Effects of Recirculating BEMF to Coil

[wattsup]

@gotoluc

Don't take it personally. I think a lot of guys are on their work benches these days. I have myself lots of catching up to do after losing the last few months to this Mylow wheel thing. I am planning slowly to make a Thane Heins variation but from now on, I want to spend most of my time on TPU studies.

So don't lose heart. Whatever you are showing does help us in ways we don't know yet and every angle of effects is important to discover.

Regarding what you have shown, you know those big cranes with huge electromagnets that are used to move scrap cars in crushers. Well they must use huge amounts of electricity. I wonder if you used your pulsing method, you could probably save them loads of energy and money. May be a new business right there. Imagine the size of the pulsing circuit. lol

Added: Look at todays Google Logo.

[gotoluc]

@gotoluc

Regarding what you have shown, you know those big cranes with huge electromagnets that are used to move scrap cars in crushers. Well they must use huge amounts of electricity. I wonder if you used your pulsing method, you could probably save them loads of energy and money. May be a new business right there. Imagine the size of the pulsing circuit. lol

ya! that would take a large mosfet.

Thanks for taking the time to post

Luc

[poynt99]

Hi everyone,

I made a new video using standard resistors of equal value on the input side and on the recirculated Inductive flyback side pre entry of the coil. I have measured the heat on each but have not wet come up with a better way of measurement that will satisfy all but I am working on it.

There is no need to prove any further that the output resistor is much hotter than the input resistor. This is what would be expected.

In this test it is very very difficult for me to understand how energy could slip though the input 22 Ohm resistor and leave basically no heat and end up in the collapsing field side resistor at higher temperature then the boiling point and also do work pushing up a one pound magnet over 3/8 of an inch off the coil. We cannot consider voltage as having anything to do with the effect since it has no heat Energy. So what is doing this?

Voltage and current have to be considered together in a load of any kind when dealing with power. If you were to scope across your input resistor, you would see that the voltage on average is much higher than that across the input 22 Ohm resistor, and since the two are of equal value, you can surmize which one will be hotter. See the attached sim scope shot.

However, the power in the input resistor is not an indication of how much power is being taken from your power supply. You have more power dissipated in the output resistor than the input one, but the total power from the power supply has to be taken into account. If done so, it will reveal that indeed a lot of power is being used.

To do this you need a small current sense resistor of about 10 milli-Ohms so you can measure the current from the power supply. Then you multiply that by the supply voltage and you have instantaneous power. Now integrate it and you have a direct readout of energy. See below Joules plot of the 170V supply (red) vs. that of the 22 Ohm output (bemf) resistor (green).

Are you surprised?

Below is the schematic, voltage plot across both 22 Ohm resistors (illustrating why the output R (green) is much hotter than the input R (red), and finally a plot of the energy in Joules dissipated by the output resistor vs. the energy taken from the 170V power supply. This is measured directly, even though it indicates mV it is mJ.

Think about a car stereo aftermarket amplifier that puts out a true 100W rms per channel. How do you suppose that can be achieved with only a 12V battery source? With Ohm's law it becomes obvious that this is not possible unless the battery voltage is boosted up somehow

Sorry for jumping in again. I'll back off and let some others explain it if they wish.

.99

[gotoluc]

Hi .99,

I do appreciate your help but I'm sorry I don't understand conventional EE

Instead can you please answer the questions below. This will help me much more.

Let us take the fine filament 12 volt bulb I used in test 3

We know that if we place that bulb in series in a circuit we can pass several hundreds of volts through it and it won't necessarily light up, correct?

If this is correct then voltage is not Energy!... correct?... however I do understand that voltage and current can be present together in a circuit if the voltage source has the ability to provide Energy.

Current is a heat Energy, correct?... and if so then voltage is not a heat energy, correct?

So if voltage cannot create heat at whatever voltage it is at then why would one need to consider it to calculate Energy used in a circuit.

Lets use the bulb in this example:

For Energy to partake in a circuit I believe we need two conditions, one is, it has to be available from the voltage supply and two, the circuit would need to create a wast (resistance) of some kind. Now if our bulb which is in series in the circuit starts to glow it is now dissipating heat because of resistance somewhere in the circuit. The bulb is a visual display or indicator of Energy now present in the circuit. This Energy is dissipating in the bulbs filament as heat, correct?... so if we have 10 volts in a circuit under load and the filament starts to glow and we reduce the duty cycle since the circuit is a pulse circuit and we then raise the voltage to 100 volts and the bulb starts to glow to the same level as when 10 volts was running through the circuit, do you really believe that we now have more Energy at the bulb because the voltage was raised 10 time higher then before???

If you believe this then can you please explain to me how voltage can contain Energy since something is not making any sense to me anymore

Thanks for your time and sharing

Luc

[rensseak]

Hi .99,

I do appreciate your help but I'm sorry I don't understand conventional EE

Instead can you please answer the questions below. This will help me much more.

Let us take the fine filament 12 volt bulb I used in test 3

We know that if we place that bulb in series in a circuit we can pass several hundreds of volts through it and it won't necessarily light up, correct?

If this is correct then voltage is not Energy!... correct?... however I do understand that voltage and current can be present together in a circuit if the voltage source has the ability to provide Energy.

Current is a heat Energy, correct?... and if so then voltage is not a heat energy, correct?

the heat depends also of the profile of your wire, if it is big enough then there is less or nearly no heat.
think of current as the quantity of electrons moving in a wire, the more elctron the more heat because of friction and voltage is just the pressur. Hope this helps.

Norbert