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Overunity Machines Forum



Marko Rodin Coil -- 007 Device

Started by Dog-One, June 02, 2016, 12:26:11 AM

Previous topic - Next topic

0 Members and 11 Guests are viewing this topic.

MileHigh

Quote from: 3Kelvin on June 17, 2016, 03:11:54 PM
Back EMF
VL(t)= -L di/dt

My Problem is, the infinitive sharp Edge from the input Signal.
From 0 to 4 Volt in 0 Time. For that assumption we will also get infinity.
So the Back EMF will be also infinity.

Is this related to di/dt or dv/dt?

3Kelvin

On my timezone it is late now.
My wife hates me for the mess with papers in the office.

She ask me:
Hey 3K, did you carry the garbage from the kitchen to the container into the back yard?
No, ma'am i do some math to eliminate infinity.
I try to understand the beauty of Physics.

Whaaat? I think you need a shrink at 0 time
,
says my wife.  :o

So far, see and read at next sunrise. ;)

Peace and Love
3K

tinman

Quote from: picowatt on June 17, 2016, 01:24:40 PM
Perhaps this thought experiment will help...

An audio power amplifier is, within its drive capabilities, a very close approximation of an ideal voltage source.  A professional PA amplifier capable of driving several hundred watts or more into a 4R load with a damping factor of 400 will have an output impedance around .01R and will have many amps of current source or sink capability.  As would an ideal voltage source, the power amplifier will source or sink current even when its output is zero volts.

Connect an FG to such an amplifier's input and set the FG and amplifier so that the amplifier output is a 60VPP 1kHz sine wave.  Now, connect a 5mH inductor with a .1R DC resistance directly across the amplifier's output.

As the sine wave on the amplifier output reverses from its +30 volt peak towards its -30 volt peak, do you expect to see a huge current spike commensurate with the amplifier's .01R output impedance and the inductor's .1R DC resistance or do you expect to see current flow more so in line with the 31.4R reactance of the 5mH inductor at 1kHz?

If the FG is switched to produce a 1kHz square wave of 60VPP at the amplifier's output, do you expect a huge current spike as the square waveform at the amplifier output rapidly transitions from +30 to -30, or again, do you expect to see current flow more so in line with the inductor's 31.4R impedance at the 1kHz fundamental (the impedance/reactance will be greater for the harmonics added to the sine wave to produce the square wave). 

In both examples above, how will changing the DC resistance of the inductor affect the outcome?  Will a 5mH inductor with a .01R or .001R DC resistance significantly affect the observed current flow?

What if the 5mH inductor had zero ohms of DC resistance?  Would that have a significant effect on the amplifier's output current in the examples above?

How will lowering or raising the frequency of the applied waveform affect the results?

As stated, this is just a thought experiment, consider the questions rhetorical...

PW

(With regard to the above, please assume the amplifier to have zero volts of DC offset)

But that is nothing like the test conditions stated in MH question.
There is no sine wave,as the voltages applied to the circuit(being the coil and ideal voltage source)are of a square wave pattern.The transition is not starting from a low point and ramping up to a high point,the transition is instant,and there is no resistance to sink the existing current flowing through the circuit.

Here is what you have explained.
We have a small ideal 12 v DC PM motor with a flywheel on it(ideal meaning it has no resistance in the windings). We apply a rising voltage that starts at 0 volts,and reaches a peak value of say 6 volts over 6 seconds,and then decrease that voltage back down to 0 volts over the next 6 seconds--so as the voltage wave form is a sine over 12 seconds. The motor/flywheel combo will ramp up for the first 6 seconds,and then ramp down over the last 6 seconds. During the last 6 seconds,the flywheel will return-or sink its stored energy back into the voltage supply(we are assuming the motor/flywheel combo is near frictionless).
But now we change the voltage wave form to a square wave-as in MH question.
We also have an ideal voltage source which has no internal resistance.
We apply an instant 6 volts across the motor,and the motor ramps up to a set RPM. We now reduce that voltage to a value of 0,and the current will continue to flow through the motor and ideal source due to the stored energy in the flywheel,and no resistive losses.
We then apply an instant voltage of 6 volt to the motor that is opposite in polarity to the previous 6 volts that has caused the rotational direction of the motor--what happens to the current value at the terminals when this negative 6 volts is applied instantly,when there is no place to sink the existing current already flowing through the system?.

The inductor is just like the motor/flywheel combo,in that both are energy storage devices,and in both cases,the applied negative voltage is in opposition to that stored energy,and in both cases,the circuit has no current sink for this stored energy to be dissipated into-as all are ideal.
As you stated in your example,the amplifier can be a current sink or source,as it has resistance,and that being the reason it gets hot. In MHs example,all is ideal,and has no resistance,and there for cannot sink or dissipate energy. The stored energy in the inductor is not just !gone! as MH put it,when the voltage polarity is reversed across the circuit-that still has stored energy from the previous applied voltage.That stored energy has to be dissipated,but there is no where for this stored energy to be dissipated,as there is no resistance in the circuit.


Brad

picowatt

Tinman,

In my thought experiment I also discussed using a square wave.  Do you believe the amplifier will have a problem driving the 1kHz square wave across the 5mH inductor as discussed?  Do you believe the squarewave's transition from +30 volts to -30 volts will create a large current spike somehow related to the amplifier's output impedance and the inductor's DC resistance?

Or, do you believe the amount of current that flows will be more so determined by the frequency content of that edge transition and the reactance/impedance of the 5mH inductor at those frequencies?

PW

tinman

Quote from: picowatt on June 18, 2016, 02:09:06 AM


PW


QuoteIn my thought experiment I also discussed using a square wave.

Quote:

QuoteAs the sine wave on the amplifier output reverses from its +30 volt peak towards its -30 volt peak, do you expect to see a huge current spike commensurate with the amplifier's .01R output impedance and the inductor's

???

QuoteDo you believe the amplifier will have a problem driving the 1kHz square wave across the 5mH inductor as discussed?  Do you believe the squarewave's transition from +30 volts to -30 volts will create a large current spike somehow related to the amplifier's output impedance and the inductor's DC resistance?

No and N0,as both are not ideal.
Having resistance enables power to be dissipated as heat,and having no resistance(ideal),means that power cannot be dissipated.

Here is what you are not looking at,and i have stated this a number of times now--but it keeps getting over looked--the current that is flowing through the ideal voltage source.
Switch things around,and think about this. We have an ideal voltage source(lets say an ideal cap)that has 6 volts across it. We now place an ideal current source across that ideal capacitor,but so as the current is opposite that which will flow from the cap when a load is place across it-->what happens?.


Brad