Solid States Devices > solid state devices

extra electric power / power amp.

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ckm:
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:
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:
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:

--- Quote from: ckm on April 06, 2005, 04:11:11 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.

--- End quote ---

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:
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.

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