why biasing an emitter follower with a transistor will cause the following transistor falling into active region?

Question

My textbook says Q2 is not in the saturation region. I just think the Q2 transistor can be in the saturation region as long as R3 is small enough to provide the saturation current.

The author explained that is "because the signal on Q1's collector is always within the range of the power supplies, Q2's base will be between Vcc and ground, and therefore Q2 is in the active region(neither cut off nor saturated), with its base-emitter diode in conduction and its collector at least a few tenths of a volt more positive than its emitter." The book didn't include any values for the resistors, the transistors, and the power supplies. No matter what value these components have, Q2 will always stay in the active region.

I am doing self-study and hope someone could help me with this question. This is on page 83' upper right corner of the book named "Art of Electronics Third Edition".

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

You know that driving enough base current will fall the transistor in saturation region.

But recall that for an NPN transistor, if \$V_C < V_B > V_E\$ --or simply, if \$V_B\$ is greater than both \$V_C\$ and \$V_E\$-- then the transistor will be in saturation region.

Let's test this in your circuit:

\$V_{C_{Q2}} = V_{CC} \$

Even if Q1 is saturated, \$V_{C_{Q1}}\$ will be between \$0\$ and \$V_{CC}\$. Because, as you can see, there's a voltage divider formed by R1, R2 and C-E resistance of Q1.

\$V_{B_{Q2}}=V_{C_{Q1}}\$, thus

\$0 < V_{B_{Q2}} < V_{CC} \$ which means \$V_{B_{Q2}} < V_{C_{Q2}}\$.

\$V_{B_{Q2}}\$ will never be greater than \$V_{C_{Q2}}\$ so Q2 cannot be in saturation region.