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### Author Topic: extra electric power / power amp.  (Read 82453 times)

#### ckm

• Newbie
• Posts: 14
##### extra electric power / power amp.
« on: April 03, 2005, 07:35:44 PM »
Think this is the right forum for this, please move it if I am wrong.

Not built nor tested, just idea stage.

From any power source availible, split the circuit into a parallel circuit -

in one put in a high voltage transformer

in the other put a high current transformer ( usually called a meter )

then feed both circuit paths back into one.

The ratings will need to match up of course with whatever the power is for, it'd probably work best in an off-the-grid type of home (or commercial...) power station / bank-supply set up.

Fun-est projected possible set-ups could include the farm powered by one small battery, or a tapwater minicell.

#### irisher

• Newbie
• Posts: 4
##### Re: extra electric power / power amp.
« Reply #1 on: April 04, 2005, 04:25:54 AM »
You can try it, but , you will probably find that your parallel cicuits will average out to be equal. Remember "EI=P". If you have 12watts of power beimg fed into the parallel nexus ,then, you will have 6 watts in each circuit (all loads being equal).one cicuit leg has a 1000:1 ratio xformer and the other has a 1:1000 xformer. As a result in one circuit you will have 6000 volts at 1 ma. In the other you will have 6000 amps at 1 mv. total of 6 watts in each circuit.You will have the same watts as you put in minus the xformer losses,i.e.,less power than you put in.

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #2 on: April 04, 2005, 06:21:17 PM »
P=IE and E=IR

In any transformer P in = P out.  That means that as Voltage is stepped up, Current steps down.

At the point of convergence (after your transformers) you would have 3 legs (input1 input2 and Output)
Lets say the following.

In1 = 6000V @ 1mA
In2 = 1mV @ 6000A

Here lies the problem...in order to keep the power from traveling back up to the transformer, diodes would have to be used.  The problem with that is that the bias voltage would never be reached, because In1 would feed 6000V into the junction point, but the diode at In2 would never achieve forward bias(.5-.7V)  because there is only 1 mV upstream of it.

What a mess.....without the diodes, irisher is probably right.  Convention says that they will average out.  But you will have some serious back current on one transformer...where does the Current equalize, how much heat will be created, will something smoke(the directions and warnings about putting unlike batteries in parralell seem to point that way) ?  These are all questions I am left with.  No,  don't think that the currents and voltages will simply add, more likely the Voltage will average;  and then add the currents in proportion to your voltages....but this is without any strange fluctuations, frequency induced harmonics....all of which might supprise you.

Give 'er a shot, play with the freqs, but all behind some safety glass !!
Let me know how it goes.

#### betajim

• Newbie
• Posts: 26
##### Re: extra electric power / power amp.
« Reply #3 on: April 06, 2005, 05:31:10 PM »
I forget what E signifies there,

do you have the same kind of triangular forumula for V I and R, and any other formula with V and I in?

What you say about 'P in = P out' is what I am thinking can be exploited in some way - step up the voltage and the I decreases, step up the I and the V decreases - this is suggestive that an increase of V and I will indeed increase P.

About E: some folks and text books use E instead of V, they are the same thing.

What the other person said about Pin = Pout is just that you can't get more power out of a transformer
than you put in. In reality though, Pin > Pout because of loses in the transformer.

Take care, Matthew

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #4 on: April 06, 2005, 05:57:55 PM »
Here is what I was talking about with the diodes.

Point A and C are Electrically the same point, ie. with the exception of line loss, you have the same voltage and current readings at both points.  If T1 is your High Voltage Transformer, and T2 is your High Current Transformer, then at point C you will have a higher voltage then at point E, and your diode will be reverse biased.  which means that if you don't smoke it all together, there will be no current flow through d2, as you need forward bias of .3(GaAs) to .7(Si) to allow flow.  The only way to get High Current, High Voltage is to play with the phase angle so that when there is positive voltage at E, the Voltage at A is negative. But even then, you will alternate between High current, low voltage spikes and High voltage, low current spikes, rectified to halfwaves.  I'll draw that for you.

I appologize for the quality of the drawing, but its the best I can do at work.

(http://)

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #5 on: April 06, 2005, 06:02:51 PM »
Here is the drawing of the wave form that I mentioned.

#### betajim

• Newbie
• Posts: 26
##### Re: extra electric power / power amp.
« Reply #6 on: April 06, 2005, 06:24:20 PM »
Whoa! This is way off - in high school our physics class undoubtedly used both E and V in electrical-based equations, but they were in no way ever given as the meaning the same thing.

Well, here is what Kysmett posted above:
Quote
P=IE and E=IR
Replace E with V and you have Ohm's familiar law. E and V represent the same thing: electromotive force.
Any way, they are just symbols and in this context they do mean the same thing.

The power losses per transformer still look to be additive if both are fed back into one -

if V is stepped up considerably in one circuit - taking into account the original power is diminished because the circuit is split,
and I is stepped up in the other circuit,
and both those stepped-up values can be measured as those values (the current transformer being sold as such a meter in many cases), where do these stepped-up values go, if re-joining them ( after they are stepped-up ) will not equal both their values added together?.

A power loss is just that, a loss. You don't get that energy back. Some of the electrical energy flowing in the transformer is
converted to heat through resistive and eddy current loses. The setup you discribe creates no additional power.

Take care, Matthew.

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #7 on: April 06, 2005, 06:25:55 PM »
Here is the shorthand I was taught and use:
I=Current
E=Voltage
R=Resistance
Z=Impedence
Xl=inducltive reactance
Xc = capacitive reactance
P=Power

The diodes are to keep backflow from the other transformer. ?It was just an idea, to protect the transformers. ?The reason I think it might smoke is that there will be a short of different voltages at the convergence point. ?This causes heat....

