Batteries and Door Bell Transformer

[Sahib]

You all are amazing.

That shows how nice you are as you yourself are an expert (on code).:thumbsup:

 

[Besoeker]

You all are amazing. I actually got most of it. Thank you

Another question... How does one determine the number of turns of wire needed to get the needed transformed voltage. Can I assume the number of turns on the primary and secondary are different?

As I recall, the primary turns are calculated to get the required flux density. Typically 1.6 Tesla.
It is usual for the primary and secondary turns are different That's to get a different voltage out than what you put in. That is the most common function of a transformer - to transform the voltage from one value of voltage to another. There are other purposes like electrical isolation of phase shift.

 

[LarryFine]

Can I assume the number of turns on the primary and secondary are different?

To add:
Yes, except for 1:1 isolation whose sole purpose is to provide an ungrounded power supply.

Again, there may actually be a few extra turns on the secondary to compensate for losses.

 

[Dennis Alwon]

To add:
Yes, except for 1:1 isolation whose sole purpose is to provide an ungrounded power supply.

Again, there may actually be a few extra turns on the secondary to compensate for losses.

So for an ungrounded system you would use that system and you are also saying you can go in with 480v and still come out with 480v if the windings are the ~same.

Is the 1/2v per turn pretty standard or were you just using that as an example?

 

[RumRunner]

I was thinking that if I used acircular magnet as the core and then wrapped a wire around opposite sides- One the primary and the other the secondary-- what would happen. Would I get nothing?

By having a circular (magnet) core you will have the semblance of a toroidal transformer. The behavior of the magnetic flux would be the same with dough nut-shaped as a square- shaped core.

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Both of these shapes however would have tremendous losses since magnets are solid materials.


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Cores should be laminated.


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The eddy current is proportional to the square of the diameter or cross section of the core. So, the bigger the core-- the more losses are encountered.

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By using small pieces in the form of laminated core we reduce the eddy current loss. Without these small pieces as in the case of solid one-piece core as you had envisioned—more energy will be converted in the form of heat. That's not what we want in this particular instance.


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This scheme would be useful if you are making a soldering gun or a smelting furnace.


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There will still be losses with the laminated core but those are something we can tolerate.


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Some new materials can be used for solid cores that can be manufactured like those ferrite cores but they are limited in sizes.

They are mostly used in miniaturized components in computer technology.


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And if you mention-- why can't they be made for bigger transformers. . well. . . we would simply be herded back to the Laws of Physics which says: “The larger the cross section of the core the larger the eddy current loss.”
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[gar]

190101-1304 EST

Volts per turn is very dependent upon frequency, core area, and saturation flux level of the core material.

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[LarryFine]

So for an ungrounded system you would use that system and you are also saying you can go in with 480v and still come out with 480v if the windings are the ~same.

Yes, I think, at least to the second half of your question.

Is the 1/2v per turn pretty standard or were you just using that as an example?

Made up just to make the point. It could be more or less.


Are you aware that a soldering gun is a transformer with a 1-turn secondary?

Click:

 

[Dennis Alwon]

Are you aware that a soldering gun is a transformer with a 1-turn secondary?

I did not. I assume so the point if broken doesn't hurt someone but in this case are we looking at <1v on the secondary (when you say it can vary are we talking 1-2 volts or much more). Certainly doesn't seem like enough for the element.

 

[LarryFine]

I did not. I assume so the point if broken doesn't hurt someone but in this case are we looking at <1v on the secondary (when you say it can vary are we talking 1-2 volts or much more). Certainly doesn't seem like enough for the element.

The typical Weller soldering gun is a dual-power unit, 100w and 140w, selected by a tap on the primary, depending on how hard you pull the trigger switch. Let's use 120w for this discussion.

Remember how a transformer works. To produce 120w on 120v, it requires only 1a. To produce 120w on 2v, it requires 60a, which is enough to heat a #10 or so conductor tip quite nicely.

 

[gar]

190101-19219 EST

A Weller 8200 gun is about 0.23 or 0.27 V measured across the two nuts. That ratio when squred is about 1.38 amd corresponds with power ratio of 140/100 = 1.4 .

Approximation of the number of turns on the primary 125/.23 = 544. But likely it is more in the range of 544*1.5 or 820 from internal impedance.

