Why is it harder to turn a motor/generator with shorted terminals?24V H-Bridge DC motor, why voltage drop to 3v if motor connecteddriving a synchronous motor as a generatorRotate BLDC motor with Clark and Park transformWhy does the stator field rotate at the same speed as the rotor field in a synchronous generator?How to allow a motor to free spin?Torque control of a BLDC motor acting as a generatorCan a BLDC generator power another identical brushless motor?gear motor can be used as generatorDc motor as torque transducer for BLDC motorWhy does shorting the phases of a BLDC motor cause it to lock?
Quotient of Three Dimensional Torus by Permutation on Coordinates
Does the talk count as invited if my PI invited me?
Will this series of events work to drown the Tarrasque?
French equivalent of the German expression "flöten gehen"
Why are there five extra turns in tournament Magic?
Windows reverting changes made by Linux to FAT32 partion
Why does a table with a defined constant in its index compute 10X slower?
When did Britain learn about the American Declaration of Independence?
Largest memory peripheral for Sinclair ZX81?
Why does the U.S military use mercenaries?
Why does string strummed with finger sound different from the one strummed with pick?
FIFO data structure in pure C
How come Arya Stark wasn't hurt by this in Game of Thrones Season 8 Episode 5?
Prints each letter of a string in different colors. C#
Does a windmilling propeller create more drag than a stopped propeller in an engine out scenario
What's is the easiest way to purchase a stock and hold it
Good examples of "two is easy, three is hard" in computational sciences
Why is Drogon so much better in battle than Rhaegal and Viserion?
Referring to a character in 3rd person when they have amnesia
What color to choose as "danger" if the main color of my app is red
Why does the setUID bit work inconsistently?
What technology would Dwarves need to forge titanium?
Save my secrets!
pwaS eht tirsf dna tasl setterl fo hace dorw
Why is it harder to turn a motor/generator with shorted terminals?
24V H-Bridge DC motor, why voltage drop to 3v if motor connecteddriving a synchronous motor as a generatorRotate BLDC motor with Clark and Park transformWhy does the stator field rotate at the same speed as the rotor field in a synchronous generator?How to allow a motor to free spin?Torque control of a BLDC motor acting as a generatorCan a BLDC generator power another identical brushless motor?gear motor can be used as generatorDc motor as torque transducer for BLDC motorWhy does shorting the phases of a BLDC motor cause it to lock?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between.
Why is this? (I'm using a BLDC motor.)
motor dc-motor generator
asked May 12 at 20:24
abcabc
320315
$endgroup$
add a comment |
$begingroup$
The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between.
Why is this? (I'm using a BLDC motor.)
motor dc-motor generator
asked May 12 at 20:24
abcabc
320315
$endgroup$
4
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36
add a comment |
$begingroup$
The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between.
Why is this? (I'm using a BLDC motor.)
motor dc-motor generator
asked May 12 at 20:24
abcabc
320315
$endgroup$
The shaft of an unconnected motor is easy to rotate relative to a motor with shorted terminals. If a resistive load is connected to the terminals, the turning difficulty is somewhere in between.
Why is this? (I'm using a BLDC motor.)
motor dc-motor generator
motor dc-motor generator
asked May 12 at 20:24
abcabc
320315
asked May 12 at 20:24
abcabc
320315
edited May 13 at 15:13
abc
asked May 12 at 20:24
abcabc
320315
asked May 12 at 20:24
abcabc
320315
asked May 12 at 20:24
abcabc
320315
320315
4
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36
add a comment |
4
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36
4
4
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
$endgroup$
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
add a comment |
$begingroup$
"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
$endgroup$
add a comment |
$begingroup$
As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
answered May 13 at 11:20
i336_i336_
16210
$endgroup$
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("schematics", function ()
StackExchange.schematics.init();
);
, "cicuitlab");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "135"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f438150%2fwhy-is-it-harder-to-turn-a-motor-generator-with-shorted-terminals%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
$endgroup$
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
add a comment |
$begingroup$
I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
$endgroup$
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
add a comment |
$begingroup$
I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
$endgroup$
I have to start with some terminology -- sorry if it's esoteric, but this will bring things into line with how folks talk about this subject.
When you turn a permanent-magnet DC machine*, the armature generates a voltage internally. This is called the "EMF"** of the armature, or the "back EMF" if the machine is running as a motor. This EMF is always generated when the machine turns.
When you run current through a DC machine, it generates a torque. This torque is always generated when the machine turns, regardless of whether it's a motor or a generator.
When you put a resistance on the terminals of a machine and turn its shaft, it generates that EMF. With the resistance connected, this EMF causes a current to flow that's proportional to the EMF divided by the external resistance plus the machine's armature resistance. This current, in turn, generates a torque that resists motion (due to conservation of energy, it must be in a direction to resist motion).
