why this auto motor turn off circuit needs?

Thread Starter

parksangtae

Joined May 23, 2021
5
KakaoTalk_20210608_000809000.jpg(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.
KakaoTalk_20210608_001238721.jpgi 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.
 

Alec_t

Joined Sep 17, 2013
12,064
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.
 
Last edited:

Irving

Joined Jan 30, 2016
1,496
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...
 

MrAl

Joined Jun 17, 2014
8,239
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.
 

Thread Starter

parksangtae

Joined May 23, 2021
5
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.
 

Thread Starter

parksangtae

Joined May 23, 2021
5
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
 

Irving

Joined Jan 30, 2016
1,496
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.
 
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