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.

Pretty close.

You should give the PNP a lot more base current to make sure it saturates. Use a forced beta of 10-20, which means that the resistor R2 should be more like 470R.

The value of R3 is not critical, 10K would be fine, so would 20K.

You don't need a Darlington to drive the PNP transistor, a regular NPN will be fine. Use a base resistor of more like 10K. The base resistor limits the current- it is determined by the desired base current, not by the gain of the transistor- you *do* always need it.

Keep in mind these Darlington things won't switch instantly and if you don't allow some dead time in the drive firmware you'll get 'ghosting'.

## Best Answer

Based on the numbers you have, and

assumingyou want the collector to rest at 50% of the supply voltage. then Ohm's Law:E = I x R

R = E / I

R2 = (Supply/2) / 0.005 A

R2 = 5 / .005

R2 = 1 K

Note that the circuit will not actually work. Biasing the base with a single resistor is called dangle biasing, and is severely unstable. The gain of the transistor changes with temperature, with changes in collector current, and from part to part.