How this circuit works?

sgardner025

Joined Nov 5, 2009
79
The diode is there to protect Q3 when the relay coil is de-energized, look up "back EMF". Switch 3 will cut the power going into the circuit. The capacitors are for filtering and decoupling. Look at a data sheet for a 7805, or any linear voltage regulator. The bottom circuit is a reversing circuit. When the relay is de-energized the motor will turn in one direction and when the relay is energized the motor will turn in the other direction. Look at how the relay contacts are wired and how the current path changes when the relay is switched on and off.
 

Thread Starter

zener_16

Joined Feb 1, 2010
9
Ty.
but i still do not quite understand the second circuit.
For example, when S2 is on, what will happen to the two relays, and how it changes the motor direction from the action of this two relay...
 

BMorse

Joined Sep 26, 2009
2,675
Ty.
but i still do not quite understand the second circuit.
For example, when S2 is on, what will happen to the two relays, and how it changes the motor direction from the action of this two relay...

The relay shown in the bottom of the pic is a double pole double throw relay, when S2 is closed it enables the motor to turn in one direction (if power is applied to J6), when the relay is activated, the relay will swap the 2 power wires on the motor, causing it to turn in the opposite direction.....

as for the top circuit, power to the circuit is provided via J7, when S1 is closed it provides power to the 7805 regulator.... the 7805 then provides the positive rail for the relay and the relay is actuated via the NPN transistor. when +V is applied to pin 3 of J8, this will turn on the transistor which will sink the relay to ground and activate it. D1 will prevent any back EMF from damaging the rest of the circuit when the relay power is turned off....

My .02
 
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Thread Starter

zener_16

Joined Feb 1, 2010
9
Ty.

Regarding "when S2 is closed it enables the motor to turn in one direction (if power is applied to J6), when the relay is activated," -BMorse

So, if (power is applied to J6) and my pin 3 of J8 is grounded, 0. Will my S2 be open, resulting the relay to not activated and makes the motor to turn in the opposite direction?

So the motor would still turn in one direction as you said (power is applied to J6), regardless of the value to the pin 3 of J8?

Am i correct?..
 

Thread Starter

zener_16

Joined Feb 1, 2010
9
sry s2 nids to be close inorder for the motor to work! yea i got it ^^
ty man! :)
 

JoeJester

Joined Apr 26, 2005
4,074
Both circuits make up the functional motor controller. A switch is connected to J8 to activate and deactivate the relay. The Motor power source is connected to J6. S1 in the motor power source positive terminal path is a safety disconnect.

AC power is connected to J7, most likely from the secondary of a transformer providing 9 volts to 24 volts reduction of the main power lines.
 
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