(citcuit1)
hello everybody, i want to ask about this circuit need.

To put it simply, when the switch is closed, motor turns on and when the switch is opend, motor turns off.
i think this circuit's function is same with that circuit.
(circuit2)

isn't it right?
To make an inference, when the circuit 1 is opend and charged capaciotor is dischared. And jfet(Q2) is on. and remained electric current can go out.
so if i use circuit 1, i can use motor long.

 

It appears a normal suitably rated DC switch should do it, with arc suppression across it.
.

 

I think circuit-1 is intended to provide a braking effect on the motor when the switch S1 is opened. This occurs because the charge on C1 turns on Q2, shunting the motor's back-EMF. Apart from leakage current, C1 has no obvious discharge path.

Edit: 'back-EMF' would be better stated as 'motor-generated EMF' because the motor acts as a generator while slowing down.

 

I think circuit-1 is intended to provide a braking effect on the motor when the switch S1 is opened. This occurs because the charge on C1 turns on Q2, shunting the motor's back-EMF. Apart from leakage current, C1 has no obvious discharge path.

Exactly so, I was about to post the same...

 

There is just one question.
Most power MOSFET's have a reverse diode drain to source so when the switch is turned off and the motor develops some back EMF that reverse diode would conduct thus shunting the back EMF current to ground.
So what do the MOSFET's do then...do they actually do anything.

 

I think circuit-1 is intended to provide a braking effect on the motor when the switch S1 is opened. This occurs because the charge on C1 turns on Q2, shunting the motor's back-EMF. Apart from leakage current, C1 has no obvious discharge path.

No, there is a way for C1 to discharge. If Q1's gate has 0V, Q1 will not work and C1 will be discharged, and Q2 will operate by putting a voltage on Q2's gate. Then the remaining electricity in the motor escapes through Q2.

 

I think circuit-1 is intended to provide a braking effect on the motor when the switch S1 is opened. This occurs because the charge on C1 turns on Q2, shunting the motor's back-EMF. Apart from leakage current, C1 has no obvious discharge path.

I heard there's a brake effect on circuit 1, isn't that the same for circuit 2? Eventually, if the voltage doesn't go off, the motor won't turn

 

If Q1's gate has 0V, Q1 will not work and C1 will be discharged

C1 charges via D1, but there is no discharge path through D1, Q1 or Q2 gate.
In circuit 2, opening the switch will allow the motor to coast to a halt, but without any braking effect provided by the circuit.

 

If Q1's gate has 0V, Q1 will not work and C1 will be discharged

Why do you think Q1 when it's off has a low resistance? Most MOSFETs, when off, are very high impedance drain to source (several megaohms).

Eventually, if the voltage doesn't go off, the motor won't turn

When you remove power from a motor it continues to spin due to inertia of the rotor and attached load (flywheel, etc). In many environments this can be dangerous. While spinning the motor acts as a generator but with nothing connected to it as in circuit 2 only friction is absorbing that rotational energy. By putting a braking resistor across the motor terminals current flows in the motor windings generating power which absorbs the rotational energy by converting it to heat in the braking resistor. If the resistor is a low enough value the motor will stop very quickly and safely. This is standard practice on almost all industrial machinery.