how to construct a DC motor circuit by changing the motor polarity using relay?

[alexan_e]

I assume that you are asking for the real circuit because the 555 model in proteus doesn't turn off.
Your relay can only control the direction of rotation , how do you expect to stop the motor?
It can only be stopped using the switch close to the motor supply and this is not controlled by the 555, it is a manual switch.
The motor voltage should be the operating voltage of your motor, is should have some specification, this voltage is irrelevant to the relay voltage.

Alex

 

[EESeeker]

actually wat i plan to design is once there is no light the motor wouldnt run as the 555 timer do not trigger any output. therefore, no voltage produced which sufficient to rotate the motor instead of adding the manual switch there to stop it. So, my ques is : can i connect the output of the NPN to the motor part ? as i know the output voltage from NPN that i observe in protues is just around 5 V ...so how to increase the voltage so it is sufficient to run a 12 V dc motor? will the motor stop rotating if the light intensity is not strong enough?
(becoz the proteus couldnt stop the 555 timer as it triggered once, so asking is it applicable to real board)

thx=)

 

[alexan_e]

In a real board the 555 is triggered to turn on when the voltage at pin 2 is below 1/3 Vcc, if it goes above that it will not trigger and it will urn off after the delay period.
Your transistor in your circuit can only control the relay and the relay is wired to invert the motor polarity, not stop the motor.
If you want to stop the motor then you should remove the power supply from one position of the relay so that it can stop the motor.
You can either use the relay to control the motor or the transistor directly, with the relay you will have lower losses.
If you are going to use the transistor then depending on the motor current you will need more current to the base of the transistor.
A third solution is a mosfet, which will give the lower losses possible with no moving parts (relay)

One more note, you need to use a resistor in the base of the transistor to limit the base current

Alex

 

[EESeeker]

wat do u mean by use a resistor in the base of the transistor ? i thought the R4 in the schematic is the one u mean?

 

[alexan_e]

The base resistor I mean should be connected between pin 3 and the base to limit the base current because you can damage the transistor without the resistor.
If you are using a small relay that uses up to 200mA then you don't need a transistor at all, connect the relay directly to the output of the 555.

Take a look
www.jaycar.com.au/images_uploaded/relaydrv.pdf

Alex