Electronic – Calculating base current and resistor required from transistor datasheet? Do I need a different transistor

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.

The typical gain of a 2N3904 is 200-300 at a couple mA collector current (more as it warms up due to not being saturated)

Even with 560K, that's half a mA or so, which will give plenty of light from a modern LED, but you should be able to see that it's not as bright as when a 10K resistor is used.

Do not use the hfe for this calculation if you want the transistor saturated hard on, use a forced beta of something like 20 to 50, if the typical hfe is 200 or so and the minimum 100 or so. If you use, say 30, in your equation you get a resistor value of 8.8K, so you might use 10K or 8.2K.

The reason is that you won't likely have a guaranteed hfe for the current you're using, and the hfe decreases at temperature extremes. It's still only "wasting" a few percent of the LED current, so no big deal.

To prove this to yourself, take a voltmeter and measure Vce of the transistor when it is on. If it is something like 50-100mV it is saturated.