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I need to determine the base current of the circuit I have attached below.
Given:

current amplification \$B = 500\$
\$U_{CC} = 12V\$
\$U_{BE} = 0.7V\$
\$R_C = 3.2k \Omega\$
\$R_E = 9k \Omega\$
\$R_1 = 47 \Omega\$
\$R_2 = 19 \Omega\$

transistor circuit

What we need to get: base current \$I_B\$ in $\mu A$ (2 decimals)

My attempt: (feel free to correct me if I use the wrong vocabulary to describe my attempt, thank you!)

At first I introduced a new current \$I_q\$ which flows through \$R_2\$. Having that done I know that \$I_q = \frac {U_{BE}}{R_2}\$. Since \$U_{BE} = 0.7V\$ and \$R_2 = 19 \Omega\$ are given I calculated the value for \$I_q \approx 0.03684210526A\$.
Now I looked at the top left part of the circuit. We know that \$R_1\$ must be \$R_1 = \frac {U_{CC} - U_{BE}}{I_q + I_b}\$. Well solve the equation for \$I_b\$. We then receive \$I_b = \frac {U_{CC} - U_{BE}}{R1} - I_q\$. If we fill the equation with the given values and \$I_q\$ we get: \$I_b = \frac {12V - 0.7V}{47} - 0.03684210526 = 0.2035834267A\$. Now we need to convert \$I_B\$ to \$\mu A\$ which should be \$203583,43 \mu A\$ (rounded).

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    \$\begingroup\$ I think you forgot to take into account the voltage drop acress $R_E$. The Voltage drop across $R_2$ is $U_{BE}$ + $U_{R_E}$ \$\endgroup\$ – RedGrittyBrick Nov 27 '12 at 23:26
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At first I introduced a new current Iq which flows through R2. Having that done I know that Iq=UBE/R2.

This is incorrect; the voltage across R2 is \$U_{BE} + I_E R_E\$

Also, I suspect that the values of R1 and R2 should be in \$k\Omega\$ and the value of \$R_E\$ is suspiciously high.

Regardless, there's a step by step approach to finding \$I_B\$.

Form the Thevenin equivalent circuit looking out of the base:

\$U_{BB} = U_{CC} \dfrac{R_2}{R_1 + R_2}\$

\$R_{BB} = R_1 || R_2\$

Now, write the KVL equation around the base-emitter loop:

\$U_{BB} = I_B R_{BB} + U_{BE} + I_E R_E\$

Using the relationship:

\$I_E = (\beta + 1) I_B\$

Substitute and solve:

\$I_B = \dfrac{U_{BB} - U_{BE}}{R_{BB} + (\beta + 1)R_E}\$


You can ignore this if you like, but you ought to, before turning in or publishing an answer, do a sanity check to make sure that, on the face of it, your answer isn't hopelessly, impossibly wrong.

For example, consider the answer you give for the base current and the implication of it. If the base current were 0.2A, as you've calculated, the emitter current, which is 501 times the base current, would be an enormous 102A.

It's always good to do a sanity check on your answer. Even if \$U_{CE}\$ were zero, the emitter current could not be any larger than:

\$I_{E_{max}} = \dfrac{U_{CC}}{R_C + R_E} = 984\mu A\$

This places an upper bound on the base current which is:

\$I_{B_{max}}= \dfrac{I_{E_{max}}}{\beta + 1} = 1.96\mu A\$

So, by making a very quick calculation, you have a good sanity check for any answer you may come up with.

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  • \$\begingroup\$ I am sorry @Alfred Centauri that I dissapointed you. For my further approaches I will perform the sanity check before posting here. I never heard of the Thevenin equivalent before but Wikipedia gave me a brief overview which in my opinion is satisfying. Again I would like to thank you that you put effort in my problem. You helped me to understand the approach. Sadly I lack a lot of physical and electronical practice which is mirrored in my poor topics. However the correct $I_B$ must be $0.6108975 \mu A$. Hope I got it right this time :) \$\endgroup\$ – optional Nov 28 '12 at 17:52

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