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Transformer Theory

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StudentSA

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Good Day,

I need help understanding transformer operation, I have been working on a Microwave Oven Transformer conversion commonly found on YouTube and the likes where the secondary coil is replaced with 2 turns of thick gauge wire (50/75mm^2). The idea is that voltage is stepped down and current is increased.

If the secondary side is "shorted" onto some metal, the high current transfer can be used to spot weld.

Anyhow, Im super confused on the theory... how is the current through the secondary calculated. I want to understand how secondary turns affects current and how resistance of the secondary (i.e from using thicker wire or shorter length) affect current. Can we actually calculate a value for current given the turns and resistance.

From my experimentation, I believe the primary coil (original, not tampered with) is 240 turns and measures (1.6 Ohm). This makes sense as two turns on the secondary with thick wire has produces a secondary voltage of 2V from 240Vac on the primary (for my side of the world).

I guess the basis of my question is "how do I maximize current at the secondary and derive a formula to calculate the expected current?"

Thanks,
StudentSA
 

JLNY

Active Member
I'm not an expert on spot welders, but my understanding is that it's not necessarily the resistance of your coil that you need to be thinking about, but the resistance of your load and the current that your transformer can deliver. Ideally, the piece you will be welding will be higher resistance than the other parts of your system (ideally, your coil should have as little resistance as possible).

A spot welder is effectively a low voltage source with huge current handling capability. That large surge of current through a slightly resistive load is what creates the weld.

For an ideal transformer (i.e. no core losses) the power at the input will equal the power at the output:

Ppri=Psec,
or
Vpri*Ipri=Vsec*Isec

and the current through your load (assuming the rest of your system is effectively 0 Ohms) will just be the voltage divided by the load resistance:

2Vac/Rload=Iload

For example, if your secondary voltage is 2V and you clamp it across a load with a series resistance of, say 0.005 Ohms, it will draw about 400A at 2V (800W), so at your 240Vac primary, you will see a current draw of 3.33 Amps (800W/240V=3.33 Amps). I don't know what a typical spot welding resistance would be, but this is the basic theory of an ideal transformer.
 
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chemelec

Well-Known Member
Transformer.jpg Here is One I made for HIGH CURRENT.
Flattened out, 1/2 inch Copper water Pipe as the Secondary wire.
 

dr pepper

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For smaller trannys the current is multiplied by the turns ratio, ie if you pull 10a from the sec and the sec voltage is 24v with a primary voltage of 240v, then 24 x 0.1 = 1a from the primary.
However with your tranny there are 2 other factors, one is loss in the windings resistance, both pri and sec, this means there will be a little more current flow in the pri that the current in the sec / turns ratio due to the losses.
Increasing the copper area like chemelecs copper tube windings will decrease losses.
The other thing and this is a factor with mot's is magnetic coupling between the windings, the coupling on a mot is loose, as they have magnetic shunts, you can remove these to increase it a little, however the core still has all its E's and I's aligned, this increases leakage flux and reduces coupling, this is done to increase the impedance of the transformer to match that of the magnetron.
One thing to consider is mots will saturate badly with no load, and even a bit with a load so cant run for long, no problem with a spotter.
When calcing a tranny current is not necessarily used for working out the turns, the core is concerned more about volts x seconds than amps, however the windings obviously need enough copper area to carry the current, 5a per mm2 is a ball park, it depends on the how hot you want to run it.
 

StudentSA

Member
Thanks, the points mentioned shed some light and I think I understandas follows...
A transformer will primarily do voltage transformations based on the turns ratio.

So in my case I get roughly 1v per turn on the secondary. If I have 2 turns I will induce 2vac at my secondary. The current is thus according to ohms law, I=v/r. So current will be primary driven by the resistance of the load(material being welded). You would then be able to increase current by increasing turns on the secondary (and thus voltage), However in this case it is important to ensure ur cable can carry this high current.

So in essence what I gather is its more of a debate around available space on ur core for thicker wire and turns. The more turns u put the higher the voltage and current, however the cable must be up to the job.
 

ericgibbs

Well-Known Member
Most Helpful Member
Hi RSA,
A rough rule of thumb is to measure the CSA cross sectional area of the transformers core.
Then its inversely 6 turns per volt for every square inch of core.
So a 1 inch CSA is 6 turns/volt
a 2 inch CSA is 3 turns/volts
a 3 inch CSA is 2 turns/volt and so on.

What do you measure the CSA on your transformer.??
E
 

JimB

Super Moderator
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A rough rule of thumb is to measure the CSA cross sectional area of the transformers core.
Then its inversely 6 turns per volt for every square inch of core.
I am at the moment in the process of re-winding a small transformer with a 0.68 x 0.68 inch core, (0.46 sq in) which would give 13 turns/volt.

