New answers tagged

0

It is the reverse voltage generated when you power a motor. Motors & generators are almost the same in principle. The difference between applied voltage & back emf is what supplies power to the motor. If you stop feeding power to a motor, it'll keep spinning and generate a reverse voltage. That is back emf, and directly proportional to motor speed. ...


0

Universal motors are often started and stopped with a switch and not speed controlled at all. It seems unlikely that the speed controller or motor would be damaged by switching it on and off. However the linked web page contains the statement "If you need knowledgeable advice about your product we urge you to call." Starting is stressful for any motor and ...


1

If the unit really uses 18.5 Amp, it cannot be used on a normal 15 Amp circuit - it would need a 20 Amp or higher rated circuit, and and would require a heavy-duty switch or contactor rated for that current or motor horsepower. A normal light switch would not handle that current.


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Not the easiest project you could find. From my point of viwew it should be as follows: use an unipolar stepper motor driver with a chopper. Complementary motor phases can be grouped, so you need only one current sense resistor for each pair, total two current sense resistors. The driver has to accept ABCD input sequence, not a step/dir interface The hall ...


0

A typical electromagnetic (EM) circuit, like most systems occupying our environment tend to resist change. For mechanical systems we have Newton's laws of motion describing their nature, and his First Law basically states that a system at rest or moving at constant velocity and not acted on by an outside force will remain at rest or at its constant velocity. ...


0

The induction motor generates the currents in the rotor by induction (hence the obvious name), these currents then generate torque by interaction with field generated by the stator. The magnitude of the emf generated in the rotor is dependent on the frequency it sees - the difference between the speed it is rotating at and the speed the rotating field ...


6

This answer was given prior to the op changing the question. Originally, the question referred to the motor as being a squirrel cage type. Subsequently, after several comments, the op changed it to a brushless motor. You just can’t get the staff any more! When you stall a 3 phase motor it becomes a transformer and, because the rotor is the secondary AND is ...


1

The circuit you're building is ... well, kind of complicated for what you're trying to do. Answer? Cheat! Use an n-FET instead, which doesn't have the drive requirements of the TIP41. A logic-level FET like this one: https://www.onsemi.com/pub/Collateral/FQD7N10L-D.pdf. Rds(on) at 5V is a mere 0.3 ohm. Bonus: no Vce(sat), more swing for your motor, less ...


2

So, what do you guys use to make your steppers to behave to meet CLASS B? I use a closed frame (faraday shield) with filters on the power to meet class B. If the motor cables need to go outside the frame, I use shielded cable. One thing that may be an advantage to knock out 100Mhz+ signals is to short them out with capacitors. X2Y capacitors are ...


1

As to why, the rotor current is induced BY the stator current. So when acting as a motor, it is ALWAYS going to be taking place AFTER the stator induces it. Power Factor is just the measurement of that lag.


3

The fan is not a resistor, so you cannot use Ohm's law to find out its resistance. Furthermore, even if you do know the winding's resistance you cannot measure it on the wires. This is brushless motor, so there is switching electronics between the power wires and the motor windings. One very common cause of the fan failure is dirt in the bushing, and ...


2

We noticed it requires 12 V and will run at 200 mA. I wasn't sure about that at first. Does that mean it has an internal resistance of : 12 V / 200 mA = 60 ohms? No, but when running it behaves like one. A motor isn't a resistor. If so that means I could supply it with 12 V and it would consume 200 mA and spin. It may draw a higher current when starting ...


5

This sounds like a bearing is seizing or a winding has gone open circuit. It needs replacing.


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Motor inductance basics current ALWAYS lags voltage. PF is only negative when operated as a generator.


3

You can not connect BLDC motors in series. It is possible to connect the armatures of two commutator type DC motors in series with the shafts mechanically coupled together essentially driving the same load. Torque sharing is accomplished by separately adjusting the field currents. There is not likely any other type of motor that will work with any type of ...


0

The PID output is basically the required increase in POWER to plot a path to the required set point - think of a cruise control on a car, as speed drops below setpoint, more accelerator is fed in. The torque curve on most DC motors follows a parabolic shape, so at high and low RPM's your "accelerator" input needs to be scaled up accordingly. Torque curves ...


1

The requirement for NRTL listing of electrical equipment is not universally applied; it is actually a State by State addition to the National Electric Code. But the only reasonable way to deal with it from the standpoint of a manufacturer wanting to export to here is to attain an NRTL listing from someone like UL or ETL (Intek) since you don't know where it ...


