Comment améliorer le couple et le régime d'un moteur à courant continu?

J'ai un moteur FA-130 (DC) avec aimant permanent, ma source d'alimentation est 2 piles AA (rechargeables) donc un total de 2.4v.

Supposons que tous les cas commencent à partir de la même spécification, théoriquement, que se passerait-il si je fais ce qui suit?

Cas 1 : augmentation / diminution de la force des aimants permanents. Qu'arriverait-il au couple et au régime? Pourquoi?

Cas 2 : augmentation / diminution de la taille des fils magnétiques. Qu'arriverait-il au couple, à la consommation d'énergie et au régime? Pourquoi?

Cas 3 : augmenter / diminuer la taille de l'armature. Qu'arriverait-il au couple, à la consommation d'énergie et au régime? Pourquoi?

Cas 4 : augmentation / diminution du nombre de tours (bobine). Qu'arriverait-il au couple, à la consommation d'énergie et au régime? Pourquoi?

En général, comment puis-je augmenter le couple et le régime de ce moteur avec une tension constante?

S'il vous plaît, expliquez-le comme si vous parlez à un enfant de 6 ans, je ne suis pas bien informé dans ce domaine mais je veux connaître le concept.

dc-motor dpp
la source

peut-être que ce modèle sera utile, robotics.ee.uwa.edu.au/courses/embedded/tutorials/tutorials/…

Standard Sandun

il le fait généralement, mais pas toujours, car il existe différents paramètres dans ce modèle. Vemf représente sa tension de retour. Je pense donc comment je pourrais l'expliquer à un enfant de 6e année.

Standard Sandun

@sandundhammika merci, je pourrais peut-être mettre un peu plus d'effort, vous pouvez me considérer comme un enfant de 12 ans maintenant qui ne connaît rien à l'électronique ...

dpp

Je pensais que vous demandiez comment enseigner cela à un enfant de 6e année, je suis confus, désolé.

Standard Sandun

Je pense que je comprenais en quelque sorte les CEM maintenant, je l'ai vu, "Chaque fois qu'un inducteur (dans ce cas, la bobine) traverse un champ électrique, il crée une tension. C'est ainsi que les générateurs fonctionnent. C'est toujours vrai lorsque le moteur tourne sous propre puissance. Mais, cette tension va dans le sens opposé à la tension que nous mettons dans le moteur pour le faire tourner, donc il soustrait. C'est ce qu'on appelle la tension de retour ou EMF arrière. À une certaine vitesse, la tension de retour est égale à la tension nous mettons dans le moteur, et (dans un monde parfait), lorsque le moteur est au maximum en tr / min, et aucun courant électrique, donc pas de courant. "

dpp

Réponses:

Je vais supposer que cet enfant de 6 ans a au moins un peu d'expérience en physique. Je vais commencer par répondre à la raison pour laquelle chaque résultat se produira avec beaucoup de mathématiques pour décrire la physique derrière tout cela. Ensuite, je répondrai à chaque cas individuellement avec les mathématiques fournissant le raisonnement derrière chaque résultat. Je terminerai en répondant à votre question «en général».

Pourquoi?

La réponse à tous vos "Pourquoi?" questions est: la physique! Plus précisément la loi de Lorentz et la loi de Faraday . D' ici :

lorentz and faraday

Le couple du moteur est déterminé par l'équation:

τ = K_{t} \cdot I (N \cdot m)

$\tau = K_t \cdot I~~~~~~~~~~(N \cdot m)$

Où:

$\tau = \text{torque}$
$K_t = \text{torque constant}$
$I = \text{motor current}$

La constante de couple, , est l'un des principaux paramètres du moteur qui décrivent le moteur spécifique en fonction des différents paramètres de sa conception tels que la force magnétique, le nombre de tours de fil, la longueur de l'armature, etc., comme vous l'avez mentionné. Sa valeur est donnée en couple par ampère et est calculée comme suit: $K_t$

K_{t} = 2 \cdot B \cdot N \cdot l \cdot r (N \cdot m / A)

$K_t = 2 \cdot B \cdot N \cdot l \cdot r~~~~~~~~~~(N \cdot m / A)$

Où:

$B = \text{strength of magnetic field in Teslas}$
$N = \text{number of loops of wire in the magnetic field}$
$l = \text{length of magnetic field acting on wire}$
$r = \text{radius of motor armature}$

La tension Back-EMF est déterminée par:

V = K_{e} \cdot ω (v o l t s)

$V = K_e \cdot \omega~~~~~~~~~~(volts)$

Où:

$V = \text{Back-EMF voltage}$
$K_e = \text{voltage constant}$
$\omega = \text{angular velocity}$

