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more noob questions - transformers

Ok ... try not to laugh when i ask... tho i only have a few questions on designing a home made transformer.. And of course i dont want to burn my house down.. so lets start from there.

I would like to make my own hv transformer and thought i would start by pulling a apart a regular 12 volt ac / dc 12 volt 1.5 amp transformer and removing the rectification.

From what i see of ac to ac transformers on design is that mains

come into and go directly to the core however i would think there is some kind of resistor before or hiden around the primary windings. First questions is this dont make alot a sense to me as it is esentially a short curcuit. However if it does actually work this way then the windings are wrapped so that itself is a good resistor to precent the short circuit. however that the main reason why I am writing all this. So my question is if there is a resistor what should it be. ( or how can i calculate what it shoudl be) . Now i know there is a billion formulas for making good transformers so that why i want to start from a premade dc to and remove the rectification so i can run it into another transformer to step it back up again. this shoudl limit what i shoudl have to calculate for resistors if needed.

. this is again why i like to know why it works that way from the mains as i will have to pass the secondary winding thru the primary of the home wound step up transformer which again would be a short circuit .. or at least what looks to be. i will put fuses ( and i would expect resistors ) between the 2 transformers to prevent too much current on my hv winding . I plan on striping another laminated core . or at least the secondary windings on a second core to wind my prefered step up

now if i use a 110 ->12vdc transformer and its a 1.5 amp . I will choose my second core to be rated around 1 amp and my hv amperage should end up in small ma rating for light usage without burn out.

I reviewed many diferent transformer designs and all they ever give is is a direct input from the mains to the windings or there just way to complicated as there beyond the scope of anythign basic.

im looking for such a basic design it just keeps the same frequency so that why i thougth stripping the rectificatin should do from a dc supply. I can figure out the insulation needed for the layers on my core to prevent it from shorting out to the core or itself due to the hv so lets not worry about that either.

If anyone could please let me know if this should really be more complicated if it really is that simple. Im looking simplest as possible design that will not blow up when I plug it into the mains.

I appreciate any imput reguarding this. I have some HV experience with whimhurst and those types of machines however not much when it comes to current limiting. Im certianly not going to make something without making sure i understand the safty parts of it first.

Thanks,

Caltus

Reply to
caltus
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There is no resistor in a transformer. There are 2 coils. The one that the mains comes in contact with is called the primary. The secondary does not make any contact at all with the primary, and therefore not the mains either. They are 2 separate coils wrapped around a piece of metal (iron I think). Since the mains voltage is AC, and therefore moving rapidly from one direction to the other, it induces a current into the secondary. The voltage you get out of the secondary depends on the ratio of the # of turns in the primary to the # of turns in the secondary. So if you have 10x as many turns in the primary than the secondary, there will be 12v (assuming 120v at the primary) on the secondary. As far as choosing the type of wire or number of wraps, that's a little out of my league, so I'll leave that to teh more experienced in the group.

As always, be very careful when working with mains voltages......

Reply to
tempus fugit

Thanks for your reply. Thats the way it looked to me without a resistor and i just wanted to make sure as primary going into the mains still is not clear to me why it dont short out. If thats the case does transformers come some power all the time? they must. And if what im doiung is essentially chaining a few together . i would think they even comsune more power without a load? or is it the other way. because there is no load it prevents current on secondary as the equal and opposite fields react as a balanced resistor until the current is used? Is that right? if this is the case the ammount of efficency loss in transformer i would think would be measureable to the constant power drain of a transformer with no load? Like i said im a noob to this so I just want to make sure i understand this the best i can befire i start with this. and again thanks for your help.

Caltus

Reply to
caltus 

--
First off, what you have is _not_ a transformer, it\'s an AC to DC
converter or, simply, a "power supply"

The transformer is the device in the power supply with two leads
connected to the mains terminals and the other leads connected
either to four rectifiers (two leads) or to two rectifiers and to
the output common terminal (three leads)
Reply to
John Fields

Hello! Th reason the coil or inductor does not short out is beacause the voltage i AC. Inductors are like capacitors, they store energy. However real world inductors do have some small resistance allso. In the inductor the energy is converted to a magnetic field. So when you turn of the power, the magnetic field collapses and is converted back to current. But very importently the current is now going in the reverse direction! With a capacitor the current will not change direction. It`s more like charging a battery. So when you have two inductor close to each other the magnetic field in one(primary) will be converted a current in then other(secondary). But if the secondary inductor has no load, no power will be used there. But like i wrote before some of the power will be lost in the inductors because of the internal resitance in the wires and thus generate heat. And heat is the problem. When you buy a transformer it is rated in VA[Voltage x Ampere]. You might think why not use Watt instead? That V*I too. True, but only for resistance. For reactive elements like inductor and capacitors there is a phaseshift between voltage and current. The "resitance" you see in such a element is called impedance Z and is dependent on the frequency of the AC signal.

So when you connect a reactive element in a circuit you must watch out because the the current can become quite high. A typical reactive device is a electrical motor or transformer. You often have to compensate this by inserting a capacitor so the phaseshift is minimal. In big factories that use alot of electrical engines this is requirement or the power company will cut the power.

Hope this helped a litle. If you want to know more about reactive elements and such you should take a look at laplace transform of electrical circuits. Its much easier than using diffrental equations :)

Good luck

Anders

Reply to
Anders N.Vinje

Sorry, no. The current does NOT change direction - in fact, that is one of the key things to learn about inductors. (The voltage ACROSS the inductor is another story, however.)

Two simple rules:

- Capacitors oppose changes in voltage (i.e., a capacitor will "try" to maintain the voltage across it at the same level despite changes in the current through it), while...

- Inductors oppose changes in current.

If you try to interrupt the current flowing through a conductor, energy will come back out of the magnetic field (where it has been stored) in such a way as to try to keep the current going at the same level. This will result in a (possibly very large) voltage spike across the inductor, if that's what it required to try to keep the current at the same level (as would be the case if you suddenly switched a large resistance into a series connection with the inductor, or simply tried to open the circuit with a switch - which is why switches that are switching large inductive loads very often "arc over").

This is expressed by the equation:

V = L(di/dt)

If you aren't familiar with calculus, all this is saying is that the voltage across an inductor is proportional to the time rate of the change of the current. Note that if the change is negative, so is the resulting voltage.

However, an ideal transformer with a purely resistive load does NOT present a reactive load at the primary, which may be a source of some confusion here. Any apparent reactance seen at the primary (and the resulting phase shift) would come from the characteristics of the load itself and from imperfect coupling between the primary and secondary windings (i.e., not all of the field from the primary winding "links" to the secondary, and so that portion of the field accounts for some inductance apparent from the primary side).

But again, a transformer per se is not NEARLY as reactive as you might think it would be from just looking at the primary winding. Neither is a motor. (Can you see why? Think about where the energy in the field is going...).

Bob M.

Reply to
Bob Myers

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