Answer 2—There are two answers to this question, because R3 supplies a small amount of positive feedback in the circuit that creates a hysteresis effect. The output switches at one input voltage for a low-to-high transition, and a different lower voltage for a high-to-low transition. This provides a snap-action effect that reduces the effects of noisy inputs. 

Q1 will start/stop conducting when its base reaches 3.7 V. R1, R2, and R3 form a voltage divider that has a ratio of 0.625. The action of R2 and R3 together determines whether the bottom end of this divider is at 0 V (ground) or 0.767 V, depending on whether the output is low (0 V) or high (4.6 V). 

If you consider R2 and R3 together as a Thevenin source, when the output is low (0 V, ignoring R5 because it is only 1% of the value of R3), together they look like 166.7 KW to ground. And when the output is high (4.6 V), they look like 166.7 KW to 0.767 V. Turn this around and treat R1 and the parallel combination of R2 and R3 as a voltage divider (166.7 K/266.7 K = 0.625) whose bottom end is connected to either 0 V or 0.767 V. 

Therefore, if the output is low, it will go high when the input voltage rises to 3.7 V/0.625 = 5.92 V.

Similarly, if the output is high, it will go low when the input voltage falls to (3.7 V – 0.767 V)/0.625 + 0.767 V = 5.46 V.

Contributor: David Tweed

Published January 2004