Shorting the machine puts the smallest possible resistance on it -- you can't get lower than 0 without resorting to active circuitry. The back torque in this case is purely a product of the EMF and the armature resistance. Increasing the resistance by putting a resistor on there means less current for the same machine speed, which means less back torque. In the extreme, you have no resistor at all, which means infinite electrical resistance -- this means that the back torque will be from mechanical effects such as friction (and windage, if you're turning it that fast), and possibly mechanical and electromechanical effects as the field magnets work against the iron in the armature.
* I'm calling it a "machine" instead of a "motor" because it can be a motor or a generator, depending on how you use it. But you don't have to change anything internally to change how it's used -- hence, "machine".
** EMF stands for "electromotive force", which is just and older term for "voltage". It seems silly to have two terms, but sometimes it's useful.
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
answered May 12 at 20:42
TimWescottTimWescott
8,4441718
8,4441718
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
add a comment |
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
1
1
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
I appreciate the fundamental explanation. I find a lot of information regarding the "whats" of DC motor operation, but the "whys" are harder to come by.
$endgroup$
– abc
May 12 at 21:14
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
$begingroup$
You mention active circuitry--are there examples of motor drives that actively introduce current in response to a back EMF to provide better braking than shorting the terminals can provide?
$endgroup$
– Andrey Akhmetov
May 12 at 23:44
3
3
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@AndreyAkhmetov yes. In fact, it's possible to build an amplifier whose output impedance is negative and equal in magnitude to a motor's armature resistance. Then for the purposes of motor dynamics, the system comes close to acting like a motor with zero-resistance winding. Speed regulation is much (but not perfectly) improved, including regulating down to speed = 0. I'm not sure if it's been used for motor braking, but it was used for a while in the 1970's to regulate the motor speed of cassette tape drives in sorta-high-end consumer audio equipment.
$endgroup$
– TimWescott
May 12 at 23:52
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
$begingroup$
@TimWescott Neat, thanks!
$endgroup$
– Andrey Akhmetov
May 13 at 0:00
1
1
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
$begingroup$
@AndreyAkhmetov If you want a fine level of control, you could do what Tim said, but for a quick-and-dirty method, you could just drive the motor in the opposite direction. (staying mechanically in sync, of course) This also ends up with regenerative braking, so make sure that your upstream circuitry can handle that.
$endgroup$
– AaronD
May 13 at 4:34
add a comment |
$begingroup$
"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
$endgroup$
add a comment |
$begingroup$
"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
$endgroup$
add a comment |
$begingroup$
"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
$endgroup$
"applying a resistive load" to a running motor is essentially how an electric brake works. As a first approximation, the torque produced by the motor is proportional to the current, that's turning the motor is harder as the load resistance gets smaller. When you short the terminals, there's only the internal resistance of the motor which limits the current.
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
answered May 13 at 11:11
Dmitry GrigoryevDmitry Grigoryev
18.9k22978
18.9k22978
add a comment |
add a comment |
$begingroup$
As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
answered May 13 at 11:20
i336_i336_
16210
$endgroup$
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
add a comment |
$begingroup$
As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
answered May 13 at 11:20
i336_i336_
16210
$endgroup$
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
add a comment |
$begingroup$
As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
answered May 13 at 11:20
i336_i336_
16210
$endgroup$
As I read the accepted answer my brain came up with the following simplification, which I think is loosely accurate (?):
Motors are both dynamos and electromagnets.
Turning a motor invokes its properties as a dynamo.
Because the motor's terminals are shorted together, the generated voltage is applied to the motor coil windings, invoking the motor's properties as an electromagnet on its own axle.
answered May 13 at 11:20
i336_i336_
16210
answered May 13 at 11:20
i336_i336_
16210
answered May 13 at 11:20
i336_i336_
16210
answered May 13 at 11:20
i336_i336_
16210
16210
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
add a comment |
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
$begingroup$
Every motor is also a generator. Put a mechanical drag on it, and it draws electrical power. Apply torque (nagative drag), and it delivers electrical power. The first law of thermodynamics is in control.
$endgroup$
– richard1941
4 hours ago
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f438150%2fwhy-is-it-harder-to-turn-a-motor-generator-with-shorted-terminals%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
$begingroup$
Normally electrical power sources are constant voltage, so a load with a smaller resistance is considered to be a larger load. Could you edit your title, please?
$endgroup$
– TimWescott
May 12 at 20:32
$begingroup$
Not in my my experience, shorted terminals is more difficult with a permanent magnet DC brushed or BLDC motor. Be specific about the type of motor you are using.
$endgroup$
– Neil_UK
May 12 at 20:33
$begingroup$
@Neil_UK I agree with you. I think that's what I stated in the description.
$endgroup$
– abc
May 12 at 20:36