Try running a couple small transformers...power adapters for example. ?Run one forward and one backwards--- so that they are step up and step down respectively.
Keep it in a fire safe environment, as you are shorting two seperate voltages at the convergence point, and stay behind some plexiglass or something. ?If it doesn't blow or smoke, take a scope shot and let me know. ?I just realized that Voltage will lead the current through the transformers, and I'm not sure what effect, if any that will have. ?The diodes should be able to stand up to the voltages and currents. ?

Kind of skepticle, but dying to be supprised

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #8 on: April 08, 2005, 02:47:29 PM »
Try it.

Use simple power adapters as your transformers.  See what happens.  (Note: Do not read any sarcasm here...I'm truly interested)   Try it even without the diodes.   Right now I cant get my head around what the phase shift caused by the transformers will do to the wave form (voltage leads current by 90 degrees)

Here is the information I think you were looking for.  Voltage adds only in series, otherwise averages.  Current adds only in parallell, otherwise averages.  I think thats right.  If there is a way that you could get the voltage and the current to add differently, then you may be on to something.

Look forward to hearing about the results.  I may even rummage around to find my stash of spare adapters to try it.  If I do, I'll let you know.  The only thing is that I don't have a frequency generator, so any work I do will be limited to either DC or 60 Hz.

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #9 on: April 11, 2005, 04:37:36 PM »
Here is the relevance of voltage and current adding/ averaging, depending on configuration.

At the convergance point, you are putting the two outputs in paralell.? So, the currents would add, but the voltage will (or should if the experiments I've done in the past with bateries hold true for transformers and AC) equal 1/( 1/Voltage1 + 1/Voltage2)

So if you had 2 ten-to-one transformers and you started out with a 100V 10A signal(and assuming no loos in the transformer--which there will be) on one output you would have 1000V 1A and on the other you would have 10V 100A.? Where they meet, you would add the current so there would be 101A, but the voltage(Vout) would follow the above equation:

Vout=1/(0.001 + 0.1) = 9.901

Now if you multipy the Current and the Voltage you will still have 1000W

#### rlm555339

• Jr. Member
• Posts: 73
##### Re: extra electric power / power amp.
« Reply #10 on: April 12, 2005, 03:26:47 PM »
Huh?

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #11 on: April 12, 2005, 10:58:11 PM »
Here is what I think you are talking about.  This is what is known as the ohms law wheel.  It has all the relationships between voltage current resistance and power.  Remember though that these equations are only valid when all of the measurements are taken at the same point.

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #12 on: April 13, 2005, 05:32:12 PM »
You are right...IF you can simultaneously increase the current and the voltage at the SAME POINT.

The reversed transformer setup will actually reduce the voltage downstream of where they merge...even as they ever so slightly increase the current, maintaining a constant P throughout.

#### BushWacker

• Guest
##### Re: extra electric power / power amp.
« Reply #13 on: April 13, 2005, 09:01:42 PM »
Thats exactly right Kysmett. I want to point out an interesting note which you may or may not be interested in hearing. I modified the latest design for the MEG unit using a nanocrysaline core I purchased from Elna Magnetics. I added two extra coils across from the original input coils, and crossed them. The first set of input coils criscross kitty corner to the second set of coils in an X fashion. The first set is wound with 200 winds of 26 gauge tesla wire, the second set is wound with 300 winds of the same wire. The output coils are each wound with 3000 winds of 27 gauge wire. I then used SmCo magnets in the center of the unit. My readings average a better than 11 to one ratio on the output Voltage, but can't measure any Amperage at all.
Now hears what I found most interesting and I believe that this is where the true potential is, in my humble opinion at least. I was curious to see what effect if any audio input would have when passed through the modified MEG unit. I didn't have anything else to use for the experiment besides my Toshiba laptop and free tone generator software which I downloaded off the Net, so I plugged in the audio cables to the MEG, and turned the volume up as high as the poor pittiful laptop would go, which is not much to say the least, and began running different audio frequencies through the first stage input coils. When I hit 1600Hz, the MEG began to register 24v AC on the digital multimeter. Now consider the old school crystal radio technology? What do you think would happen if we were to build a double wound field coil to resonate at the Earth's own frequency/Shumann resonance, and modifiy/change the output to correspond to 1600Hz and route it through the modified MEG unit? This is just an idea now, and I have not had the free time to try this, nor the finances to by the neccessary materials, but I sure am curious, lol. Perhaps this is something to note anyway, and maybe someone out there does have the free time and extra cash to give it a go. I would be willing to help in any way possible. Again guy's, just a notion.

#### Kysmett

• Full Member
• Posts: 101
##### Re: extra electric power / power amp.
« Reply #14 on: April 14, 2005, 10:42:09 PM »
When I say 'reverse the transformer' I am refering to the fact that all transformers are reversable.  They work on a ratio.  a 1:10 transformer, for example, will put out 10 volts for every volt you put in, but at 1/10th of the amperage.  If you ran the transformer backwards it would be a 10:1 transformer, granted giving you 10 times the amps, but one tenth the voltage.  You can do both with identicle transformers.

So, saying that, and under the assumtion that you were using identicle transformers, one would be configured 'in the reverse' of the other.  One being a step-up, one being a step-down.

As for joining the bateries together, that would be a great illustration of what I am talking about.
Take a 9 volt batery, and from the negative place a 10000 ohm resistor, then take a D cell and run a 100 ohm resistor.  This will give you high volts and small amp on one and lower volt and more amp on the other.  tie the output from the resistors together and attach to common (positives).  Let us know what you find.