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Let's pretend we have a standard 4 MOSFET H-bridge that controls a 12V brushed DC motor. When driving the motor forward, we can keep SW4 on continuously and pulse SW1 on and off (see below image). If I understand correctly, in this case during the off-state, the motor will be essentially open-circuited, and no current will flow through it, in what is called "coasting".

enter image description here

However I stumbled upon a motor driver IC (DRV8812) which in one particular mode of operation (EN/PH mode, see below image), forces you to switch between "forward" and "brake" when using PWM (assuming the PWM frequency is faster than the tSLEEP timeout of 1ms or so). When EN goes low (the datasheet says to connect the PWM to EN), it turns both SW2 and SW4 on, causing the motor to "brake". A first glance this seems like a bad idea, why would you want to brake and waste more energy than you need to when driving the motor (and slow it down). Is there a benefit to this style of control?

enter image description here

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2 Answers 2

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A first glance this seems like a bad idea, why would you want to brake and waste more energy than you need to when driving the motor (and slow it down).

When the motor is momentarily forced into a braking situation, the current remains largely the same throughout that small part of the cycle. The alternative is to leave the MOSFETs open and let the energy release back to the main power supply bulk capacitor so that it can be used again.

One problem in doing the latter is that you have to draw that current into the motor all over again and, that wastes energy in the switching transistor (significantly higher peak currents) so, in many applications, it's better to momentarily short out the motor and keep that magnetic energy where it could be useful.

In case you didn't know this method of control is called phase shifted PWM and is used extensively in higher power applications (not just motors) but, in series resonant converters for instance.

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  • \$\begingroup\$ So motors behave much more like inductors do then I first thought? I was under the assumption that if you left the MOSFETs open, because the back EMF would always be less than the supply voltage (unless it was sped up), it wouldn't be able to return current to the source. If however motors behave mostly like an inductor, the inductance would be able to generate a high-enough voltage to keep the current flowing in the same direction. This also correlates with you saying the current remains largely the same (i.e. the motor has enough inductance to behave like a buck or boost converter). \$\endgroup\$
    – gbmhunter
    Jul 20 at 9:07
  • \$\begingroup\$ ti.com/lit/an/slva321a/slva321a.pdf helps explain it also \$\endgroup\$
    – gbmhunter
    Jul 20 at 9:19
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    \$\begingroup\$ Precisely: In the very short time you might leave a motor open circuit, the inductance will dominate that short time period and return its energy to the bulk capacitor on the bridge supply via the body diodes of the relevant MOSFETs. The back emf from the inductance will exceed the supply by exactly two diode drops in returning this stored energy. \$\endgroup\$
    – Andy aka
    Jul 20 at 9:19
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It is all in the description column. What do you think "low-side slow decay" means?

When you short-circuit the motor, current does not simply "disappear". The inductance of the winding will try to keep it going in the same direction, and the bridge provides a path for it. As a result, the motor will continue running in the same direction, but with gradually reducing speed.

If the duty cycle is short (i.e. motor running slow) or if you turn EN off then eventually the bridge will go into Hi-Z mode, quickly removing current. However with long duty cycle it will hit the next active period and continue with increased current right away.

The end result is that it is much more efficient than coasting, not less.

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  • \$\begingroup\$ I didn't realize motors behaved like so much like inductors :-O I thought they had only a small parasitic inductance, and were mostly modelled by a back EMF proportional to the rotational speed, and a series parasitic resistance. I thought that the current during the off part of the duty cycle would switch direction through the motor because of the back EMF (current would not be able to switch that fast if it had a large inductance). It sounds like current continues in the same direction during the off part? \$\endgroup\$
    – gbmhunter
    Jul 20 at 9:18
  • \$\begingroup\$ ti.com/lit/an/slva321a/slva321a.pdf helps explain it also \$\endgroup\$
    – gbmhunter
    Jul 20 at 9:19

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