Ready Steady Eddy

[Jonathan]

Please forgive my ignorance but could someone explain how this will work in basic terms? Why is Eddy's first circuit not fail safe and how does the next circuit achieve that?

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.

[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."

[irving2008]

Just to clarify Jonathan's answer, the BOB ENable signal is Active Low i.e. enabled = 0v, disabled = 5v. In Eddy's original circuit the relay would be OFF when the BOB says Enabled and ON when disabled. But a BOB failure, a supply voltage failure, or burn out of the transistor would all turn the relay OFF, an enabled state, so not fail safe. As J says, the addition of the second transistor inverts the logic so relay is ON only when system is enabled and voltages are present.

J's later solution achieves the same result by using a PNP transistor to invert the logic.

[edit] typed this ages ago but forgot to hit submit.... :roll: and now its out of date lol

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.