Regenerative/Dynamic Breaking with minimal components.

Thread Starter

anishkgt

Joined Mar 21, 2017
387
Hi All,

I was looking for a simple braking circuit for a DC motor like the 775 rate 12-24V. here is a circuit that i had come up with. The components names are generic, yet measure the reverse current from the motor.

Would appreciate if somebody could advise how ideal this would be ? Or should i go with an H-Bridge that has the breaking built into it ?
Capture.PNG
 

Thread Starter

anishkgt

Joined Mar 21, 2017
387
JP1 here is DPDT switch which would connect pins 1 and 2 to switch ON the motor and pin 2 and 3 would short the motor terminals with a resistor in series to dissipate the heat. I could send it back to vcc rail but this is powered by another PWM controller with a soft start so not sure if i should be doing it that way.
 

kubeek

Joined Sep 20, 2005
5,647
If you want to push the energy back into battery, then the voltage coming from the motor needs to be higher than the battery voltage. You will need a step up converter.
 

LesJones

Joined Jan 8, 2017
2,417
I think the simplest way to add braking (Not regenerative.) is to to use a single pole double throw relay and a resistor. The relay coil would be directly across the supply. One side of the motor. (We will say negative.) is connected to the supply negative. The other side of the motor is connected to the common on the relay contacts. The positive supply is connected to the NO contact on the relay via a diode rated for the motor current. (Cathode to the relay contact.) The NC contact is connected to the supply negative via the braking resistor.

Les.
 

LesJones

Joined Jan 8, 2017
2,417
Not quite right. Diode D2 should be between the power supply negative and connection S (NO contact.) on your schematic. As it is in the negative supply the anode should be connected to S on your relay. The only thing connected to terminal O on your relay should be the 25 ohm load resistor, The diode is to prevent the EMF from the motor which will still be rotating from holding the relay in. Diode D2 would also provide reverse polarity protection so you do not need D1.
When the power is first removed from the motor it will be producing almost the supply voltage at first. So if the supply is 12 volts the current will by 12/35 amps = 342 mA. If the supply voltage is 24 volts the the initial current will be 24/35 = 686 mA. With a 24 volt supply the initial resistor dissipation will be about 16 watts. As this will be for a very short time I think a 5 watt rating resistor will be OK. (The 2 watt resistor should be OK with a 12 volt supply.) The relay coil voltage rating needs to be the same as the supply voltage.

Les.
 
Last edited:

Thread Starter

anishkgt

Joined Mar 21, 2017
387
Thank you Les.

Since I don’t have an H bridge, how fast would it brake compared to this design ?

Would it be like this or an H bridge would give better braking ?
 

Thread Starter

anishkgt

Joined Mar 21, 2017
387
When you said ‘between the negative supply and connection S’ did you mean in series with S where anode to S and cathode to negative ?
 

LesJones

Joined Jan 8, 2017
2,417
I can't understand how the circuit in post #15 works. The circuit in post #13 should work but as you now say you motor takes 5 amps you would need larger current rating diodes. You don't need both diodes. You can remove either one. This will reduce the voltage drop to your motor by about 0.7 volts. Using a 35 ohm braking resistor will not give very good braking. I would use a 1 to 2 ohm resistor. (Or even a direct connection between relay terminal "O" and the motor positive.) One thing you need to be aware of is that if the blade is held in place with a nut then it will tend to unscrew during braking.

Les.
 

Thread Starter

anishkgt

Joined Mar 21, 2017
387
I can't understand how the circuit in post #15 works. The circuit in post #13 should work but as you now say you motor takes 5 amps you would need larger current rating diodes. You don't need both diodes. You can remove either one. This will reduce the voltage drop to your motor by about 0.7 volts. Using a 35 ohm braking resistor will not give very good braking. I would use a 1 to 2 ohm resistor. (Or even a direct connection between relay terminal "O" and the motor positive.) One thing you need to be aware of is that if the blade is held in place with a nut then it will tend to unscrew during braking.

Les.
I initially did try without the resistor but it took some time to stop.

In the video in post#15 he explains.
 

LesJones

Joined Jan 8, 2017
2,417
Are you sure that it is a permanent magnet motor ? It will not work with series wound (Universal.) motors. It should be posible to make it work with shunt wound motors but the field winding would have to be powered while the armature was shorted. (Either directly or via a resistor.)

Les.
 

LesJones

Joined Jan 8, 2017
2,417
If you use a single pole double throw switch to switch the motor on and off you do not even need a relay or diode. If you don't need to prevent it from stopping too quickly you dont even need a resistor.

Les.
 
Top