Motor triggered by a vibration sensor

[julienfhd]

Hi, my project consists of a vibration sensor that, when enabled, will trigger a motor which will function for a certain amount of time, then stop, and stay in this position as long as the sensor is detecting vibrations.
When the vibrations stop, I need the motor to reverse polarity and function in the opposite direction,for the same amount of time as before then stop, and repeat...
I'm a first year electrical engineer so please consider me a rookie in this field.
If someone could provide me with a suitable electrical circuit, with names and values for the part, i think i might be able to draw the corresponding board.
Thanks

[pebe]

You need to decide on a vibration sensor in order that you know what output it gives. Only then can you design a circuit for the motor. An electret microphone would be suitable if the vibration is in the audio range. If not, then the frequency and amplitude of the vibrations would need to be known.

Incidentally, if you design for the 'time' of motor run, rather than its position, there would be no guarantee that the motor would return to its original position. Would that matter?

[julienfhd]

The vibrations I need to detect, are of mechanical nature, more specifically, caused by a smartphone when one receives a notification (I mean the physical vibration of the phone, not the notification's sound).
And no, it doesn't matter, I'd rather stick with the motor run time.

[pebe]

One of these may do the job. They are as cheap as dirt on ebay from HongKong, so are worth experimenting with.

http://shop.moderndevice.com/products/m ... elerometer

[julienfhd]

Oh okay, thanks...
And I know I should use an SR latch to control the motor, but I have some trouble combining it with the timer, and how to reverse the polarity of the motor...

[pebe]

OK. If you can sort out the sensor and let me know what output you get, I'll do you a circuit for the motor.

What is the motor voltage and current?

[julienfhd]

hi pebe, I'm sorry for the delay... Here's the sensor I found:
http://www.dfrobot.com/wiki/index.php?t ... KU:DFR0027)
It requires an input of 5V, and delivers a constant 5V also, unless it detects a vibration, in which case its output briefly drops to 0.
I have yet to find a suitable motor, but i should get back to you in a couple of days.
Thank you for the help

[julienfhd]

Okay so the motor's voltage 12V, and its maximum current is 1 Ampere.
And I need it to run for about 7 seconds.

[julienfhd]

Okay, so the motor's voltage is 12 V, and the maximum current it can support is 1 Amp.
Also I want its runtime to be of approx. 7 seconds

[pebe]

Just to clarify. Can I assume the vibrations from your smartphone will always last longer than 7secs?

[julienfhd]

Yes...

[pebe]

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.

[julienfhd]

Okay great, thanks a lot Pebe, I owe you one.
Actually, the sensor's output will flicker with the vibration, so i will only have an open circuit for a fraction of a second, then a closed circuit for another fraction of a second, and so on... Meaning i don't have a constant open circuit during the vibrations, Will that be an issue?

[pebe]

No, that won’t be a problem. When the sensor goes o/c the 10K base resistor will make TR1 conduct hard to discharge the 1uF cap. But when A closes again and TR1 cuts off, the cap can only discharge relatively slowly through the 100K, so the voltage at the collector stays low – certainly not high enough to operate the gate.

Since I did the circuit, though, I realized that unless the period when the sensor is active is much longer than 7secs, the 10uF will not have enough time to fully charge via the 1M. That would result in the 2nd pulse being shorter than 7sec.

If it looks like that could be a problem, then I’ll redesign the circuit.

[julienfhd]

No it's okay, it's just a prototype of a project, so I can control any external conditions...
But I do have a question though, is the NF2EB a suitable DPDT Relay?