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I read in an online book at https://www.learncse.com/images/Lessons/Lesson_14.pdf following reason for using a voltage regulator to power DC motors. While I understand the importance of using a regulator in certain situations, I am not convinced of the reason given. Please let me know if the following is correct?

The lesson as posted at https://www.learncse.com/images/Lessons/Lesson_14.pdf is to control a DC motor using an H-bridge and interfaced with an Arduino.

One characteristic of the DC motor that is very important for the purposes of this lesson is the motor's electrical resistance. The slower the motor turns the lower its electrical resistance and, therefore, the greater its demand for electrical current.

Reason

The reason this is important is that in these lessons the motors are powered by batteries. Batteries, in turn, have an internal resistance. As the resistance of a motor goes down, the voltage across the motor also drops as the proportion of the battery's internal resistance to total resistance increases. The result is all motors being driven by the batteries slow. To the observer, the DC motors just seem to be running rather poorly.

Outcome

To minimize that effect, this lesson uses a battery with a higher voltage than what was intended for the motor and a voltage regulator to maintain the voltage delivered to the motor at a lower but constant value. This approach works because the internal resistance of the battery impacts the voltage being delivered to the regulator, not to the motor.

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  • \$\begingroup\$ I's hard to say this this is incorrect ot not without knowing the nature of the experiments to be performed. \$\endgroup\$ – Jasen Jan 15 at 9:23
  • \$\begingroup\$ I have added a link to the lesson information in my question. \$\endgroup\$ – John Galt Jan 16 at 7:04
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    \$\begingroup\$ Those quotes are a perfect example of why "a little" knowledge is a dangerous thing, and how the Internet enables sharing one's misconceptions. Voltage regulators are not used for motors. PWM drives are. That author then goes on to compound their ignorance by recommending a horribly lossy darlington bridge chip. Best to ignore their uninformed "advice". \$\endgroup\$ – Chris Stratton Jan 16 at 7:48
  • \$\begingroup\$ @ChrisStratton, I agree with you. Did you mean that the L293 is a bad choice here? How about L298 H-bridge IC? \$\endgroup\$ – John Galt Jan 19 at 6:44
  • \$\begingroup\$ L293, L298, etc are all darlington chips, and as a result all horrid. \$\endgroup\$ – Chris Stratton Jan 19 at 6:51
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It is important to consider that the first line of the quoted material contains the words "for the purposes of this lesson." The explanation that follows may be adequate for that purpose, but it is overly simplified for a general understanding. The phrase "As the resistance of a motor goes down" uses the term "resistance" to describe the combination of the motor's resistance and reverse electromagnetic force or "back EMF." The reason for motor speed decrease is not mentioned. The reason would probably be increased load torque. Also not covered is the performance of the motor from the time it is switched on until it reaches a stable operating speed,

In general, rather than a voltage regulator, it is more desirable to use and electronic speed controller (ESC) with a DC motor, and brushless DC motors can not function without an ESC. If an ESC is used, the ESC performs the voltage regulation in a manner that is more suitable for a motor, and an additional voltage regulator is not necessary. The quoted source may cover those points later in the lesson or in subsequent lessons.

Re: Comment & Link to Lesson

The lesson is a crude oversimplification. The motor used is very tiny when viewed in the universe of motors that includes motors propelling electric vehicles, ocean-going ships and industrial machinery. A brushed DC motor should be modeled as a resistor in series with a generator with polarity opposing the supply voltage. The voltage (back EMF) generated by that generator is proportional to motor speed. After accelerating to the maximum speed with no load, the back EMF is close to matching the supply voltage. Any load tends to reduce the motor speed and the EMF allowing increased current so that the current multiplied by EMF is equip to the mechanical power produced. The series resistance of an efficient motor may dissipate only 10 or 20 percent of the power taken from the supply with the remainder converted to mechanical power.

Find a lesson that uses the resistor - back EMF model. The one you have found does a terrible injustice to electrical engineering.

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  • \$\begingroup\$ Charles, please see my comment to Puffafish. I have also provided a link to the chapter [link] (learncse.com/images/Lessons/Lesson_14.pdf) \$\endgroup\$ – John Galt Jan 16 at 7:16
  • \$\begingroup\$ @John Galt: See material added to answer. \$\endgroup\$ – Charles Cowie Jan 16 at 9:52
  • \$\begingroup\$ "important" makes no sense unless you say "important for what". If the "what" is "maintain the motor speed at the requested speed despite changes in power and load", then an ESC is appropriate -- but I don't think that's what the chapter author was talking about. \$\endgroup\$ – Scott Seidman Jan 18 at 20:38
  • \$\begingroup\$ This makes sense Charles. Thanks for clarifying this. If you can point me to a good source to read about this for a non-EE guy, that would be great. It can be an online source or a book. \$\endgroup\$ – John Galt Jan 19 at 6:45
  • \$\begingroup\$ My references are all EE text books. I bought an older edition of Fitzgerald, Kingsley, Umans Electric Machinery on eBay. Overall, I believe it is the best I have. I believe there are sources online that are not EE texts but better than the one that you found. You may need to look at several to find one that suits you. \$\endgroup\$ – Charles Cowie Jan 19 at 16:23
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It is correct.

