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I have an old phone and I want to connect it with Relay 12v so that I can control a pump that is always working

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zwawitotti10

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I Need help
I have an old phone and I want to connect it with Relay 12v so that I can control a pump that is always working
When I call her, the pump turns off
I brought a Relay ,old phone, a transistor BC547 , a reset button, and a MOSFET IRF540N in order to maintain the state even after the excitement applied to it had stopped.
I designed the schema and tried the project on the Proteus software, it works normally, but when I delete the ground and rempalce it by connect it alone to the negative, the project does not work
**broken link removed**
**broken link removed**
 
R3 is redundant/useless, it is between two ground points.
You need a back EMF diode across the relay.
Phone only has 12V and ground connection. It cannot trigger anything the way it is drawn.
 
R3 is redundant/useless, it is between two ground points.
You need a back EMF diode across the relay.
Phone only has 12V and ground connection. It cannot trigger anything the way it is drawn.
I want a proper schema to cut off the current on the car pump using the electric signal of the phone vibrator that excites the transistor and which stimulates the relay coil and changes its state from normal close to normal open
But the problem is that as soon as the vibrating signal stops, The pump will work, So for that I used the MOSFET RIF540N because it will maintain the state unlike the transistor, Because I do not want it work until I come to it and restart it manually via the restart button like this video
 
I want a proper schema to cut off the current on the car pump using the electric signal of the phone vibrator that excites the transistor and which stimulates the relay coil and changes its state from normal close to normal open
But the problem is that as soon as the vibrating signal stops, The pump will work, So for that I used the MOSFET RIF540N because it will maintain the state unlike the transistor,

Why would you imagine that?, it certainly won't - it won't latch. Easiest solution is to use a double pole relay, and use the other pole to latch the relay permanently ON.
 
Simply connect the second pair of N/O relay contacts across source and drain of Q1, and make sure to fit a flyback diode across the relay coil (as already mentioned).
 
Simply connect the second pair of N/O relay contacts across source and drain of Q1, and make sure to fit a flyback diode across the relay coil (as already mentioned).
What I want is only a schema to control the relay with a transistor, but until I remove the exciter on the transistor, the relay remains in its state until I come to it and press the restart button after that change the state
 
Simply connect the second pair of N/O relay contacts across source and drain of Q1, and make sure to fit a flyback diode across the relay coil (as already mentioned).
you mean like that?
but the pump return active when excites transistor left , i need to steal in the same state when i click on the button of restart change the state like the initialy time
b1.PNG
 
If you've removed the FET, then you need a second pair of N/O relay contacts across collector and emitter of Q2, so it shorts the transistor out when the relay is activated.
 
How you connect the phone's vibrator terminals to the transistor will depend on whether the vibrator has a high-side switch or a low-side switch. Have you determined that?

Edit: If you intend the phone for general use, how will you prevent the pump switching off whenever there is an incoming call?
 
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Well, it looks like the OP has thrown his toys out of the cot.

Time to just close the thread?

JimB
 
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