Wesley,
Nice vid 200 watts out.
You know us.................
DETAILS PLEASE!!!!!!!!!!!!!!!!!!!!!!!
PS Well done
Here is the tough one.
Results are expected
Results might wait to be delivered in easy to absorb form
There is number of factors related.
-concerns about recent happenings... messages received, say warnings,approaches absolutely unexpected, and unwanted,Do not want to go to details.When it increases than I will go loud very loud and if that not would be me any longer than someone will..I'm not afraid.Data is being collected, saved and distributed on daily basis. lets
deliver it different way.
Preliminary ( at maximum brightens)expanation of used terminology:
apparatus- caduceus based device with all components
resistive load - (light bulb)
fundamental - frequency of primary oscillation
sinus - shape of signal
bandwidth of active sweep - is active region of frequency sweep at which results of interaction between power delivered and power dissipated have been visible as significant change.
effective energy dissipated - is energy consumed
impedance mismatch- is difference between impedance of input and impedance of output
that causes energy reflected back to port 1
port 1 - point#11 at schematic diagram.
port#2 - light bulb connection
network- is the set of devices under analysis representing properties of analysis.
our network is set that starts from port 1 and ends at port 2.
reversed attenuator - is an apparatus that is relative to amplifier but of highly non linear nature
harmonic frequency-
A harmonic frequency is a multiple of a fundamental frequency.
example:
A fundamental frequency of 500Hz has a first harmonic frequency of 1000Hz, double the fundamental frequency. Its second harmonic is 1500Hz, the third harmonic is 2000Hz and so on. A musical instrument produces both fundamental and harmonic frequencies, which allows the human ear to discern the differences between instruments even if they are playing the same note.
SWR-
Standing Wave Ratiocan we have SWR=1
yes we can be close to that.
can we have and S12=S21?
read this
Properties of lossless networks.For a network to be lossless, all of the power (or energy) that is incident at any one port has to be accounted for by summing the power output at the other ports with the power reflected at the incident port. None of the power is converted to heat or radiated within a lossless network. Note that an active device is not in the same category as a lossless part, since power is added to the network through its bias connections.
Within the S-parameter matrix of a lossless network, the sum of the squares of the magnitudes of any row must total unity (unity is a fancy way of saying "one"). If any of the rows' sum-of-the-squares is less than one, there is a lossy element within the network, or something is radiating.
Why are we looking at sum of the squares instead of sum of the elements themselves? Because the S matrix is express in terms of voltage, and as we said, we are accounting for power. Power is proportional to voltage squared[/i], get it?
Guess what? You can never make a lossless network. But you can come extremely close
In this point one might raise doubt that I have omitted amplification factor and a comparison S21 to S12 is not reflecting and/or correlating with factional experiment.
That might be good point but as we do not deal with traditional network that is assigned to particular group of network devices such as
-filters
-amplifiers
we may assume that we could compare the apparatus to "reversed attenuator" of non linear nature.
Amplification factor of such apparatus should be taken to consideration strictly based on its performance at given frequency within given spectrum.
One might also find VNA analysis to be close to impossible with the presence of spark gaps.
Yes it is not easy ..but VNA analogy could be applied at any time.
HV delivered to #11 of diagram.12.9 kV
generator power supply,set at max 30W
SWR mismatch
Reversed power 28W (losses)
2W effective energy dissipated by the apparatus including resistive load (light bulb) light bulbs in series
28W loss because of impedance mismatch.
Frequency bandwidth occupied 100 to 124 KHz
Limitation:
-at present time no ability to check how device react below 100KHz due to limitation of HV driving circuitry.
- Suspected that investigation was provided at third harmonic frequency up, not primary resonance frequency
1st=56000Hz
2nd=84000Hz
3rd=112000Hz
the fundamental should be 28kHz
signal delivered to Abramienko fork sinus
the same situation as in Lithuania Experiment at secondary of flyback we have had sinus of even amplitude at 16.8KHz
Energy dissipation Only fundamental frequency current can provide real power. Current delivered at harmonic frequencies doesn't deliver any real power to the load. When current of a single frequency is present in a system, you can use the measured values in Ohm's Law and power calculations
However, when currents of more than one frequency are present, direct addition of the current values leads to a summed value that doesn't correctly represent the total effect of the multiple currents. Instead, you need to add the currents in a manner known as the “root mean square†summation.
due to spark gap implementation we assume that we dealing with nonlinear load.
nonlinear load is our apparatus including 2x100W / 110V
in Lithuania experiment:
we have used generator that supplied the impulse to the transformer at secondary of transformer we have had even sinus at frequency of 16.6KHz
at recent experiment lets call it NY Experiment
we have used frequency of 122kHz in within bandwidth of active sweep 24KHz
explanation:
bandwidth of active sweep - is active region of frequency sweep at which results of interaction between power delivered and power dissipated have been visible as significant change.
