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I have a chinese dc motor controller (YK31C, 350W) which has a connector for the brake. This connector will only interrupt the PWM Signal from the controller and does not short the motor for braking. I have access to the port that can interrupt the PWM Signal and to the Motor port and I try now to figure out a circuit for braking. Actually just a switch that would short both ports is not a good idea, since the dead time should be considered. Anyone an idea how this is normally done?

Edit:

I designed a circuit, but I am not sure if this really can work. On the first image you see the motor controller. It might not work because I cannot interrupt the supply voltage. PWM_OFF will deactive the PWM on the N-Channel MOSFET when grounded.

enter image description here

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  • \$\begingroup\$ Shorting the motor is only practical for very small motors/loads. A big motor needs to be braked by absorbing the power from it properly, either by regernative recharging the battery, or by dumping power in a resistor. \$\endgroup\$
    – Neil_UK
    May 14, 2017 at 11:32
  • \$\begingroup\$ Ok, so I will use a resistor array for dissipating the power. But I don't see how I actually could implement the brake at all. The motor itself will produce a negative Voltage to counteract it's movement. So I have to switch this Votlage to my resistor array in the right moment, that is when the PWM signal from the controller is turned off. I would like to use a mosfet to do so, but I think that's only possible with P-Mosfets, which requires a charge pump and make everything complicated. Anyone a idea how to do it simpler? \$\endgroup\$ May 14, 2017 at 12:10
  • \$\begingroup\$ What is the supply voltage? \$\endgroup\$ May 14, 2017 at 14:24
  • \$\begingroup\$ The supply voltage consists of two 12V car batteries in series. \$\endgroup\$ May 14, 2017 at 15:10
  • \$\begingroup\$ I am pretty sure now that it will work. If I turn of the PWM signal and short the motor ports, the torque to turn the motor increases significantly. Will transistor Q1 turn on until Vds gets too low and then it stops? \$\endgroup\$ May 14, 2017 at 15:30

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Your circuit is on the right track. Applying a resistor in parallel with the motor will work. The voltage produced by the motor depends on the direction it spins. It won't reverse just because you stop applying power.

The motor will keep the switch node at a voltage between the supply voltage and ground. At this point, shorting the motor or applying a resistor across its terminals will allow it to drive current opposite the direction it normally flows dissipating energy into the resistor and/or its own internal resistance.

Using an N-channel MOSFET is problematic. It should be either on (Vgate=10V) or off (Vgate=0 volts) as you have it now, the circuit applies whatever voltage is across the motor to the gate. As the motor slows down, this voltage will decrease and the MOSFET will eventually enter it's linear region. At that point, it will dissipate more and more of the braking power and will POP. You need some way of supplying a voltage higher than the supply voltage to keep the MOSFET fully turned on. This can be done with a bootstrap circuit possibly with a charge pump depending on how exactly the brake will be used. The alternative would be to use a P-channel MOSFET on the high side to switch the braking resistor into the circuit. This simplifies gate drive.

Of note, the triangle wave generator in the controller generates a square wave internally that could be used to drive a charge pump.

Here's what a charge pump based solution would look like: enter image description here

Note: the Zener diode should be 10-12 volts. It limits the voltage produced by the charge pump. The capacitor should be large enough to supply the current required by the turn off resistor while keeping the gate drive voltage at a reasonable level.

Here's what doing the same thing with an P-channel Mosfet would look like. Notice how much simpler the gate drive is. The resistor ratio determines the voltage ratio between the signal and the gate drive voltage.:

enter image description here

P-Channel MOSFETs may be a little more expensive than equivalent N-channel MOSFETs but not that much more. It's worth considering them for this application.

Here's how to do it with a relay:

enter image description here

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  • \$\begingroup\$ I thought there is no way to overcome the P-Channel Mosfets. I think I will first try out a relais before I implement a charge pump, because if the relais does not work properly, I will anyway design the whole controller by myself. The relais switching time should act as a low-pass filter, right? So the PWM signal will faster turn off over the BC547 than the relais can switch on? \$\endgroup\$ May 14, 2017 at 16:30
  • \$\begingroup\$ The relay switching will be delayed so when the brake is turned on, the PWM will turn off very quickly and the relay will take 3-4ms to turn on. This will keep the controller and relay from short circuiting the battery. \$\endgroup\$ May 14, 2017 at 17:18
  • \$\begingroup\$ Thank you very much, that looks fantastic. The P-Channel approach especially is interesting, since no charge pump is required and it allows variable braking by using a PWM braking signal, right? But I think a low-pass filter is required to ensure the PWM signal is turned off. \$\endgroup\$ May 14, 2017 at 17:29
  • \$\begingroup\$ Another question about the P-Mosfet circuit, why is the input voltage 5V? Would it be enough to use a 1kOhm resistor at the gate to VCC to use Vcc as input voltage? \$\endgroup\$ May 14, 2017 at 17:42
  • \$\begingroup\$ Driving MOSFETs with PWM signals is tricky. Every time they switch, they waste energy (ti.com/lit/an/slyt664/slyt664.pdf#page=2). Higher frequency means more switching and more heating, which can destroy the transistor. That's why the motor controller has the extra transistors for driving the MOSFETs. To turn them on and off faster so they heat up less during the transition. Additionally, layout is important when switching at higher speeds. Try to keep the PWM frequency as low as possible. \$\endgroup\$ May 14, 2017 at 17:44

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