201 \cdot 3.3k)}=3.9uA\$

We know that \$\dfrac{I_c}{I_b}=b\$, then \$I_c = b \cdot I_b \$

\$ = 3.9 \times 10^{-6} \cdot 200 \approx 0.8mA \$

  • \$\begingroup\$ I have formatted your equations using MathJaX, but I couldn't understand the last step, so I left it as it was. Maybe you could learn MathJaX from my edit and format the last step yourself. How does that sound? \$\endgroup\$ – Ricardo Jan 5 '16 at 17:40
  • \$\begingroup\$ It's actually 3.9x 10(power)-6 (divided by) 200 is approximately equals 0.8 mA. Okay,I'll try the software \$\endgroup\$ – Aadarsh Jan 5 '16 at 19:14
  • \$\begingroup\$ You meant this: \$\dfrac{3.9 \times 10^{-6}}{200}\$? That's \$19.5 \times 10^{-9}\$. That's why I didn't get it. I think you meant to multiply by 200. Then the result is \$0.78mA\$. \$\endgroup\$ – Ricardo Jan 5 '16 at 19:24
  • 1
    \$\begingroup\$ @Ricardo ,Oops....sorry for that,my mistake. \$\endgroup\$ – Aadarsh Jan 6 '16 at 5:45

It's simple. You are ignoring \$R_E\$. Here is the correct solution:

\$I_B = \dfrac{V_{IN} - V_{BE}}{R_B + 201 \cdot R_E}\$

\$I_B = \dfrac{5 - 0.6}{440k + 201 \cdot 3.3k}\$


\$I_B = 3.988{\mu}A\$

\$I_C = 200 \cdot I_B = 0.8mA\$


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