The voltage out of the battery will vary depending on the load out of the battery. This is due to the battery's internal impedance. This means that the higher the current out of the battery, the lower the voltage of the battery.

In an ideal world, the battery will provide whatever current is drawn at a fixed voltage. However a battery could be seen as a resistor in series with a voltage source. So as current goes up voltage drops.

The regulator here is going to draw current out of the battery so that it's output is the required voltage. This is because the regulator actively adapts it's drive to give a constant voltage. This should give you a more consistent performance out of your motor.

Clearly the regulator is still limited; draw too much current from the battery and the voltage will be too low for the regulator to operate. Draw too much out of the regulator and something will fail (depending on the design of the regulator).

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  • \$\begingroup\$ As the load at the motor increases, the current requirement would go up. So, using your logic, if a battery is seen as a resistor in series with a voltage source, then the voltage across the source would drop. This would necessitate a voltage regulator to maintain constant voltage. I understand this much. However, I still don't understand how a slower motor causes less resistance in motor. If by resistance, they mean back EMF, then that would be higher due to higher load. The resistance in the armature is always fixed. Is n't it? \$\endgroup\$ – John Galt Jan 16 at 7:09
  • \$\begingroup\$ Back EMF is related to the speed, not the load - so if anything, high load which means low speed means low back EMF, high current, and more of the supply voltage dropped elsewhere. But that author's proposal to use a linear voltage regulator is an impractical synthesis from two pieces of their partial knowledge. If you want constant speed from an ordinary motor, you need feedback control, and you throttle the motor with an efficient switching drive, not the feedback controlled heater that is the author's recommended linear voltage regulator. \$\endgroup\$ – Chris Stratton Jan 16 at 8:13
  • \$\begingroup\$ @ChrisStratton, did you mean by feedback control using an encoder wheel with a sensor and PID control to monitor and regulate speed? What do you mean by "an efficient switching drive"? Examples? \$\endgroup\$ – John Galt Jan 19 at 6:47
  • \$\begingroup\$ No. A Linear voltage regulator is a feedback controlled heater which produces the amount of waste heat necessary to result in the nameplate output voltage at the current being drawn. Conversely, a switching drive applies power only the fraction of time needed, wasting little. \$\endgroup\$ – Chris Stratton Jan 19 at 6:50
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From your comment to Puffafish,

I still don't understand how a slower motor causes less resistance in motor. If by resistance, they mean back EMF, then that would be higher due to higher load. The resistance in the armature is always fixed. Is n't it?

You need to be aware that (as others have already commented) the lesson you're referring to is very badly done. Specifically, you can think of a motor having an "effective resistance" which varies, which having a fixed armature resistance. The key is to include the back EMF. In effect, a DC motor can be modelled as a resistance (the winding resistance, which is pretty much constant, although copper does change a good deal with temperature) in series with a generator, which provides a voltage (the back-EMF) which is proportional to the speed of the rotor. So a circuit with a voltage source will look like

schematic

simulate this circuit – Schematic created using CircuitLab

For a fixed voltage in, the current drawn by the motor will be affected by the back-EMF, and therefor the speed, of the motor. Maximum current will be drawn when the motor speed is zero, and this can be very high. It's called "stall current", for obvious reasons. As the motor speeds up, the current will be determined by the difference between the input voltage and the back-EMF. The lesson you linked to ignores any attempt to measure the motor speed, and explains that change in current as saying that the "resistance" of the motor increases with speed. If you change that to "effective resistance", the concept is not entirely unreasonable. At least, for the purposes of the lesson.

The lesson uses a voltage regulator to ensure that, for a PWM H-bridge, the motor speed will be more-or-less proportional to the PWM duty cycle. If it doesn't, the motor will appear to be unaccountably weak at low speeds, since a battery voltage will drop at high current levels, which will occur when the motor speed is low, back-EMF is low, and "effective resistance" is also low.

Note that this effectively presumes what is called "open-loop" speed control. that is, there is no feedback on what the motor speed actually is. Any real system which actually cares about what the motor is doing will include a shaft sensor and feedback to adjust the duty cycle of the H-bridge. Once you do this, there is no need for a voltage regulator - the motor controller will adjust the PWM to produce the speed required.

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  • \$\begingroup\$ thanks. I understand it now. The only real way to keep the motor speed constant is to use feedback control. \$\endgroup\$ – John Galt Jan 22 at 4:43

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