Single Polarity Power Supply

This circuit uses a PNP Power Transistor TIP2955, you can use any other according to your current and voltage requirement.

Look at R2 a 10 Ohm resistor, when the current in your load to the power supply is less than 70mA the voltage across R2 is less than 10E * 70mA = 700mV right. The base emitter junction of Q1 will be biased or turned on around 700mV, less than 700mV the transistor just does nothing.

When the current in your load goes over 70mA the voltage across R2 goes above 700mV and a small base current Ib flows from emitter to base of Q1 turning on the transistor. Now a collector current Ic flows from emitter to collector and then to your load supplying the excess demand. The Ic = Ib * hfe where hfe or beta is the DC gain value.

From my Power Electronic Circuits

Some transistors will have only AC gain specified which is lower than DC gain. TIP2955 has a gain of 20 so for an Ib of 50mA the Ic will be 1 Amp which saves the regulator from heating up or shutting down as the main current flows thru the transistor. Q1 should be provided with a good heatsink.