TRUTH TABLES AND EXERCISES
Left input is - off -. To determine this, the status of the gates feeding the left input must
Looking at the OR gate (2) above it
The right input to the OR gate is - off - because the hand control switch
is in the AUTO position. The OPEN position contacts are not made up,
resulting in an off signal.
The left input to the OR gate comes from the AND gate (1) above it.
Looking at the three inputs to the AND gate. The bottom input
is - on - because the hand control switch is in the AUTO position
and the AUTO contacts are made up, resulting in an on signal.
The middle input to the AND gate is - on - because the NOT gate
reverses the off safety signal.
The top input is - off - because the valve is not fully open,
resulting in the generation of an off signal. Note this is the signal
that, once the valve has traveled to the fully open position, allows
the valve to remain open after the hand switch is allowed to
spring return to the AUTO position.
Now that all the inputs are known, we can work back through the circuit to determine
the status of the left input to the AND gate (4).
Because the one input, the top, to the AND gate (1) is off, the output of
the AND gate is off. Therefore, the left input into the OR gate (2) is off.
Therefore, because both the left and right inputs to the OR gate (2) are
off the output of the OR gate (1) is off.
Yes, de-energized is correct because the left input of the AND gate (4) is off and its right
input is on. But because it is an AND gate and both its inputs are not on, it will not pass
an on signal to the solenoid to energize it.
It can be energized one way - the hand switch can be momentarily placed in the OPEN
It can be de-energized two ways - the hand switch can be placed in the CLOSE position,
or, if the valve is open and a safety signal is received, the valve will automatically close.
Yes, the valve can be opened, but it will not remain open when the hand switch is
allowed to spring return to the AUTO position. This is because the safety signal's NOT
gate removes the on signal that allows the AND gate (1) to output an on signal and
energize the solenoid.