by pebe » Mon Jul 01, 2013 4:22 pm
OK. Here is the circuit.
There is not much detail in your link to the vibration sensor, but I think it may be a simple trembler that opens up a contact when it is vibrated. This circuit assumes that.
‘A’ is normally closed so TR1 is cut off and its collector will be at +12V. Both inputs to IC1 are high so its output will be low. So TR2 is cut off and the relay is de-energised.
Both input to IC2 are low so its output will be low and the FET, which switches on the motor, will be cut off and the motor will not run.
When the phone vibrates and ‘A’ goes open circuit, TR1 will conduct. Its collector will go low and be held low with the 1uF cap. The two inputs of IC1 are not now the same so its output will go high and energize the relay. As the two inputs to IC2 are now not the same, its output will go high. The FET will switch on and the motor will run forwards.
There is now 12V across the 1M resistor so current will flow through it and charging up the 10uF. When the charge on that cap reaches ½ of the supply voltage, that input of IC2 will have gone high. With both inputs high the output will go low, turning off FET and stopping the motor. Time of travel will be 0.7 x C x R = 7secs.
Nothing more happens until the vibrations stop. ‘A’ will then close and the 1uF will charge through the 100K taking the input high. IC1 output goes low and the relay de-energizes, changing over the motor polarity. The inputs to IC2 are not now the same so its output will go high again for 7secs as before, running the motor in the reverse direction. Then the circuit will wait for the next time….
For clarity I haven't shown the supply to the IC or the supply decoupling, but I'm sure you can fill that in.
I hope you find that OK.
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- Vibration sensor motor drive.GIF (6.53 KiB) Viewed 60461 times