# Calculating base current

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?".

• Vrb is not (Vin - Vbe). You haven't taken Vre into account. – 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 $I_e=(1+b) \cdot I_b$

Proof:

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

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

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 $I_c = b \cdot I_b$

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

• 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? – Ricardo Jan 5 '16 at 17:40
• It's actually 3.9x 10(power)-6 (divided by) 200 is approximately equals 0.8 mA. Okay,I'll try the software – Aadarsh Jan 5 '16 at 19:14
• 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 $0.78mA$. – Ricardo Jan 5 '16 at 19:24
• @Ricardo ,Oops....sorry for that,my mistake. – Aadarsh Jan 6 '16 at 5:45

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

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

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

So,

$I_B = 3.988{\mu}A$

$I_C = 200 \cdot I_B = 0.8mA$