Effects of Recirculating BEMF to Coil

[poynt99]

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.

Ok Luc, I'll try.

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?

This depends on what the total circuit resistance (or impedance) is and what the RMS value of the wave form is.

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?

Voltage alone is not necessarily energy, that is correct. A 12V battery sitting with open leads is not supplying much if any energy external to itself. So it depends on the circuit that the "voltage source" is connected to.

Current is a better indicator that energy is being supplied by a source, but voltage and current are closely tied in a relationship that together amounts to power output or input. If either the voltage or the current are zero, there will be no power, no matter how large the other parameter is. You have to have at least some voltage AND some current simultaneously in a circuit in order to "make power".

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.

I'm getting to this as it is related to the previous paragraph. Power in Watts does equal heat if that power is dissipated in a dissipative element such as a resistor.

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.

More or less, yes. The circuit is such that it draws power from the source because of its loading effect on it.

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?...

This is not necessarily correct. As you know you can connect the bulb by itself across the supply and it will glow. No resistance is required in the circuit to allow the bulb to glow. If you used superconductor wires (i.e. zero resistance) in the circuit, the bulb would glow, and glow brighter. The bulb is it's own resistance, and yes it is this resistance which gets hot (dissipates power) and glows as a result.

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???

No. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as 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

Getting back to the relationship between voltage and current in producing power in a circuit element, it is necessary to be aware of and understand a few simple equations.

Power (P) in Watts = V x I (voltage times current)

From this you can see that if either the voltage-across OR the current-through a circuit resistor is 0, the resulting power dissipated in that resistor will also be zero. So for power to manifest in the circuit and be dissipated in any circuit resistance, there must be some voltage-across the resistor AND some current-through it.

Hope that helps.

.99

[gotoluc]

Hi .99,

Thank you for taking the time to answer the questions.

This question was the most important to me and you answered it as I understand it.

My Question:

    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???

Your Answer:

No. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as before.


Your answer is worth Gold to me and I will make a new video to show you why.

Thanks again for sharing.

Luc

[poynt99]

No. Assuming that you were able to set the bulb intensity roughly the same as before, then the average current and voltage in the circuit would also be measured as roughly the same as before.

Luc,

Since it was important to you, I'd like to expand a bit.

A pulsed-DC supply can be replaced by a linear (straight DC) supply of the same average value.

So a supply that is pulsed at 10V peak with a pulse width of 10ms and a period of 100ms (10% duty cycle) will have the same average voltage output as a pulsed-DC supply with 100V peak pulses and 1ms pulse width and the same 100ms period (1% duty cycle).

The average voltage output in both pulsed-DC cases is 1V.

So using a plain DC supply set to 1V output is the same as either case using the pulsed-DC supplies above.

The difference in applying that 1V source to a load is evident by how much power can be drawn from it.

Remember the car amplifier I mentioned? Are you familiar with power inverters? You can buy them at many automotive stores etc. They convert automotive 12VDC to 110VAC for obvious uses. Some of these units can supply 1000's of Watts to a load.

Let's use a household incandescent bulb as an example. A 100W bulb will have a hot resistance of about 120 Ohms. If we were to connect this bulb up to a automotive 12V battery, the most power we could obtain from this bulb would be (from P = V² /R) about 1.2 Watts. This is also the maximum we can draw from this 12V source, unless we begin to parallel many of these bulbs together.

But if we now use one of these 12VDC to 120VAC inverters, we can draw 100W from the battery with a single bulb connected. The inverter is able to step up the RMS voltage (by a factor of about 10) from the battery which allows more power to be drawn from it.

When doing the type of tests that you are, it is very important to always know how much power is being taken from your source. The best way to do this is to use a low resistance shunt in series with the supply (either + or - side) and measure the voltage across this shunt. Your good meter should be able to handle this (use "AC voltage" setting, but try DC as well). That will be a direct current readout in RMS current, which you will then multiply by the supply voltage (170VDC?) to get total power taken from the supply.

I don't recommend using resistors in the legs of your MOSFET.

.99

[wattsup]

@poynt99

Thanks also for your explanation which is the basic volts x amps that we know very well.

The difference Luc is showing is what you can do more with the pulsed DC in that your are now creating a steady flyback condition that he capitalizes to make the magnet rise with very low overall wattage, when compared to the standard straight DC supply method that does not favor any flyback production.

In the pulsed method the bulb which is the overall wattage consumption visual indicator was barely lit, while in the straight DC method, to lift the magnet to the same height, the overall wattage consumed and shown by the very brightly lit bulb indicates that more energy is consumed.

Again, this is not intended to show any OU effect but to simply show that the same work (raising and holding the magnet levitated) can be accomplished with drastically lower power consumption. This in my view is a perfect differentiation that could give a certain direction for further experiments leading to OU. But that's my opinion only, for what it's worth.

wattsup

[gotoluc]

Hi wattsup,

your explanation is exactly correct

I have just completed 2 videos to demonstrate what has been interesting me so much in this simple circuit and will start uploading them now.

Stay tuned

Luc