Hi rogue bird, here’s the circuit.
The motor is switched by two single pole change-over relays in an ‘H’ arrangement. Normally, the two ends of the motor are connected to 0V. When one of other of the relays is energised, the motor will turn one way or the other.
IC1 is a 555 timer wired in a bistable mode circuit. When IC1 is triggered, it switches on REL1. It is triggered ‘on’ by taking TR to 0V, and reset to ‘off’ by taking RES to 0V. When TR and RES are both at 12V, the IC stays in its last commanded state. IC2 is wired in an identical circuit.
This is how it operates. Assume the arm is at the top. The limit switch S4 was closed when the arm hit the top stop. The time switch has just put 12V on to the supply rails and the 555s are powered up.
The capacitors C1 will hold the TR pins low for a split second and IC1 will be triggered on. The same would apply to IC2, but it cannot be triggered on because S4 is closed, keeping it in its reset mode. REL 1 energises and S1 changes over and puts 12V on the LH side of the motor while S2 stays unchanged, so the RH side of the motor remains connected to 0V. The motor starts to run, taking the arm down. The limit switch S4 opens up but IC2 can’t trigger on because C1 has charged up via R1 to 12V, so TR is no longer at 0V.
The motor will continue to run until the arm is at the bottom. Then S3 will close and reset IC1. REL1 will de-energise, S1 will change over, and the motor will stop. There it will remain until the timer ends and the 12V supply to the 555s is disconnected.
Next time the timer comes on it will be IC2 that is triggered (IC1 cannot trigger because S3 is closed). REL2 energises and S2 changes over, turning the motor in the opposite direction. The arm will rise until the top limit switch, S4, closes and resets IC2. Then it will stop, ready for the next timed period. So now we are back at square one.
I hope that all makes sense. Just shout if you have any questions.
Good luck with your project.
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