Electronic – The advantage of using base biasing over emitter biasing

Here's an overview of the design process to get you started. I'll let you work out the exact calculations.

I would replace \$R_{\text{load}}\$ with an independent current source \$I_{\text{load}}\$ for your simulation (you can use your CircuitLab schematic for simulation once you add resistor values). Set \$I_{\text{load}} = 25\$mA since that is your worst case.

Pick a relatively large emitter resistor \$R_3\$. This simply provides a load to the transistor if the actual load isn't connected (e.g. \$I_{\text{load}} = 0\$). For example, use \$R_3 = 10\$k\$\Omega\$. If \$V_{\text{out}} = 5\$V then the current through \$R_3\$ is \$0.5\$mA and \$I_{E} \approx 25.5\$mA in the worst case (\$I_{\text{load}} = 25\$mA).

Next you need to determine the worst case (highest) \$I_B\$. Use the lowest \$\beta\$ in the transistor's datasheet (worst case) and then calculate

$$I_B = \frac{I_E}{\beta + 1}$$

Now in order to make the resistor voltage divider "stiff" you need to make sure that the unloaded bias current through the resistors (call it \$I_{\text{div}}\$) is at least 10 times the load current (in this case \$I_B\$ is the load for the voltage divider). Otherwise the load current draws too much current away from \$R_{2}\$, which causes the voltage at the output of the voltage divider decrease too much. This puts a constraint on the maximum value of \$R_1 + R_2\$ since

$$I_{\text{div}} = \frac{15}{R_1 + R_2} > 10I_B$$

This equation plus the voltage divider equation

$$\frac{R_2}{R_1+R_2}15 = 5.6$$

gives you two equations and two unknowns.

The book's approach is an estimation, on the assumption that the base current is negligible - which turns out to be not so great of an assumption here, as you found out.

your approach is more precise, but made the same mistake right here:

The maximum current through the divider without connected load is: Idiv=VccR1+R2=15280⋅103≈54μA

you didn't factor in the equivalent resistance on the base side: at 8.1v @ 10ua, that's equivalent to a 810K resistor (approximately Re * beta - see note below).

So the lower resistor R2 is paralleled by a 810K resistor. Once you factor that in your calculation, it will be alright.

Most people don't take that approach. for example, I typically set the current through R1/R2 to be 10x of the base current. That yielded R1 + R2 = 15v / 100ua = 150K. and go from there for R1/R2 individually. the 10x is picked to make sure that the base current is indeed negligible.

what it shows you is that 1) don't put too much stock in any book; and 2) don't take estimation too seriously. many times, good enough is indeed good enough.

edit:

note: for better approximation, some people would assume that the lower resistor is by-passed by a equivalent resistance of beta * Re -> 750K in this case, vs. 810K the real one calculated earlier. This approach works fairly well as an approximation.