Motor Generators

[bistander]

You have proven nothing because you have never made a controlled bridge rectifier. If you had, you wouldn't be writing this nonsense.

But I have not only done it, but also taught others. For one mini-hydropower plant.
The voltage at idle (no load) was 60-65V, and the battery voltage at which the charge occurred was 24V. So, the thyristor was given a switch-off signal when the voltage reached 27V, which corresponds to a full charge of the 24V battery. Before that, where did the rest of the voltage, or rather the EMF, go?

Mr. Rakarskiy,

One of your blunders involved a FWBR, full wave bridge rectifier, not a controlled bridge rectifier, so you have not addressed the issue.

91466693 (2).jpg
But I'll respond to your new question about the controlled bridge rectifier using the thyristor charging a battery if you supply a circuit diagram and define the "switch-off signal".
bi

[Rakarskiy]

You clearly have a perception problem. You are confusing the whole thing (isolated circuit with the household network via distribution transformers). Under load, on the capacitor will be the voltage that remains after voltage drop and current generation in the isolated circuit I am considering.

This sketch is based on motifs from Soviet technical specialist magazines, variants of adjustable rectifiers for permanent magnet generators, which I did for an amateur group on building mini hydroelectric power plants.

925982836.jpg

[bistander]

You clearly have a perception problem. You are confusing the whole thing (isolated circuit with the household network via distribution transformers). Under load, on the capacitor will be the voltage that remains after voltage drop and current generation in the isolated circuit I am considering.

This sketch is based on motifs from Soviet technical specialist magazines, variants of adjustable rectifiers for permanent magnet generators, which I did for an amateur group on building mini hydroelectric power plants.

925982836.jpg

Mr. Rakarskiy,

91466693 (2).jpg
This is the circuit where you blundered. What is the voltage V across the load R which is also across the capacitor C1?
https://www.patreon.com/posts/emf-current-112414335
It is in your published essay found at the address above. You conclude and state:

As a result, we get a constant voltage of 220V on the load.

That is incorrect. Connected as per your diagram, the load voltage is NOT 220V when the circuit input is 220VAC.
Address this issue. Or perhaps you do not care about the quality and correctness of your publications, which you offer for sale. How do your customers feel when you ignore uncorrected errors and mistakes the likes of which you've been informed?
bi

[Rakarskiy]

bistander, What are you looking at? Your socket? If so, you don't understand how the power grid works, from the alternator at the power station to your socket.

The case I am considering is very clear: when the load is switched on, the voltage across the capacitor terminals will be equal to the voltage to which the total induction emf falls from the generator winding to the load. The voltage at the capacitor terminals will be equal to the voltage measured at the load terminals.

[bistander]

bistander, What are you looking at? Your socket? If so, you don't understand how the power grid works, from the alternator at the power station to your socket.

The case I am considering is very clear: when the load is switched on, the voltage across the capacitor terminals will be equal to the voltage to which the total induction emf falls from the generator winding to the load. The voltage at the capacitor terminals will be equal to the voltage measured at the load terminals.

I am looking at your example here:

91466693 (2).jpg

From:
https://www.patreon.com/posts/emf-current-112414335

You say

we simulate a circuit with an alternating generator, the current of which is rectified through a diode bridge, we also plan a capacitor C1 in the circuit, which acts as a storage device for smoothing pulses. As a result, we get a constant voltage of 220V on the load.

That is incorrect. Look it up. I gave you 2 references explaining FWBR calculations. You insist you are correct and millions and millions of us are wrong. It is a simple circuit. Hook it up and measure it for yourself. I have used and measured FWBRs hundreds, likely thousands, of times. The text books are correct. At the load you'll see approximately 310Vdc depending on the ripple (size of cap and current).
bi

[Rakarskiy]

You are referring to your socket, where the connected phase is separated from the three-phase network, not the network I am considering, where there is only a generator phase and a load phase. Therefore, in a three-phase network, it will be calculated differently. Secondly, I gave the formula for calculating the capacitor below in that post.

And if you do the full calculation, you will have more places where you will not understand why. All electricians in power grids consider a source with zero internal resistance value.

I really doubt your professionalism. It would be easier for you to act as an observer. It is very difficult for me to change my mind about something, especially with references to impractical training materials, when I have experienced everything first-hand.

[Rakarskiy]

MAGMOV LOW COST ELECTRICITY GENERATOR PRODUCTION - YouTube

The company from GEORGIA has been offering a domestic power source for five years. The rotor of the generator is driven by a magnetic motor of a special design. Unfortunately, I have not been able to find any information about the happy owners of this source.
The presentation video personally raises questions for me.
The three automotive generators are designed for a rated speed of 5000 rpm. At what speed must the linear magnet switch move to provide at least 1000 rpm of the drum of magnets that is connected to the generators. Second, a contact of two points on the magnetic pole of the drum, about 1 cm in diameter, unclosed by magnetic interaction, can develop the appropriate torque to turn the three rotors of the generators in the boost mode?

https://magmovenergy.com/

[bistander]

You are referring to your socket, where the connected phase is separated from the three-phase network, not the network I am considering, where there is only a generator phase and a load phase. Therefore, in a three-phase network, it will be calculated differently. Secondly, I gave the formula for calculating the capacitor below in that post.

And if you do the full calculation, you will have more places where you will not understand why. All electricians in power grids consider a source with zero internal resistance value.

I really doubt your professionalism. It would be easier for you to act as an observer. It is very difficult for me to change my mind about something, especially with references to impractical training materials, when I have experienced everything first-hand.

91466693 (2).jpg

This is the network of your example. That's it complete.

brcap.gif
Vac = Vdc x 0.71

That is the correct relationship between AC voltage and the resulting rectified filtered DC voltage



Full-Wave-Rectifier-6.png
Look it up.
bi