Ready Steady Eddy

[cncJim]

The transistor is used as a switch. When positive current is applied to the base of an NPN it switches on. In the original circuit, the transistor is simply connected in series with the relay, so when the current is applied the transistor switches on and so does the relay. The problem is, if the signal to the base is broken (e.g. wire accidently cut), the transistor and thus the relay will switch on. We want the relay to be off in this situation, so another transistor is used to invert the signal. When the current is applied to 'new' transistor, it switches on so connects the base of the second transistor to ground, which in turn switches it off - hence the signal is inverted.

Thanks for taking the time to reply Jonathan. Very helpful. I was also having hard time understanding/visualising the "active low" part of the puzzle.

This also helped me understand:-

"If the input signal is high there will flow current through R2 and the transistor's base-emitter junction (base, not gate). This current will be amplified, and the collector current through R1 will cause a voltage drop so that the output will be low. Input high, output low.
If the input signal is low there won't be any base current, and no collector current. No current through R1 means no voltage drop, so that the output will be at +V. Input low, output high."