Current biasing in npn transistors
I would be happy if you could help me to understand the schematic.


That much I can attempt. Let's look at the schematic, drawn somewhat less jazzy:







simulate this circuit &ndash; Schematic created using CircuitLab

The current sink part is pretty easy.

\$Q_3\$ is arranged into an emitter-follower. The pair of base biasing resistors, \$R_1\$ and \$R_3\$, form a simple voltage divider. Since their values the same, the mid-point voltage will be half the difference between ground and \$V_\text{EE}=-15\:\text{V}\$, or \$-7.5\:\text{V}\$. The emitter of \$Q_3\$ will follow that base voltage, one base-emitter diode drop below that point. Since this is \$700\:\text{mV}\$ as from your text (and I think higher, but who is counting?), then the emitter voltage will be \$-8.2\:\text{V}\$. This leaves \$6.8\:\text{V}\$ across \$R_2\$, meaning that the emitter current (which is the collector current in your case with \$\beta=\infty\$) is sinking \$6.8\:\text{mA}\$. It is a current sink.

The rest is a differential pair or so-called long-tailed pair using two collector resistors. (Other arrangements are often made for the collectors, different from this.) The emitters of \$Q_1\$ and \$Q_2\$ are tied together, so they are always at the same voltage as each other. If both bases are tied to the same low-impedance voltage source value (say, \$0\:\text{V}\$), then the emitters will be \$700\:\text{mV}\$ below that.

Since the collector current of a BJT is determined by its base-emitter voltage, and if both base voltages are the same (and the emitters are tied together, too), then both BJTs will have the exact same collector current. The only way this happens is if the current determined by the current sink splits two ways, equally. So this means each collector current is the same and will be \$3.4\:\text{mA}\$, apiece. (Which totals to the current sink value computed above.)

If both base voltages move upward or downward a little bit, but do so exactly together with each other, then the collector currents remain equal to each other and they split the current sink value. It is only when the two base voltages are different, that there is a difference in the two collector currents.

I don't think I noticed anything saying that there was a phase difference in your two signal sources. If not, then the two collector currents will remain the same even with these signals applied. All that happens is that the two base-emitter junctions of the two BJTs move up and down together with the signal. But there is no difference in their base voltages so there will be no differences in their collector currents.

If, however, your two signals sources had a different phase or were operating at different frequencies, then something else would happen. But apparently your problem specifies quite clearly that the two signal sources have the same phase and frequency. So that's all there is to it.