This circuit generates a current that is constant. The 1k resistor in the top left serves to bias the Zener on the left side. The current through the left side is simply equivalent to the voltage across the Zener (6.3V) divided by the 630 Ohm resistor. This means the load above the BJT will always have 10 mA assuming that one condition is met. The voltage at the top of the right most BJT must be greater than the Zener Voltage and the BJT drop (approx. 7V). In this case the Voltage on the right side minus the current (10mA) times the resistor (200 Ohms) is in fact greater than the required 7V. Play with the circuit!
And the current through the base is neglected for the most part if the right BJT subscribes to the above rule of Voltage
