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I want to use the motor of an electric vehicle as a brake. It is a 36V ~27A motor.

I looked up 500W+ resistors and they are quite pricey, I figured the cheapest alternative would be 100W resistors in parallel.

schematic

simulate this circuit – Schematic created using CircuitLab

This is my slightly complicated schematic, I couldn't find a motor symbol, excuse the ignorance.

I don't need to reverse the motor so for the drive I'm just going to use a bunch of MOSFETs (yes I will use MOSFET driver ICs). PWM1 controls the amount of drive current while PWM2 controls the amound of braking, it will only be one or the other never both. I would probably need to add a few more P channel MOSFETS in parallel but I'll think about it later.

So anyway getting back to the problem, in order to achieve 27A of braking current I need a 36V/27A = 1.3 Ohm load. I will order ten 12 Ohm 100W resistors and wire them in parallel this should give me a resistance of about that and still leave some headroom for the resistors because we all know they probably can't live up to what's said on ebay (or at least for long).

What will happen if the resistance is lower than that? The vehicle is moving and is generating plenty of mechanical rotation on the motor shaft, we apply PWM2 100%. I suspect the motor is going to generate more than the rated 27A? If my assumption is correct then why is that? I mean it is clear that the torque on the motor shaft from the vehicle's inertia will be greater than what is required to generate 27A but why does that not just limit the amount of braking torque that the motor can apply?

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  • \$\begingroup\$ Doesn't you electric vehicle have a normal brake? If not, why not? \$\endgroup\$ – Andy aka Aug 6 '20 at 12:11
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    \$\begingroup\$ Most electric vehicle manufacturers prefer to dump the power back into the battery, rather than wasting it as heat. But that would require a more complicated charging circuit. \$\endgroup\$ – Simon B Aug 6 '20 at 12:34
  • \$\begingroup\$ Where is the 36V for the calculation 36/27 coming from? Is it guaranteed to be 36V? What is the lower and upper bound on that number? \$\endgroup\$ – AJN Aug 6 '20 at 12:58
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    \$\begingroup\$ "500W+ resistors and they are quite pricey," Me, looking at his 10€ hair dryer that's rated 1.8 kW and contains a lot more than just a resistor: "nah." Then, walking over to my electrical stove, rated 2.3 kW: "not really". Finally, sitting down, pondering whether to buy a 1 kW or a 500 W halogen floodlight to illuminate the wall I want to paint: "huh, pricey?". \$\endgroup\$ – Marcus Müller Aug 6 '20 at 13:21
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    \$\begingroup\$ @MarcusMüller My halogen worklight went through bulbs constantly; you don't want your brakes to fail when a bulb blows... Price will go up considerably with precision and reliability. \$\endgroup\$ – mbedded Aug 6 '20 at 15:55
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What will happen if the resistance is lower than that? The vehicle is moving and is generating plenty of mechanical rotation on the motor shaft, we apply PWM2 100%. I suspect the motor is going to generate more than the rated 27A?

Yes.

If my assumption is correct then why is that?

Because the rated current is not the maximum current the motor can actually generate. Maximum current is limited by armature and brush resistance. Since this is very low, the current can be very high. In motor mode this corresponds to applying full voltage while the motor is stalled. In generator mode it occurs when the motor is running at full speed and a short is applied across the terminals. In both cases current = voltage / resistance, and all the power drawn or generated is dissipated inside the motor. Unless the motor is grossly oversized for the job, this will quickly overheat it.

A PMDC motor's rated current is generally somewhere between 2 to 5 times less than the stall current. Why? Maximum possible power output is achieved at approximately half the stall current, when half the input power is dissipated inside the motor and the other half is delivered to the load (ie. motor efficiency is 50%). There is no point rating a motor higher than this because the power output at higher current is lower. To achieve higher efficiency the operating current must be less than half the stall current, and then the motor doesn't have to dissipate as much power so it can be made smaller and lighter (important for a traction motor).

What this means in practice is that you can draw several times the rated current for braking by applying a low resistance shunt, so long as the vehicle slows down quickly to limit motor heating. If the vehicle can accelerate from a standing start at full throttle (with no current limiting) without overheating the motor, then it should be safe to brake it with a low resistance because the current curve should be similar. If the throttle has to be applied slowly or current limited to protect the motor, then you should tailor the braking to draw similar maximum current.

voltage, current, torque and speed in DC motors

enter image description here

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What will happen if the resistance is lower than that?

There will be a higher current when the MOSFET is on.

I suspect the motor is going to generate more than the rated 27A? If my assumption is correct then why is that?

The motor will not generate more current than the load is drawing. To do that it would need to produce more voltage.

I mean it is clear that the torque on the motor shaft from the vehicle's inertia will be greater than what is required to generate 27A but why does that not just limit the amount of braking torque that the motor can apply?

No. That is not clear. The torque will be limited to what is produced by the electrical and mechanical losses plus the torque that the motor produces when generating 27 amps.

Perhaps the following is the answer to what you really want to know:

The basic principle of dynamic braking with a DC motor is that you shut off the supply and put a resistor across the armature. The resistor will draw current and cause the motor to act as a generator thus producing braking torque.

You can control the torque by selecting the resistor value. As the motor slows down, it will produce less voltage resulting in less current and less braking torque. Since changing the resistor is difficult, you start with the lowest possible value that is tolerable and reduce the initial voltage by using PWM.

You can brake with the highest current that is safe for the motor. That might be 150% to 200% of the rated motor current. I am not very familiar with the failure modes of DC motors, but I did experience flashing a commutator once. That is an arc that extends over several commutator segments or even all away around to form an arc between the brushes. That can occur with excess current. I believe that armature reaction at high current can demagnetize a permanent magnet motor.

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  • \$\begingroup\$ Well basically the question was why can a motor generate more current than rated for, rather than capping the maximum braking torque at the rated current and not generating more than that. It's kind of more of a fundamental question I guess. You have given valuable information, thank you! \$\endgroup\$ – php_nub_qq Aug 6 '20 at 20:01
  • \$\begingroup\$ See addition to my answer \$\endgroup\$ – Charles Cowie Aug 6 '20 at 20:55

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