Counting the turns on the original 9v winding, I found 115 turns.
This equates to 12.77 turns/volt.

Good information Eric.

JimB
 

nsaspook

Well-Known Member
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Larger transformers usually have a 'percentage impedance' (voltage drop on full load due to the winding resistance and leakage reactance ) that's used to calculated short circuit currents.

http://www.eaton.com/EGDRCUS/US/Lea...Released&Rendition=Primary&dDocName=CT_154545

I'm replacing this transformer (chinese power) with one designed for a US delta power feed so I would like to
match the 'percentage impedance' for electrical safety reasons from the main panel to equipment feed.

 

MaxHeadRoom78

Well-Known Member
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A typical example akin to a spot welder is the Weller style soldering gun where the secondary is one turn and ~100+ secondary amps flowing.
Max.
 

Colin

Active Member
The easiest way to work out if a transformer is suitable is to work on "watts" VA. If it is a 1,000 VA transformer, you will be able to up to 1kW into a spot-weld. Of course the secondary winding and the leads have to be correct to allow the high current to flow.
 

nsaspook

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dr pepper

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Hi studentsa, to answer that a little more simply, the turns ratio on an ideal transformer transforms the voltage and current, ie if the secondary has 1/10 the turns as the primary, the secondary voltage will be 10 times less and the current 10 times more than the primary, multiplying I and V then is still the same primary and secondary.
Parasitics also apply as we've all mentioned.
Because your using an mot in a different application there will be factors increasing the losses, all you can do is try it.
 

MrAl

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Hi,

If the primary has 10 turns for every 1 secondary turn, then if we have 1 ohm secondary resistance and 1 amp secondary current we loose 1 volt in the secondary. Since we loose 1 volt in the secondary and the reverse turns ratio is 1:10, that is equivalent to loosing 10 volts in the primary. Since the primary current is 1/10 of the secondary current, the primary current is 0.1 amp. To loose 10 volts in the primary at 0.1 amp we need 100 ohms because 100 ohms times 0.1 amps is 10 volts.
Thus, the resistance of the secondary is reflected to the primary by the square of the reverse turns ratio. In this case the ratio was 10 so that is 10^2 and since the secondary winding resistance was 1 ohm that means the equivalent primary resistance is:
100*1 ohm=100 ohms.

The primary also has resistance, so when this 100 ohms is added to the primary resistance we get the total equivalent primary resistance, and can then figure that the secondary has no resistance. This allows us to apply the voltage to the primary, calculate the voltage drop due to the primary current, and apply that new voltage to an ideal transformer with 10:1 turns ratio and that gives us the true secondary voltage.

So for example, if the primary was 10 ohms and secondary 1 ohm and the turns ratio 10, then we have total resistance:
100+10=110

and if the current in the primary is 0.1 amp we have a drop of 11 volts, which leaves us with 109v with a 120v input. With the 10:1 turns ratio that gives us 10.9v on the output. So instead of getting 12v out we only get 10.9v out.

That's just one example.

We did ignore the leakage inductance in the above though, which could cause more voltage loss in the secondary output.
 

StudentSA

Member
Hi,
Im not following what you mean Eric, How is the cross sectional area relevant? my transformer measures as follows:
Transformer.jpg

Sorry Mr Al, you completely lost me.
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

To make a long story short, it is hard to calculate the exact current because we dont know the leakage inductance, but we can calculate the max current that you can get from an ideal transformer, and that will be the upper limit of what you can get in real life.

Also, the secondary resistance affects the output because the resistance reflects back to the primary as the square of the reverse turns ratio. Lets look at your likely transformer operation.

You stated that you get 2v out with 240v in. That means the turns ratio is 120:1 and this is just 240 divided by 2.
That means the secondary resistance reflects to the primary as 120^2 times the secondary resistance. That's 14400.
Now given a secondary resistance of 0.0025 ohms, which is 10 feet of #4 gauge (AWG) wire, the reflected resistance is 0.0025*14400=36 ohms.

Now if the core cross section is 1.2 by 2.5 the area is 3 sq inches. According to Eric's formula, that means we need 2 turms for each primary volt. That's how the area fits in. And that means we have about 480 turns on the primary.
Assuming the primary creates a 1/2 inch build, each turn at the start is 7.4 inches and each turn near the end is 11.4 inches, so the nominal turn length is 9.2 inches. With 480 turns that means we have 368 feet of wire. That means we have a primary resistance of about 0.94 Ohms assuming the wire gauge is at least #14 (AWG).
The primary resistance of 36 Ohms reflects to the secondary as 36/14400=0.0025 Ohms.

So the total primary resistance is close to 37 ohms.

