The trick to understanding this design is to understand the behaviour of an emitter-follower:

^{simulate this circuit – Schematic created using CircuitLab}

The "input" here is the potential \$V_B\$, produced by source V1, and the "output" is emitter potential \$V_E\$. As I sweep input V1 from zero to +12V (blue), watch what happens to output \$V_E\$ (orange):

\$V_E\$ is always 0.7V lower than \$V_B\$, hence the name "emitter-follower":

$$ V_E = V_B - 0.7V $$

In other words, we can set whatever potential we like at emitter E, by applying a 0.7V higher potential at base B. We do that with resistor R1 and diode ZD1. Let's say that ZD1 is rated for 5.6V:

^{simulate this circuit}

By doing this we know that:

$$ V_E = V_B - 0.7 = +4.9{\rm V} $$

By Ohm's law we can find emitter current through Re:

$$ I_E = \frac{V_{RE}}{R_E} = \frac{V_E - 0{\rm V}}{R_E} = \frac{4.9{\rm V}}{4.9{\rm k\Omega}} = 1mA $$

Since collector current and emitter current are very similar (base current is assumed to be small in comparison), that means that most of this current must be coming from the collector, through whatever's connected up there; an ammeter in this case, showing this to be true.

The trick to understanding this design is to understand the behaviour of an emitter-follower:

^{simulate this circuit – Schematic created using CircuitLab}

The "input" here is the potential \$V_B\$, produced by source V1, and the "output" is emitter potential \$V_E\$. As I sweep input V1 from zero to +12V (blue), watch what happens to output \$V_E\$ (orange):

\$V_E\$ is always 0.7V lower than \$V_B\$, hence the name "emitter-follower":

$$ V_E = V_B - 0.7V $$

In other words, we can set whatever potential we like at emitter E, by applying a 0.7V higher potential at base B. We do that with resistor R1 and diode ZD1. Let's say that ZD1 is rated for 5.6V:

^{simulate this circuit}

By doing this we know that:

$$ V_E = V_B - 0.7 = +4.9{\rm V} $$

By Ohm's law we can find emitter current through Re:

$$ I_E = \frac{V_{RE}}{R_E} = \frac{V_E - 0{\rm V}}{R_E} = \frac{4.9{\rm V}}{4.9{\rm k\Omega}} = 1{\rm mA} $$

Since collector current and emitter current are very similar (base current is assumed to be small in comparison), that means that most of this current must be coming from the collector, through whatever's connected up there; an ammeter in this case, showing this to be true.