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Here you go; if the image is rotated placing the real values on the horizontal axis as is normally the case with (say) phasor diagrams you get the imaginery axis being negative: -


1

Many people here will not know what N20 are - or what you mean by "there are different RPM of". Even 4WD RC experts would need much more information to provide a useful answer. This is an electronics / electrical site. You could phrase this an EE question but at present it will be closed as being off topic and with not enough information. I strongly ...


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In particular the gimbal, I can't figure out how the manufacturer can put a gearbox in such small object considering also the axial and radial loads apply directly on the gears. They balance the load on the gimbal motor so that the center of mass runs through the axis of the motor so that it takes zero torque to hold the gimbal at every position. The torque ...


3

The commercial gimbals work like this: They have no gearbox, they're a BLDC motor in direct drive. The motors work the same way as a regular BLDC, but they have ultra low kv. The motors are driven open loop, similar to a stepper motor, since they're spinning too slow for the back-emf method that is normally used. And here's the important part: they don't ...


1

In a switching converter this is desirable because it means that it lets the current in the inductor never fall to zero which lets you run in continuous mode instead of discontinuous mode which produces less ripple and noise (I think...it's something like that). In a motor this is desirable because you don't throw away all the energy put into building up ...


1

You typically choose a PWM frequency high enough to limit the ripple in the current through the motor to prevent torque oscillation and audible noise, and so the current during the off period won't decay to zero except at very low duty cycles. At 50% duty cycle, the motor will be running at about half full speed, and the current through the freewheeling ...


2

Induction motor DC injection braking is like eddy-current braking. DC injection braking is accomplished by disconnecting the AC supply and applying a DC voltage to the stator windings. DC current in the stator windings produces a stationary magnetic field. The motion of the rotor bars through the stationary field produces currents in the rotor that have a ...


0

Let's say you design a motor to operate at a given field strength. Most of the magnetic path reluctance comes from the air-gap. Therefore the larger the air-gap, the more H field and therefore length of magnetic material you need to establish the field, so the more expensive the motor is to make. The size and weight also go up as the amount of magnetic ...


13

Air has a much higher reluctance (the magnetic equivalent of resistance) than the magnetic materials used in the motor. The smaller the air gap is, the lower that reluctance, and thus the higher the magnetic flux (which is the magnetic analog of current), allowing the motor to work more efficiently and at a higher power. Smaller air gaps also minimize ...


1

What do you mean by "loss of signal"? The SLAMTEC unit you cite mentions a lifetime for slip rings Most traditional non-solid LIDARs use slip ring to transfer power and data information, however, they only have thousands of hours of life due to mechanical wearing out. In a digital system there is no "loss of signal" due to small resistance in the data ...


0

just use a USB cable to your wifi head and reverse direction after 370 degrees. or use a fixed transmitter and movable reflector like this but smaller.


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It sounds to me that you are running an open loop "Volts per Hertz" algorithm similar to what you could run with an induction machine. The problem you are seeing, I think, is that a permanent magnet synchronous machine has very little damping (contrary to induction machines), and your rotor is oscillating pretty wildly. When your currents are high, this is ...


0

I agree, it likely has to do with the response time of your meter and the fact that when the speed is reduced by almost a factor of 12 with that gearbox, the shaft torque to the load is INCREASED by the inverse ratio; i.e. 12x the motor's rated torque. So the motor is likely able to get to it's full speed under load in a very very short time, too short for ...


2

Your expectations are wrong. The "4-7 times" statement is what is known as a "rule of thumb", used for rough estimation. It doesn't mean that every motor ever built draws a minimum of 4× its rated current at start-up. In your situation, it would seem that the motor is able to spin up fairly quickly, limiting the time duration of the current "spike". ...


1

It's probably something like this that you are after.... When this type of circuit is used for a bridge connected class AB audio amplifier (swap motor for speaker) it is important to reduce crossover distortion as much as possible by using both techniques of feedback and biasing the transistors slightly on with the diodes. Without the diodes, (bases ...


1

There are basically three approaches to running a 3-phase motor from single phase power. None of them is as convenient as buying a single phase motor in the first place. Static phase converter. This involves a bespoke arrangment of capacitors to shift the phase. Here is a commercial supplier. It does not give you full torque and has relatively low ...


-2

Due to the cost of matching PU plastic grid rated cap Impedances to motor impedances are greater than VFD, look only for the best match to your needs in current in a VFD solution. the costs in your range are ~< $10/A for 3 phase VFD’s. Up to $25/A. e.g. https://www.ato.com/single-phase-to-three-phase-vfd. Which is not to say it is a preferred source , ...


