(Assuming that the voltage from the motor when braking exceeds the motor power)
That's the problem — it doesn't. You need a way to boost the voltage coming from the motor to a level that will actually charge the battery. You can use a separate boost converter, or you can create a more tightly integrated solution that uses the inductance of the motor ...
The voltage on the terminals of your motor will be a function of its speed and the load.
It will not be higher than your power supply, unless some external force is trying to accellerate the motor.
Regenerative braking supposes that you get back some energy from the motor while braking. Therefore the braking must be performed by applying a load accross the ...
Thermal overload relays are sensitive to the heating due to current flowing in the heater elements. The thermal sensing element in a three-phase overload is calibrated to trip based on the current flowing in all three heaters. If only two heaters are used, the overload will have a lower temperature for a given current, so the trip level will be higher than ...
Now the question is, how can I connect the motor to delta connection if I have only 400V line to line voltage?
Wire the motor in star and connect it to your 400 V phase-to-phase supply.
Where can I get 230 V line to line?
Will I damage the motor in delta connection?
Yes. You would be applying 400 V to a winding rated for 230 V.
Your circuit doesn't provide any braking because the motor is running free when the switch is open, and won't produce higher voltage than the battery unless it is 'over-driven' to higher speed by an external force.
To brake the motor you must put a switch across it, like this:-
simulate this circuit – Schematic created using CircuitLab
When SW2 is ...
The concept of "real time" is a stretchable thing. If your controlled object is, say, a metallurgical oven, then, depending on the size of it the real control time may be from seconds to minutes or even hours. So you need to define more clearly what kind of sampling rate is sufficient to make a robust and stable control for your object. To determine which ...
ad 1) The current is distributed to the each half. The brushes are short circuiting the segments at GNA position, so the only current that flows to that section is a recirculating current, from one segment to the other through the brush contact.
Electric power that is converted to mechanical power by a motor is real power. The motor appears to be mostly resistive to the AC power supply, but that is the effect of the power conversion not actual resistance in the motor.
I don't have enough rep to leave a comment, so this is going in the answer section, but here are a few things to consider:
Do you have enough power capacity upstream? You might be trying to draw too much power from your existing upstream transformer.
Did you correctly match all parameters of your motor with your VFD?
Is a winding burnt open in your motor ...
Someone correct me if I am wrong.
Let’s start with just a superconducting inductor. It is a purely reactive load so you will lose no energy in it and receive it all back eventually. The energy is stored as a magnetic field and will be converted back to a current when allowed to.
If you have a normal inductor with both resistive and reactive qualities, ...
The fundamental frequency is determined by the commutator switching speed (which varies with motor rpm), but the rf produced by arcing can go up to GHz frequencies.
Here's an example:-
The blue line is before adding ferrite cores to the motor leads, the green line after.
At lower radio frequencies (below 100MHz or so) small capacitors connected from each ...
Now the question is, how can I connect the motor to delta connection
if I have only 400V line to line voltage?
If you have 400 V, use the star connection. There is no reason to use the delta connection and the motor will draw too much current and overheat if you apply 400 volts to a connection designed for 230 V.
Where can I get 230V line to line?
A spec sheet link and actual component IDs would be very useful.
Presumably these are BLDC motors.
Compared to the other configuration:
Needs higher voltage
Requires more cells.
Requires lower current so lower I^2 R losses, lower spec cells, lower current FET needed.
Requires lower voltage
Requires fewer cells
Yes, that will work fine. I would use 4.7K pull downs. Only one (at most) will be drawing current at a time.
Put the pull-down at the 4051 output rather than the gate to avoid unnecessarily reducing the gate drive voltage.
Make sure your MOSFETs are logic-level drive (eg. 4.5V for a 5V MCU). With a couple hundred ohms in series with the gate they will not ...
You are correct !
Burried magnet change the permitivity distribution inside the rotor. Thus depending on the arrangement of the magnets, asymetrising its direct and quadratic inductances.
Look at this post: Explanation of Lq, Ld and how it relates to motor inductance
And this paper (if you can): https://doi.org/10.1109/emobility.2010.5668074
As far as motor characteristics, I believe these are constant and are irrelevant to changes in voltage. [...] For the values such as stall current and nominal current, I'm assuming these are dependent to the system voltage.
Your assumptions are correct.
The stall current / peak current / starting current are the same and they are defined by the maximum ...
There is no formula to calculate the cost of a motor from the number of coils and width and height. Each motor is priced differently you will just have to search around online and see if you can find a motor that meets your speed, torque and size requirement.
A “soft start”, even if appropriate for the motor, will only lead to a stall of the compressor. You will need an inverter (and batteries) capable of handling the start-up surge.
It’s possible that there are special fridges that have less start-up surge designed for your application and similar ones such as RVs.