# Tag Info

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It is normal, because $i_c=C\dfrac{du_c}{dt}$. You are doing PWM, so $\dfrac{du}{dt}$ is very high and the current spikes go to infinity. Not an expert here, but I do think it has nothing to do with CE. You can install additional chokes on both lines to limit $\dfrac{di}{dt}$ You can use also a common mode choke that will turn into differential mode ...

1

This appnote describes, in detail, 2- and 4-quadrant motor control with regenerative braking for each direction. Link: https://www.roboteq.com/index.php/docman/motor-controllers-documents-and-files/documentation/application-notes/application-notes-1/33-an70614-1/file The short answer is you need at least 2 FETs to effect regenerative braking in the forward ...

1

Voltage is not energy. It's only power (over time resulting in energy) if it flows through some resistance. When the upper switch is closed, current builds up due to the EMF on the motor coils, in some period of time (maybe tens of microseconds) it builds up to some peak current. You then open the upper switch and the current flows backwards through a diode ...

2

I believe this should work for what you want. simulate this circuit – Schematic created using CircuitLab

-2

You need a circuit similar to this: simulate this circuit – Schematic created using CircuitLab Motor direction is selected by SW. The bumper switches BSW1 and BSW2 open as soon the respective end position is reached.

0

Your circuit is dangerous because when PWM is initially applied the brake is still on, which will cause a very large 'shoot-through' current that could damage the MOSFETs. If the brake MOSFET has lower resistance than the PWM FET then motor voltage may not ever rise high enough to switch the brake off. To prevent shoot-through you need a circuit which adds ...

3

The two capacitors complement each others’ strengths and weaknesses. Large electrolytic capacitors work better at lower frequency than they do at high. This is due to their internal resistance, referred to as Equivalent Series Resistance, or ESR. They also have relatively high series inductance, or ESL. Both of these limit the electrolytic capacitors’ ...

4

Put your current sensor in one of the supply lines that feeds the H-bridge. Either put a high-side sensor in series with the positive supply, or put a low-side sensor in series with its ground connection. Regardless of which way the motor is turning, the current flows through these points in only one direction. simulate this circuit – Schematic ...

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Okay, I find this from quora: answer from Rafal Muszynski Capacitors are connected on input and output side to secure stable (oscillation-free) operation and improve step response (behavior during sudden changes of the load or input voltage). Most of the times the required capacity is in order of 1uF-100uF and such values can only be practicaly ...

0

A DC motor produces the high pitched whine of its PWM frequency when the motor tries to start and the PWM duty-cycle is too low. A DC motor's magnets can also be permanently de-magnetized if it is overloaded and gets too hot.

1

I recommend putting the motor outside of the vacuum and use a magnet-coupled drive through the container wall. Another option is using an AC induction motor and putting the coil outside of the vaccum for better heat transfer. This is done for high-power X-ray tubes to rotate the anode. Again, the reason is better heat transfer. https://en.wikipedia.org/...

1

Interesting question! Your worst case is that you have a perfect vacuum and no heat path out of your chamber. 600 rpm at 1000 gcm is about 6 watts of mechanical power. I wouldn't choose a big motor because it would be less efficient; rather, choose a "right-sized" brushless motor and gearbox for your speed and load. You should be able to achieve at least ...

1

The current doesn't exactly alternate in the coils of a DC motor. Each time the brushes transition to different commutator segments the current reverses in part of the rotor winding. In inexpensive toys, there are only three commutator segments, so the current reverses in 1/3 of the total winding with each transition. In the kind of DC motor used to operate ...

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The relay information given indicates that is is intended to control a blower motor. A relay like that, as used in a car, may have its own fuse. If you can identify the blower motor fuse in one of the vehicles that the relay can be used in, that would tell you the maximum current required to drive the relay. You might find a seller that sells it for general ...

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Any conductive surface which can be touched should be grounded. The reason for this is that if there were some fault in your device that made a touchable surface "HOT", then a fuse or circuit breaker would blow. So, for your sander, if there is some metal that can be touched, (the screw that holds the sandpaper on? The center of the sanding disk? the shaft ...

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Only the manufacturer knows why these connectors are rated for 12V usage. If you use them at 24V you are on your own. The connectors might fail immediately or might never give you any trouble. If they fail they could fail into a safe mode, simply cutting off the flow of current, or they could catch fire. We can give you an opinion about using the connectors ...

0

For bidirectional control of a DC motor you typically need a full bridge made up of four FETs or transistors. Each lead of the motor is connected to a half bridge where the upper transistor can take it to the positive supply or the lower transistor can take it to the negative. By setting either side high and the opposite low, you can make the motor spin in ...

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I understand how half-bridges and full-bridges work. From that it seems to me that half-bridge is not an option for a DC motor as it can be used only for creating AC and not DC. [...] Am I right? The bold marked statement is incorrect. Considering a half bridge being supplied by a grounded DC voltage: If the upper transistor in the half bridge is ...

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With half bridge you can't change the polarity of the motor and thus you can't change the direction. If you don't need to change the direction a half bridge is fine, otherwise you need an H bridge.

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If you measure 6Ω, then the stall and starting current is as you calculated it 4A, If you have a PWM controller turining it down to 72% will reduce the current to 2.88 A which will work fine with your 70W power supply.

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For the sake of the power supply, the commutator and brushes, and the PWM switching devices, you should design the controller to limit the current to about 150% of the rated motor current. The controller should measure the motor current and use the measure value as a feedback signal to keep the current continuously under control. The speed command can be an ...

-1

Yet another option -- if you're designing an H-bridge driver for the motor, then limit the input current to the driver to what the supply can handle, and let the motor current rise as it will. You'll get something in between the stall current and the running current of the motor.

2

It depends what you want the motor to do. If you want a rapid start, as from a low impedance supply, then you want a supply that will give you the full stall current. If you are happy for a slow start, then you only need a supply that will limit without foldback at the running current. You'll get full running torque, just not stall torque. You choose.

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The motor needs enough current to meet the inrush current spec. If the only thing in the circuit is power supply, then it needs to source enough current to meet the inrush current spec. Inrush current can also be limited by series resistance or an NTC to limit the effect the motor has on the power supply.

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If you want to run on 12v ypu will require at least 5A supply. It is not worth it because the brushes inside the motor wont last a day and motor will be too fast for any practical application. Solution: Open the motor carefully. Pull the rotor out, Unwind and remember the pattern (Its always same direction for all coils.) Count the number of turns while ...

1

You have already discovered that speed is proportional to voltage. Current is proportional to load torque when the speed is stable. However the current require to start the motor is higher than the current required for steady speed. Attempting to start the motor with too much voltage is apparently causing your charger to shut down. As soon as it shuts down, ...

0

Your power-supplies are not capable of putting out enough power to run the motor, as a result, when you connect the motor, the power-supply goes into shutdown. Maybe add a DC-DC buck converter with a input current limit between the supply and the motor, set the current limit such that the power-supply is happy. Search "CC-CV DC-DC Buck" for some examples, ...

4

Reading between the lines of the specification, it sounds like the tacho output is the result of two 'flags' of something, perhaps magnets to a Hall sensor, or optically opaque material between a LED and photodiode. This results in two 'highs' and two 'lows' per revolution. Manufacturing tolerances will mean that the length of one high is only approximately ...

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