With the output load a complete and perfect short, the input current is:
Iin=240/37=6.5 amps
With 6.5 amps in the input we have in the output:
Iout=6.5*120=780 amps

I think i did this right but you or someone else may want to double check those numbers if i dont get to do it first [LATER: checked and corrected now].

Also remember this does not include the leakage inductance which reduces output current because it also tends to reduce the influence of the primary voltage. It's hard to say what the leakage inductance is without some measurements like the short circuit current with a small input voltage.

Another point to think about is that from what i can tell from transformers i have examined myself in the past that came from microwave ovens, the transformers are underdesigned because they are used in intermittent operation. That means the primary wire gauge may be under rated, meaning we might loose more voltage that way, and the core might be butt stacked which increases leakage inductance so we might loose more voltage (and therefore current) that way too. If you show a clear picture of the transformer from the side we can see if it is butt stacked or not.

Also, a test is worth a thousand words sometimes (also see my sig line). If you have a variac and some meters we can test it a little and see what we actually have here.
 
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spec

Well-Known Member
Most Helpful Member
Some good explanations in this thread.

As I understand it, there are three main areas of loss in a low frequency (50Hz or 60Hz) transformer:

(1) Copper loss (I↑2R in both primary and secondary)
(2) Core loss
(2.1) Eddy current loss (I↑2R)
(2.2) Coercive force loss (force required to run the core around the BH curve with an AC input)
(3) Leakage inductance (less than perfect coupling between primary and secondary)

Is the coercive force loss significant for a microwave oven transformer (bearing in mind that microwave transformers are built down to a price)?

spec

(PS: 'B'= magnetism in core and 'H'= magnetizing force [Amp/turns])

https://www.quora.com/What-is-the-shape-of-the-B-H-curve-for-air-Why
 
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MrAl

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Some good explanations in this thread.

As I understand it, there are three main areas of loss in a low frequency (50Hz or 60Hz) transformer:

(1) Copper loss (I↑2R in both primary and secondary)
(2) Core loss
(2.1) Eddy current loss (I↑2R)
(2.2) Coercive force loss (force required to run the core around the BH curve with an AC input)
(2) Leakage inductance (less than perfect coupling between primary and secondary)

Is the coercive force loss significant for a microwave oven transformer?

spec


Hi there spec,

We could probably look up some typical magnetic materials and try to find out. Grain oriented si steel laminations in every MW transformer i have seen to date.

Pardon for the slight correction, but "leakage inductance" in itself is not considered a loss because it is a storage element and storage elements in theory do not dissipate energy.
Resistance, core losses that result in heat, do result in actual energy losses.
Leakage inductance does cause a loss in voltage, but like any other pure inductance it itself stores and releases energy. It's like connecting an extra inductor in series with the primary that has no losses.
 

spec

Well-Known Member
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Hi there spec,

We could probably look up some typical magnetic materials and try to find out. Grain oriented si steel laminations in every MW transformer i have seen to date.

Pardon for the slight correction, but "leakage inductance" in itself is not considered a loss because it is a storage element and storage elements in theory do not dissipate energy.
Resistance, core losses that result in heat, do result in actual energy losses.
Leakage inductance does cause a loss in voltage, but like any other pure inductance it itself stores and releases energy. It's like connecting an extra inductor in series with the primary that has no losses.
Hi MrAl,

Thanks for info.

Thanks also for the explanation about 'leakage inductance' too. Of course, you are quite right- I had not viewed it as a storage element.

I like the analogy: 'It's like connecting an extra inductor in series with the primary that has no losses'. Am I right in thinking, that because of this, leakage inductance reduces the effective input voltage and thus output voltage so there is a loss of voltage as opposed to power?

spec
 
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StudentSA

Member
Aha, now we are on the same page Mr Al. Very enlightening explanation given, Thank you.
Now if only I can get my Triac driver to work I'll be one step closer to spot welding.
 

MrAl

Well-Known Member
Most Helpful Member
Hi MrAl,

Thanks for info.

Thanks also for the explanation about 'leakage inductance' too. Of course, you are quite right- I had not viewed it as a storage element.

I like the analogy: 'It's like connecting an extra inductor in series with the primary that has no losses'. Am I right in thinking, that because of this, leakage inductance reduces the effective input voltage and thus output voltage so there is a loss of voltage as opposed to power?

spec
Hi there,

Yes that sounds right, and i think there are diagrams on the web showing this too but i havent looked in quite a while now.
One good point about it though is that in a DC power supply it acts like any other choke would act so we get a little filtering action there too even though it is on the AC side of the bridge.
Another side benefit is some current limiting. Some wall warts (some unregulated types if not all) depend on this leakage inductance to limit short circuit current and thus power dissipation. Well, that in addition to the winding resistance.
 
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