BJT input connected to another BJT's output and loading effects

Qualitatively, yes, the second transistor will load the first. Qualitatively is not good enough, however. You have to run the numbers and see QUANTITATIVELY what is going on: how MUCH will Q2 load Q1?

Here's the quick-and-dirty sanity check analysis.

Vb1 = 1/3 Vcc = 5V

Divider current = Vcc / (Rba+Rbb) = 15/(10K+5K) = 1mA.

Ve1 = Vb1 - 0.7V = 4.3V

Ie1 = Ve1/Re1 = 1.4333... mA

Ic = Ie x beta/(beta + 1). Usually, this is approximated as Ic = Ie.

From the 2N3904 datasheet, at Ic = 1 mA, beta = 70.

Ib1 = Ic1 / Beta = 20.5 uA, which does not significantly disturb the voltage divider output.

Vc1 = Vcc - Ic1 x Rc1 = 15 - 1.4333 mA x 5K = 7.8333 V.

Vb2 = Vc1, because of circuit topology.

Ve2 = Vb2 + 0.7V (because it is a PNP transistor) = 8.5333 V

Ie2 = (Ve2 - Vcc) / Re2 = (8.5333 - 15) / 2K = -3.2 mA.

Ic2 = Ie2

From the 2N3906 datasheet, at Ic = -1.0 mA, beta = 80 (minimum).

Ib2 = Ic2 / Beta = -40 uA, which is negligible compared to 1.4333 mA.

So the load from the second transistor does not SIGNIFICANTLY affect the operating point of the first transistor.

The above numbers are not exact, but neither are the transistor datasheet beta values. Both datasheets give the stated minimum values, and state a maximum beta of 300. If the beta for Q2 is higher than the minimum value, then the load presented by Q2 to Q1 is lower, meaning even less disturbance than the already-negligible -40 uA.

In the solution it says that assuming the first resistor is still in active mode (like it was in the previous example), then VB, IB, IE, and IC are the same.

This is correct if we ignore the Early effect (the dependence of \$i_C\$ on \$v_{CB} \$).

Recall that, to a good approximation, the collector of a transistor is a voltage controlled current source with \$v_{BE}\$ as the controlling variable.

This means that the collector current is relatively insensitive to the load. However, connecting Q2 will change the collector voltage \$V_{C1}\$ though not by much.

The BJT bias equation for Q1 is:

\$I_{C1} = \dfrac{15 \frac{R_2}{R_1 + R_2} - V_{BE1}}{\dfrac{R_1||R_2}{\beta} + \dfrac{R_4}{\alpha}}\$

Assuming \$\beta = 100 \$ and \$V_{BE1} = 0.7V \$, the DC collector current for Q1 is:

\$I_{C1} = 1.28mA \$

Without Q2 connected, the entire Q1 collector current is through R3. This implies that:

\$V_{C1} = 15V - I_{C1}5k \Omega = 8.61V \$

With Q2 connected, the Q1 collector current is (effectively) unchanged but is now the sum of two currents:

\$I_{C1} = I_{R3} + I_{B2}\$

\$I_{C2} \$ can be similarly calculated:

\$I_{C2} = \dfrac{15V - 8.61V - V_{EB2}}{\dfrac{R_3}{\beta}+ \dfrac{R_5}{\alpha}} = 2.75mA \$

Then,

\$I_{B2} = \dfrac{I_{C2}}{\beta} = 27.5 \mu A \$

Thus, adding Q2 reduces the current through R3 by just \$27.5 \mu A \$ raising the Q1 collector voltage by a mere \$137mV \$.

What is different is the current through R3. Think of Q1's collector as a current source which is programmed by R1, R2 and R4.

When Q2 is hooked in, the Q2 base sources current. But that does not change the current that is drawn by the Q1 current source. What that means is that less current is pulled through R3 by Q1, since it is obtaining some from the base of Q2.

But: if extra current is pulled through R3, then less voltage develops across R3. This means that the Q1 collector is driven to a somewhat higher voltage when Q2 is present. And this means that, due to the Early effect, the Q1 collector current does in fact change: it is increased. But (again): any such increase in the Q1 collector current due to the Early effect will increase the current across R3, and lower the collector voltage again, so there is a feedback mechanism going on here opposing the lowering of Q1's voltage and the associated change in current due to the Early effect. And of course the Q2 base current reacts to all this due to the changing VBE on Q2. But these are all small effects.

The current drawn by the base of Q2 is not zero therefore it will slightly load the collector of Q1. That is a fact. But by how much?

The current gain (\$H_{FE}\$) of Q2 can be reasonably assumed to be in the order of 100 and Q2's collector current cannot be more than \$\frac{15V}{2k0 + 2k7}\$ = 3.19mA.

This assumes Q2 is a dead short (worst case scenario). Given that probably less than 3mA flows through R5 it's reasonable to say that less than \$\frac{3mA}{H_{FE}}\$ (30uA) is taken by the base of Q2.

How much could this affect Q1's collector voltage? Well, Q1's emitter will have about 4.3V on it and this develops a current through R4 of about 1.43mA. This current also flows through Q1's collector resistor and compared to 30uA (drawn by Q2's base) it is nearly 50 times bigger.

This indicates that the loading effect of Q2's base onto Q1's collector is quite small and would probably raise Q1's collector quiescent voltage by no more than 30uA x R3 = 0.15V. That's a little bit but not much in a power rail voltage of 15V.