Sonic Resonator Results and Findings, As Well As LTspice Models To Download

[D.R.Jackson]

Well folks don't let no one confuse you around here with all of the analytic speculations!

The following table is from:
"The Electronics Equations Handbook"
By Stephen J. Erst 1989
Tab Books

~ Page 4

"Voltage Conversion Factors For Three Waveforms"

waveforms 

sine wave:

     Multiply By:
     rms to average    Pi/(2(Sqr 2))
     average to peak  2/Pi
     rms to peak        1/(Sqr 2)

triangle wave:

     Multiply By:
     rms to average    2/(Sqr 3)
     average to peak  0.5
     rms to peak        1/(Sqr 3)

square wave:

     Multiply By:
     rms to average    1
     average to peak   1
     rms to peak        1
     
    "all are unity for a square wave"

* * * * * * * * * * * * *

These factors apply to current too, and so, to power.

* * * * * * * * * * * * *

Ok ~ since as was posted here by one of our forum analyst ~ that the amount of power to heat a resistor to the same amount of heat as that of a purely DC power input is ( = ) equal to the rms value of the peak of the wave form.  Well lets now consider the rms value of a square wave form?

You can see that the rms value of a square wave is the same as its peak and average values, which are all unity ( 1 ).  And so. the amount of power for a 8.32 Watt peak DC square wave to heat a resistor in rms terms is the same as 8.32 Watts peak, and it is also the same as 8.32 Watts pure DC which means that the square wave is considered to do the same as the pure DC power of the same wattage level.  Or else someone has some messed up equations in their books.

Yet the above equations are based upon a square wave with a half cycle of only 180 degrees.  Now we have one that is 8.32 Watts for 230 degrees.  +50 degrees over the half cycle and is a little more effective in duration terms at heating the resistor than the mere 180 degree half cycle.

Well you can see here that if you are analyzing these wave forms as a sine wave with sine wave rms thinking, you are way off base.

I just thought that some of you should know this.

And so, what is the peak power of 5.29 Watts DC converted to peak?  If the circuit is pulling some peak sine wave power from the power supply?  It will not add up to 8.32 Watts peak.

I hope none of you get confused with all of the comments about analytical views posted here.

And so, in turn post some questions to our panel of analyst here.

I just thought that you should know that a 8.32 Watt square wave will have to have 8.32 Watts DC to heat the resistor up to the same value.  But we are talking about a pure non periodic input of 5.29 Watts pure DC, producing a 8.32 Watt square wave in a software circuit simulation.  So 5.29 Watts DC will produce less heat than a 8.32 Watt square wave. 
See? 

Do the math!

Let them analyze onwards though.

* * * * * * * * * * * * *

Now as for the one forum analyst comment here about "what we engineers and techs call..." etc.

Ok, I have been to three schools of electronics and studied "Electronics Communications Technology ~ Associate Degree" level.  And have taken advanced technical mathematics along with advanced physics in those schools.  Have studied Newtons equations, and have done extra electronics circuit design studies off and on for 10 years afterwards.  (1976)

I have worked in the field of radio electronics since 1980.  So, lets talk electronics and wave forms if you want.  Lets talk peak power, and instantaneous power, or rms, average or whatever you want.  Or even rise and decay times of the positive going peaks, whats your poison?

But you really should go to Wikipedia and look up sine waves and square waves.  And their rms, average and peak power equations.

And you should stop thinking that without reading the principles of my circuit that you know what part of the circuit is doing what.

My transistor current is I(V1) + I(Dx) + I(R1)  and I(Dx) (diode Dx) is a restored loss back into the circuit.  And you should understand that I have up to two losses in the circuit restored back into the circuit to do work with.  And so, in some models I have Dx and Dx2.

And the low pass filter is not the means of additional energy.  L2 is that means along with Dx.  In fact the restored current of Dx is one means that the period of the half cycle is extended.

[D.R.Jackson]

Eplanation of circuit, see below diagram and following post.

[D.R.Jackson]

Circuit Design And Explanation

  In the above graphic you see that the transistor collector current is equal to I(L1) + I(L2).  You will find out now how that the better expression for this is I(V1) + I(Dx) + I(R1) where R1 equals a load resistor.

  Looking at the circuit above, L1 has a very high impedance at 1 kHz and so, chokes down the portion of current that flows through the collector of Q1 and thus limits it.  And yet, the induced voltage on the collector of Q1 that is offered by L1 ~ over comes that current limitation to the power supply by upping the voltage swing on the collector which is very useful.  As far as the current limitation (choking down of current) that L1 offers, this is a desired thing in that we do not want to draw allot of power supply current and hence V*I ~ which equals watts is limited.

And L1 working off of the 10V (volt) power supply reference as a back board or spring board (fixed voltage reference), offers an induced voltage of a higher potential ~ that can easily reach up to twice the applied power supply voltage, and the heavy inductance value above in this circuit induces a voltage peak that nearly reaches three times the applied power supply voltage; ~ with its induced voltage peaks of around +27V in the models we have analyzed in LTspice.  The way L2 and Dx is contrived to work aids the circuit in adding a branch of current to collector of Q1 that has a positive potential to the collector but is independant of the power supply V1.

Now if L1 were a mere choke, we could realize and induce voltage being added into the circuit to up the voltage on the collector.  And so, add a little bit more voltage swing on the collector.  But L1 as a mere chole would be radiating magnetic energy that is not being used for anything.  So we add L2 as a secondary winding and make L1 and L2 into a transformer.  As so, recapture the radiated magnetic energy to induce a current into L2 that we can do useful work with.

