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.

Differential gain:

If the base of Q1 moves down by \$-\Delta V_{be}\$, and the base of Q2 moves up by \$\Delta V_{be}\$, then the junction of \$R_1\$ and the two emitter resistors \$R_e\$ will remain fixed. Since no signal current flows through \$R_1\$, the signal current through Q2 will be simply

\$\frac{\Delta V_{be}\ - (-\Delta V_{be})}{2R_e}\ = \frac{V_{diff}}{2R_e}\$.

The voltage gain will then be

\$\frac{V_o}{V_{diff}} = -\frac{R_c}{2R_e}\ \$.

Common mode gain:

The simplest way to calculate this is to note that \$R_1\$ will carry both \$I_2\$ and \$I_2\$, and these currents will be equal in magnitude. It's therefore possible to split resistor \$R_1\$ for analysis purposes into two resistors equal to \$2R_1\$ in each leg of the pair and break the center connection. Then from inspection the common mode gain is:

\$ \frac{V_o}{V_{CM}} = \frac{-R_c}{R_e + 2R_1}\$.

The common mode rejection ratio is the differential gain divided by the common mode gain, or:

\$\frac{\frac{R_c}{2R_e}}{\frac{R_c}{R_e + 2R_1}}\$

or:

\$ \frac {R_e + 2R_1}{2R_e} \approx \frac{R_1}{R_e}\$.

## Best Answer