La vitesse angulaire est la vitesse du moteur en radians par seconde (rad / sec) qui peut être convertie à partir de RPM:

rad/sec = RPM \times \frac{π}{30}

$\text{rad/sec} = \text{RPM}\times\dfrac{\pi}{30}$

est le deuxième paramètre moteur principal. Curieusement, est calculé en utilisant la même formule que $K_e$ $K_e$ $K_t$ mais est donné dans différentes unités:

K_{e} = 2 \cdot B \cdot N \cdot l \cdot r (v o l t s / r a d / s e c)

$K_e = 2 \cdot B \cdot N \cdot l \cdot r~~~~~~~~~~(volts/rad/sec)$

$K_e = K_t$

$P_{in} = P_{out}$
$V \cdot I = \tau \cdot \omega$

Substituting the equations from above we get:

$(K_e \cdot \omega) \cdot I = (K_t \cdot I) \cdot \omega$
$K_e = K_t$

Cases

I'm going to assume that each parameter is being changed in isolation.

Case 1: Magnetic field strength is directly proportional to the torque constant, $K_t$ . So as magnetic field strength is increased or decreased, the torque, $\tau$ , will increase or decrease proportionally. Which makes sense because the stronger the magnetic field, the stronger the "push" on the armature.

Magnetic field strength is also directly proportional to the voltage constant, $K_e$ . However $K_e$ is inversely proportional to the angular velocity:

ω = \frac{V}{K_{e}}

$\omega = \dfrac{V}{K_e}$

So, as the magnetic field increases, the speed will decrease. This again makes sense because the stronger the magnetic field, the stronger the "push" on the armature so it will resist a change in speed.

Because power out is equal to torque times angular velocity, and power in equals power out (again, assuming 100% efficiency), we get:

P_{i n} = τ \cdot ω

$P_{in} = \tau \cdot \omega$

So any change to torque or speed will be directly proportional to the power required to drive the motor.

Case 2: (A bit more math here that I didn't explicitly go over above) Going back to Lorentz's law we see that:

τ = 2 \cdot F \cdot r = 2 (I \cdot B \cdot N \cdot l) r

$\tau = 2 \cdot F \cdot r = 2 (I \cdot B \cdot N \cdot l) r$

Therefore:

F = I \cdot B \cdot N \cdot l

$F = I \cdot B \cdot N \cdot l$

Thanks to Newton we have:

F = m \cdot g

$F = m \cdot g$

So...

τ = 2 \cdot m \cdot g \cdot r

$\tau = 2 \cdot m \cdot g \cdot r$

If you keep the length of the wire the same but increase its gauge, the mass will increase. As can be seen above, mass is directly proportional to torque just like magnetic field strength so the same result applies.

Case 3: The radius of the armature, $r$ in our equations above, is again directly proportional to our motor constants. So, once again, we have the same results as we increase and decrease its length.

Starting to see a pattern here?

Case 4: The number of turns of our wire, $N$ in our equations above, is also directly proportional to our motor constants. So, as usual, we have the same results as we increase and decrease the number of turns.

In general

If it isn't obvious by now, torque and speed are inversely proportional:

torque versus speed

There is a trade-off to be made in terms of power input to the motor (voltage and current) and power output from the motor (torque and speed):

V \cdot I = τ \cdot ω

$V \cdot I = \tau \cdot \omega$

If you want to keep the voltage constant, you can only increase current. Increasing current will only increase torque (and the total power being supplied to the system):

τ = K_{t} \cdot I

$\tau = K_t \cdot I$

In order to increase speed, you need to increase voltage:

ω = \frac{V}{K_{e}}

$\omega = \dfrac{V}{K_e}$

If you want to keep the input power constant, then you need to modify one of the physical motor parameters to change the motor constants.

embedded.kyle
la source

I think this is what I'm looking for, you gave me all I need! I nearly grasp the concept without understanding the formula. I think I should really study more, seems like one cannot create an effective motor using theories alone.

dpp

Thats a hell of a generous answer for a homework question.

Bryan Boettcher

@insta that isn't my homework, that's for my toy car.

dpp

Great answer here! I just want to clarify this: you say increasing current increases torque "at the expense of speed". But increasing current doesnt affect Ke in the speed equation, since Ke and Kt are motor constants. So speed should remain the same but torque has now increased and the overall power supplied is also increased? The inverse relationship between speed and torque only comes into play, when any of the factors of the motor constants is altered? Thanks in advance.

