Answer 1
When the output is low, Q1 is saturated and drives Q3 into conduction. The emitter resistor of Q3 functions as a current-sense resistor. When the voltage across it begins to exceed the VBE drop of Q4 (about 0.7 V), Q4 begins to conduct and reduces the voltage at the base of Q1, which in turn reduces the drive to Q3. The output current is limited to 0.7 V / 70 ohms = 10 mA. When the output is high, Q1 is cut off and only Q2 is conducting. This transistor is wired as a current source, with the voltage reference being supplied by the two diodes, D1 and D2, between its base and emitter. One of the diodes compensates for the VBE drop of the transistor, while the other limits the maximum voltage drop across the emitter resistor. If this drop becomes too high, Q2's base current is reduced. Again, a maximum drop of 0.7 V across 70 ohms limits the current to 10 mA.

Contributor: Dave Tweed

Published May 2003