Means of tuning: Frequency change to max of lightness that may stay about tuning to minimum loss of ports 2-1 to 1-2 in Vector Network Analysis
At Smith chart applied to ADVANTEST R3754B VNA
that would be represented as correlation between input and output losses due to transition of the signal( energy)
example:SWR=1 to 1 =0 energy losses
when mismatch is present than returned signal reversed power (equal to BEMF)
represent significant losses in the network.
in this particular situation it is loss of 28W
if for instance 28W was at SWR 1 to 1 and P of S21= P of S12 than 28W would become dissipated energy + 2 W of real energy utilized by apparatus
If we make assignment of ports: 1 and 2
S12 input versus output
S21 Output versus input
Purely Resistive load is our light bulb at the output
Load is than changing resistance and affecting overall impedance match of the network, upon temperature to change.That creates need of frequency adjustment at
generator power supply .
Practically on video you can see me waiting a little while tuning light bulb to maximum brightness
Delta T is causing delay of the adjustment as it takes time for resistive component of light bulb to change its resistance that means impedance of network between P1 and P2 has to be readjusted by frequency change of generator power supply.
As L and C of light bulb is minimal we may state that impedance of load =resistance at the load point.
than the Inductive Reactance and Capacitive Reactance = 0 or close to zero, at any time.
At that point we are dealing with impedance match between
- energy supply generating device
and
- light bulb
only when considering network analysis
the only change might be taken to consideration is frequency tuning at
generator power supply to compensate resistance=impedance change, affecting overall resonance frequency of the load
That is if we look from generator power supply output delivering signal to #11 and rest of apparatus.
When calculating real power utilized by device we are dealing with
- generator power supply
and
- receiving energy apparatus including light bulbs
Instead of stating data I give you little calculation example:
current is easy to find out.
make your math:
and no mf..r will be able to take advantage of me.Example: A transformer is required to supply a nonlinear load comprised of 200A of fundamental (60 Hz), 30A of 3rd harmonic, 48A of 5th harmonic and 79A of 7th harmonic. Find the required k factor rating of the transformer:
Total rms current, I = [square root of [([I.sub.1]).sup.2] + [([I.sub.3]).sup.2] + [([I.sub.5]).sup.2] + [([I.sub.7]).sup.2]]
Total rms current, I = [square root of [(200).sup.2] + [(30).sup.2] + [(48).sup.2] + [(79).sup.2]] = 222.4A
[I.sub.1] = 200 / 222.4 = 0.899
[I.sub.3] = 30 / 222.4 = 0.135
[I.sub.5] = 48 / 222.4 = 0.216
[I.sub.7] = 79 / 222.4 = 0.355
k = [(0.899).sup.2][(1).sup.2] + [(0.135).sup.2] [(3).sup.2] + [(0.216).sup.2]([5).sup.2] + [(0.355).sup.2][(7).sup.2] = 8.31
To address the harmonic loading in this example, you should specify a transformer capable of supplying a minimum of 222.4A with a k rating of 9. Of course, it would be best to consider possible load growth and adjust the minimum capacity accordingly.
Conclusions:Take (in watts)
1.
effective energy dissipated
and
2.
resistive load energy consumption (manufacturer set parameters 100Wx2 at 110V light bulbs in series
3. calculated ratio is to be used to find how much power will be present at resistive load if 28W of power will not be lost due to SWR.
4. add two values
-effective energy dissipated
and
-
reversed power loss5.use this number as total delivery without losses from
generator power supply6.take result of calculation from point 3 and use it .
try to calculate
a. what was a real gain at NY experiment
b. What would be real gain if reversed power loss would not be present.
c. How much power than will be at output of apparatus in Watts
Wesley
PS do not forget that if I operate at 3 harmonic than we are not dealing with real power of fundamental frequency.
One more test provided that will be documented
Resistive load (
light bulbs in series connected directly to power supply
- light was not present
- spark is present if interrupted.