TisteAndii

@TisteAndii Good catch. That was a poor choice of words on my parts. I've edited to hopefully be a little clearer. But you are correct. Increasing current increases input power and output torque but does not effect speed. If power is held constant, then modifying any of the motor parameters would effect both torque and speed. This is because

K_{t} = K_{e}

$K_t = K_e$ and because current is multiplied by the constant to get torque and voltage is divided by the constant to get speed.

embedded.kyle

One explanation is to consider that power $P$ is the product of current $I$ and voltage $E$ :

$P = IE$

Power is measured in watts, and is the rate of energy use. Energy is measured in joules, and a watt is convienently defined as one joule per second.

The application of a motor, usually, is to apply a force to a thing to move it. In physics, this is called work, which is equal to the product of force $F$ and distance $d$ :

$W = Fd$

You asked about increasing torque and RPM. Torque is just a rotating force, and RPM is just a rotating speed. So the definition of work is half of what you asked (it has torque in it), and speed and distance are obviously related. It seems like we are really close. You don't want to just do more work with your motor, you want to do work faster. You want to increase force and speed, not force and distance. Is there a physical term for this in a mechanical system?

Yes! It's also called power. In a mechanical system, power is the product of force and velocity:

$P = Fv$

Or to use the equivalent terms for a rotational system, power is the product of torque and angular velocity:

$P = \tau \omega$

This is just what you asked. You want the motor to apply more torque and spin faster. You want to increase power. You want to use energy faster.

The law of conservation of energy tells us that if we want to increase mechanical power, we have to increase electrical power also. After all, we can't make the motor spin with magic. If electrical power is the product of voltage and current, then increasing either voltage or current, if the other is held constant, will increase electrical power.

When you change the strength of the magnets, or add or remove turns of wire, you can't increase power. You can, however, trade voltage for current, or current for voltage, just like a mechanical transmission can trade RPM and torque. Lenz's law and other laws of electromagnetic induction explain why this is true, but they aren't really necessary to answer your question, if you simply accept the law of conservation of energy.

Given all that, your question was "How to improve torque and RPM of a DC motor". You can improve it by giving it more energy, or you can make it more efficient. Some sources of loss are:

friction in the bearings
resistance in the windings
magnetic resistance in the winding cores
electromagnetic radiation from commutators
losses in the wires, battery, transistors, and other things supplying electrical energy to the motor

All these serve to make the motor less than a 100% efficient converter of electrical and mechanical energy. Reducing any of them usually increases something else undesirable, frequently cost or size.

An interesting thought: This is why electric hybrid cars can get better mileage in the city. Stopping at a red light converts all the energy of your moving car into heat at the brake pads, which isn't useful. Because a motor is a converter between electrical and mechanical energy, a hybrid car can convert this energy not into heat, but instead into electrical energy, store it in a battery, then convert it back to mechanical energy when the light is green. For further reading, try How can I implement regenerative braking of a DC motor?

Phil Frost
la source

Hi, in my experiments I noticed that when I increase the size of wire and decrease the number of turns, I get higher RPM and better torque, why is that? I didn't increase the power source (still 2.4v). By the way, your answer helped me.

dpp

@dpp 2.4V isn't the magnitude of the power source, it's just the voltage. You also have to measure current to know how much power you were delivering. Decreasing the number of turns weakens the magnetic field per amp, which also reduces the back-emf, which improves RPM but reduces torque for a given power. But, you also used bigger wire, which has less resistance, which make the motor more efficient and allows you to deliver more current/power with a given voltage.

P = E^{2} / R

$P = E^2/R$ and

I = E / R

$I = E/R$ . If you measure the current you will find that power went up, and you drain the battery faster.

Phil Frost

@Thanks Phil, the way you answer makes it easier for me to understand, I mean by saying which causes what etc.

dpp

Although you have received very good and detailed answers, I would like to offer a very simple answer utilizing the formulas already presented:

τ = 2. B . N . l . r . I

$\tau = 2.B.N.l.r.I$ This formula clearly shows that the torque is directly proportional to the magnetic field strength, the number of turns the length of the loop, the radius of the armature and the current in the wires. So, as any of these variables increase or decrease, so does the torque.

The other formula,

ω = V / 2. B . N . l . r . I

$\omega = V/2.B.N.l.r.I$ clearly shows that the RPM is inversely proportional to the same variables. Therefore, as they increase, the RPM decreases, and vice-versa.

If you increase the wire gauge, you increase the current (I) and thereby the torque. If you also decrease the number of turns, you will decrease the torque. Whether the total torque increases or decreases, depends on which effect is larger.

Guill
la source

I decided that having brushless motor and adding ball bearings will help. I guess the friction reduces the RPM and the effectiveness of torque.

dpp