  Now as to the action of L2 and Dx.  We use L2 to recapture and conserve the magnetic energy of L1 and put it to work in a useful way so that its application adds up, into the energy equation of the circuit.  And how it is meant to do this is simple to understand, though unusual in application to say the least, but you can follow it and it will make sense.

  L2 recaptures the radiated energy of L1 and converts it straight down into electron motion energy, or current flow.  And so, this current through an inductance of 0.5 Henry in L2 at 1 kHz has a certain level of energy and some of that energy that is added into this equation is the electro~mechanical energy of the circuit which is added into the circuit.

  Now that we have an increased collector voltage being supplied into the circuit concept by the inductance of L1.  We see that with respect to our collector current, we have another branch of current flow for the recaptured forces of L1 ~ directly to L2, to help move us some more electrons.  Increasing the collector current of the transistor that has had its voltage swing increased by L1.

  Notice Dx now, Dx as a diode will rectify a current that is flowing upwards through the collector of the transistor.  And so, this direction of current flow to Dx looks like another power supply terminal since when the far end of L2 swing high in positive voltage, current is attracted upwards towards Dx which passes this current to ground.  And this current does not come from the power supply but from the ground, up through the transistor emitter to collector and on up to Dx.  And to keep the current L1 to V1 from being mixed with that of L2, the current of L1 and L2 are 180 degree out of phase.

  The current equation then for the transistors collector current is (V1) + I(Dx) * I(R1) where R1 represent the load current of a load resistor R1.

  And so, the current of Dx is an induced current from L2 that is used as a means to reduce and restore losses that a mere choke coil as L1 would loose to magnetic energy if it did not have a secondary to recapture and put the energy to work.

  The comment made in a few replies back by someone posting comment here that the energy comes from the heavy power supply filter that I have between V1 and L1 in my current models.  Is the place that the energy comes from is incorrect.  The energy in this circuit comes both from L1 and L2, where L1 induces a higher collector voltage on the transisor and L2 induces a far end voltage at Dx to attract another branch of current upwards in the positivce direction to Dx.  Making Dx look like a virtual power supply terminal of the positive potential.  And this current is 180 degrees out of phase with L1 that goes back to V1 through a very heavy low pass filter section, which you do not see in the above diagram.

  So the extra induced branch current for the collector current of Dx is a cleaver way to add extra current to the circuit that is independant of the current of the power supply V1, and Dx then has its own terminal voltage that is the voltage induced at the far end of L2 that is attached to Dx.

And so, this is how I can design the circuit to work in LTspice because I know the physics of the circuit and its concepts and principles.

  The comments made so far here by forum analyst have nothing to do with the actual working of the circuit which I have disclosed to you here.

[fritz]

Ok,

Than the problem is your understanding of a square wave vs. pulsed dc.

YOUR FORMULA YOU REPEAT ALL OVER vp==vrms ONLY APPLIES TO A TRUE AC SQUAREWAVE.

Well if its AC -you should be capable of putting a big cap in series - and the signal will stay the same (after run in periode)

But your signal has a DC offset.

If you put a cap in series you will end up with a 0.64T<5.39 Watt peak + and a 0.36T>5.39 Watt peak at the negative periode - if you average that you end up with 5.39W.
Even if you have true AC - YOUR FORMULA CAN ONYL BE USED FOR 50:50 SQUAREWAVE.

If you treat the singal as a composite waveform with DC + AC component - you have to treat it with integration (as already posted in my mails).

An AC signal is DC free.

And if some electronics book tells you that a composite waveform with a  pulse and a DC offset can be treated the same as a DC free squarewave with 50:50 dutycycle - than they are incorrect at that point.

I´m not interested in a further "discussion". Everyone can be wrong some time - standing in the middle of the woods without finding a tree.

There are lots of enthusiasts here in this forum - and my only motivation is to keep the confusion down.

If you are a rf amateur since 1976 - you can ask and discuss that with friends and other people in that scene.
In rf you have caps around everywhere - maybe thats the reason why you trapped into this pulsed-dc story.

[exnihiloest]

...
My circuit supports 8.23 Watts for a period of 0.64 wavecycles [...] with only 5.29 watts in DC...

and supports 0 watt for a period of 0.36 wavecycle.
Thus the mean power is 8.23*0.64+0*0.36=5.27watts.

Well,

Spice is a nice tool -
BUT
You need to understand a little bit more to use it.
I would start to _READ_ about the difference between Energy and Power. Thats essential for further investigation.
Its _NO_ OU to make a pulse with higher power out of a powersupply with less power.
Take a photo flash - during charging it takes few watts for some seconds to charge - and can release it with the power of more than 100 watts.
Your circuit increases the power and not the energy.
...
The energy you got is your 8.xxW times the dutycycle - thats 0.64 - so its slightly less than your input cycle.

Your output voltage is not rectangular - its pulsed dc - so the rms value is as above.
The "rms"=="peak" only applies on _TRUE_ AC.

Having great tools like spice doesn´t compensate the basic understanding of electricity.

You are perfectly right, Fritz.
D.R.Jackson is confusing instantaneous and average values, AC and DC...
He should learn electricity and electronics basis.
Time is too precious to be wasted in blah.
:-)