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[Sahib]

Is the 1/2v per turn pretty standard or were you just using that as an example?

You may use this formula to a fairly high degree of accuracy for a given (single phase ) transformer:

Primary voltage/secondary voltage=secondary current/primary current=primary no of turns/secondary no of turns.

 

[gar]

190102-0725 EST

Sahib:

Turns ratio is an approximation of voltage or current ratios, but not high accuracy.


On resisitivity relative to comments by others:

Copper about 0.017 micro ohm meter from
https://www.google.com/search?q=res....69i57j0l5.11518j0j8&sourceid=chrome&ie=UTF-8

Permanent magnetic material NQP-B powder from table 20 about 100 micro ohm meter from
file:///Users/bradra2/Downloads/7_Slusarek.pdf

A 5000 to 1 ratio.


Magnetic fields are additive taking into account sign.

Set a permanent magnet at some point, A, in space. Two meters away put an AC excited coil, point C. At a mid point, B, measure the magnetic field. This consists of the sum of a DC component and the AC component. Measurable with a Hall device. The Hall device will see both the DC and AC components. Note also that the earth's nearly DC field, it slightly oscillates at a fraction of a Hz, is there as part of the sum. Put a coil at point B and measure the voltage. There is only an AC component.

Move the AC exciting coil to overlay the permanent magnetic. To the extent that demagnetization does not occur the field at B is similar, except for a phase reversal of the AC component. But the magnetic material being in close proximity to the coil does influence the AC field intensity at B.

Within the last year I set up a coil in my living room to measure the earths magnetic field intensity. This is done with a coil, a measurable current to the coil, information on the coil, and a compass. Ran this same experiment back in the 40s for high school physics experiment. Earth's field has changed quite a bit in that time.

My home points north and south, but my compass was way off from pointing north. At the experiment time I had two speakers in the room about 10 ft from the compass. The stray field from the speakers was sufficient to shift the compass reading. Moved the speakers out of the room because my goal was to measure the earth's field. In the early 1800s surveyors in the Negaunee, MI area were getting strange compass readings. This led to the discovery of huge deposits of iron ore.

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[Frank DuVal]

Sixth page and no mention of Right Hand Rule? This explains the relationship between flux and current flow. So, the flux generated by a primary winding of a PT or CT is found by RHR, then this flux producing the current in the secondary can also be shown by RHR with "flux fingers" in correct position to show secondary current. Google it.

And yes, current reverses polarity ( and direction of your hand) at a 60 cps rate, but primary and secondary always see the same instantaneous flux force (hand direction).

Current makes heat , i.e. I^2 R =W. Voltage only important in its relationship to making current flow at a resistance. Weller 8200 is 100 watts on low, resistance of #10 (1.21 ohms/ 1000 feet) for 3 inches (1/4 foot) is .0003025 ohms, a pretty good short. 100/.0003025=330578.5, sq root of that is 575 amps. I would say the nut connections and probably closer to #12 awg wire makes current less than that in actuality. BTW, at that, E=IR E= 575 x .0003025 = 0.17 volts. So Weller will not shock you, but gives a nice burn, DAHIK!

 

[Sahib]

190102-0725 EST

Sahib:

Turns ratio is an approximation of voltage or current ratios, but not high accuracy.




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The transformer is a very high efficiency device and so its internal resistance is very small. Hence the induced voltage almost equal to supply voltage and so the turns ratio is close to voltage ratio, IMO.

 

[winnie]

The transformer is a very high efficiency device and so its internal resistance is very small. Hence the induced voltage almost equal to supply voltage and so the turns ratio is close to voltage ratio, IMO.

The larger the transformer, the more true this statement is. For a 150VA control transformer, not so much.

-Jon

 

[mivey]

As I recall, the primary turns are calculated to get the required flux density. Typically 1.6 Tesla.
It is usual for the primary and secondary turns are different That's to get a different voltage out than what you put in. That is the most common function of a transformer - to transform the voltage from one value of voltage to another. There are other purposes like electrical isolation of phase shift.

I agree.

There is a multi-step design process. Most texts aren't detailed (not really needed) but I have this detailed in a paper or book somewhere. Interesting to read but not used in my day-to-day activity.

Was it J&P? I don't have my copy handy at the moment.