1

The old-fashioned alternative to a modern VFD, as mentioned in another answer, would be a "rotary inverter". A rotary inverter is really just an electric motor driving an alternator. Using a single phase motor and a three-phase alternator would give the required result.


3

If your budget allows it you should get a VFD (Variable Frequency Drive). It can create three phase signals with different phase shifts and frequencies, giving you the ability to cintrol speed, direction, power etc. of your motor. These can be found on eBay for moderate prices, the more powerful they are the more expensive. There are probably other options, ...


1

So, we contacted the motor manufacturer. And after analyzing the situation, we came to the following conclusions. I share them here, just in case someone else might one day see similar effects and would like to understand their own motor-problem. The motors even if they are identical, i.e. from the same supplier and are of the same model type, still have ...


0

In the video link you shared, the symbol 'phi' indicates the magnetic flux. Not any angle. Whereas, 'theta-m' means an angle. The equation should be, E=2*NABcos(theta-m). Not 'cos(phi)'. (Though I couldn't find the equation in the video) I can't say the exact reason for the high values of the current you are getting. But, one thing always should be ...


4

This delay is due to inertia, not to friction. A spinning motor has rotational kinetic energy, which is equal to \$0.5\omega^2 I\$, where \$I\$ is the moment of inertia of the motor, and \$\omega\$ is its angular speed. To make a motor spin, you have to supply this energy. The rate at which you can supply energy is limited. All motors and power sources ...


1

How small is 'rather small'? Obviously any motor requires 'some amount of time' to reach operating speed. It cannot go from stationary to full speed instantly. If you want to know why, you will need to post on a physics site. But if we are talking about your typical 2cm hobby motor (which we would expect to reach speed in a time that isn't perceptible to a ...


0

Core losses are resistive due to eddy currents that rise with f^2 and iron thickness, t^2 and and hysteresis losses due to nearing the L-10% saturation threshold. Eddy current losses are the result of Faraday’s Law http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html which states that, “ Any change in the environment of a coil of wire will ...


2

you gotta start with better specs in case you hadn’t noticed yet, your BJT’s are always conducting creating a condition called ‘shoot thru’ , shorting out the supply on both sides. There must always be a dead band and thus a deadtime during transition. start with your output impedance model of a motor with back EMF as a motor/generator or simply ...


2

The synchronous motor/generator has two power ports, a mechanical power port at the shaft and electrical power port at the motor/generator terminals. Because mechanical losses and magnetic losses (core and stray losses) are (approximately) proportional to the motor’s rotational speed, they are often taken together and referred to as rotational losses. ...


1

You aren't going to get anything below 0V if your low supply Vee is at ground and not -30V. Swing around 0 can only happen through a transistor stage if total voltage from collector supply to negative supply is bounded greater than your desired swing. This is why your output stage is automatically biasing to a a +-15V swing around +15V. Also, an AB output ...


1

Core or iron losses are not mechanical. They are mainly due to frequency and voltage, so they are constant. Eddy currents (batteries set up by impurities in the iron) and hysteresis (heat created by reversing magnetic field) are the main iron losses. Eddy currents are the main reason the core is made of laminated segments.


0

If you supply mechanical power to a synchronous machine, it becomes a synchronous generator generating electrical power. Electrical power converts the synchronous machine into a motor providing mechanical power. From Electric Machinery Fundamentals. As a motor: $$V_{\phi} = E_A + j I_A X_S + I_A R_A$$ As a generator: $$V_{\phi} = E_A - j I_A X_S - ...


1

" Therefore I could driver both sides of the bridge from the same signal " Nor really.If you move Q3 down to the Q4 location, the drive requirement will be similar to but not identical to the req for Q1. Also, unless the Q3 drive goes below ground, Q3 will be acting as an emitter-follower, not a saturated switch. I don't think the net change in effort is ...


2

You could replace all BJT's by n-channel mosfets, then you don't need to invert the driving signals. However, the Law of Conservation of Misery will apply: you now need to level shift the signals for the upper n-channel mosfets to drive them correctly: it will not simplify the circuit or reduce the number of components.


1

With either type of transistor, if you connect the collector to either supply rail, you've converted it from a common-emitter configuration (saturated switch) into a common-collector configuration (voltage follower). When you switch configurations like this, the drive requirements change, and the efficiency of the circuit will be affected as well.


1

It's not obvious what you intend, but, NO you can't. The high side PNP switch needs to provide power from the +ve supply rail. The control signal needs to be negative relative to the positive rail. There are selected cases where you could use an input signal to drive the low and hjgh side drivers - but this would be unusual and not commonly possible.


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