transistors - Designing a stiff voltage source using an emitter follower

^{simulate this circuit – Schematic created using CircuitLab}

I am new to electronics, and I am trying to do Exercise 2.2, page 67 of The Art of Electronics.

Use a follower with base driven from a voltage divider to provide a stiff source of +5V from an available +15V supply. Load current (max) = 25 mA. Chose your resistor values so that the output voltage doesn't drop more than 5% under full load.

According to my understanding, a BJT in emitter follower configuration has its emitter following the base, no matter what the emitter signal is fed to. Therefore, under normal operation the emitter follows the base (with a 0.6V difference because of forward voltage drop), and is not really influenced by what's connected to it.

We need to find resistor values so that even under maximum load current 25mA, the emitter voltage never drops below 5V - 0.05*5V = 4.75V.

If I set VE(minimum) = 4.75V, I(load,maximum) = 25mA, I can compute the corresponding minimum load resistance value but I believe this is useless and is meaningless.

If there is no load, I set VE = 5V. So VB = 5.6V. I would have used the voltage divider equation to find the ratio needed between R1 and R2, but I am not sure this is valid since there is a current going through the BJT collector.

I am kind of lost on what i need to do. It's the first design question I try, it seems like there are more unknowns than constraints and so we have to set some values, and I have no electrical intution/experience to know what to set to which value and why.

Thank you for your help.

Answer

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.