# transformer theory questions

Discussion in 'Electronic Basics' started by [email protected], Dec 29, 2003.

1. ### Guest

I already understand how the turns ratio between primary and secondary
of a transformer translates into the voltage ratio. What I would like
to understand is why this works this way.

Consider a simple 1:1 transformer. I'll get the same voltage out on
the secondary as I input on the primary. Now if I add some turns to
the secondary, I get more voltage. But if I add some turns to the
primary, I get less voltage out on the secondary. This can obviously
be seen by turning the transformer around and applying power to what
was the secondary, and getting more voltage out on the primary when
the primary has more turns.

What I want to know is how it accomplishes this. I would think that
if I add more turns on the primary, I'm going to get more magnetic flux
in the core. Or is this simply not the case?

When answering this, please try to avoid falling back to turns ratio
itself, since I am accepting that already, though on faith (because it
can be observed to work). I'm trying to understand _why_ it works.

Along these lines I have other questions, and the answer to these may
even help understanding the above.

Turns ratio is one thing (establishes the voltage ratio). But how is
the actual number of turns to be determined (e.g. what factors will it
depend on)? I'm sure that if I build a transformer with a nice solid
iron core and wrap 2 turns on the primary with some nice thick 0 gauge
wire, and wrap 4 turns on the secondary, and plug it in, I'm going to
get a blown fuse or tripped breaker, instead of double the voltage out
on the secondary (well, maybe it will be for an instant).

Suppose I have a given transformer with room to add more windings to
both primary and secondary, and I add an equal proportion to both so
the ratio stays the same. What does that do to the capacity of the
transformer?

I've seen the internal windings on a 1000 KVA transformer, and it just
didn't look like much. They were thick, but not that many. It just
seemed like it would "nearly short out" whatever power was applied to
either primary or secondary. Or is it the case that this sizing is
also going to translate into the waste factor of a transformer (e.g.
the waste in a 1000 KVA transformer with a 120/240 volt primary would
be more than a normal home 15/20 amp circuit would take)?

2. ### Robert MonsenGuest

Something like faraday's law of induced voltage is probably the easiest way
to describe it. The induced voltage is equal to the number of turns times
the rate of change of the magnetic flux through the loops. Since the loops
of the transformer are around the same core, the rate of change of flux is
equal for both sides (ignoring various issues). Thus, the induced voltage
divided by the the number of turns is equal. This means that

Vin/Npri = Vout/Nsec

or

Vin/Vout = Nsec/Npri

So, it really is the ratio...

Adding more turns on the primary does increase the flux through the core.
Faraday's law, however, causes the change in flux term to fall out of the
equation. Within certain limits, the number of turns isn't important, its
the ratio of primary to secondary that is important.

Note that you can't increase the flux indefinitely by increasing the turns
of the primary, since the permeability of the core decreases with current
through the turns. Other such issues (like resistance in the turns) conspire
to cause the secondary voltage to diverge from the ideal.

I'll let somebody who has actually built transformers answer the rest of the
questions.

Regards
Bob Monsen
---

3. ### John LarkinGuest

Nope: decreases it. That's why the secondary voltage will go down when

John

4. ### Guest

| Something like faraday's law of induced voltage is probably the easiest way
| to describe it. The induced voltage is equal to the number of turns times
| the rate of change of the magnetic flux through the loops. Since the loops
| of the transformer are around the same core, the rate of change of flux is
| equal for both sides (ignoring various issues). Thus, the induced voltage
| divided by the the number of turns is equal. This means that
|
| Vin/Npri = Vout/Nsec
|
| or
|
| Vin/Vout = Nsec/Npri
|
| So, it really is the ratio...

Well, I know it is. I'm looking for an understanding of how things work
to make it so, not just reducing some equation to see what falls out.

| Adding more turns on the primary does increase the flux through the core.
| Faraday's law, however, causes the change in flux term to fall out of the
| equation. Within certain limits, the number of turns isn't important, its
| the ratio of primary to secondary that is important.

Right. But what changes when you add more primary turns? How does the
secondary end up getting a lower induced voltage as a result?

| Note that you can't increase the flux indefinitely by increasing the turns
| of the primary, since the permeability of the core decreases with current
| through the turns. Other such issues (like resistance in the turns) conspire
| to cause the secondary voltage to diverge from the ideal.

And flux saturation. But we could ignore those issues and look at just the
number of turns. It's the ratio. But what I want to know is what fundamental
physics are involved to get that.

5. ### Guest

| On Mon, 29 Dec 2003 21:51:44 GMT, "Robert Monsen"
|
|
|>Adding more turns on the primary does increase the flux through the core.
|
| Nope: decreases it. That's why the secondary voltage will go down when

How/why does the flux go down with more primary turns? Is the coil reactance
a culprit in that?

6. ### John PopelishGuest

The flux in the core is proportional to sum of the amp turns of all
two windings (total amps going around the core). If we deal with an
open circuit secondary, so we don't have to deal with any amp turns in
the secondary (and just look at the voltage) then all the amps come
from the primary. The primary RMS current is the applied voltage
divided by the impedance of the primary, which is proportional to the
primary inductance, which varies with the square of the primary
turns. So doubling the primary turns, multiplies the primary
inductance by 4, lowering the primary current to 1/4, lowering the
peak flux in each direction to (1/4)*2=1/2 (amp turns, remember?).
The volts per turn is proportional to the rate of change of the flux.
Have you seen the V=L*(di/dt) differential definition of inductance?
Keeping in mind that flux is proportional to amp turns, voltage must
also be proportional to df/dt (f being flux). So doubling the primary
turns also lowers the df/dt to (1/4)*2=1/2 also. So the volts per
turn goes to 1/2, and we are back to another way of seeing that
doubling the number of primary turns halves the volts per turn for all
windings without assuming it.
It is limited by the peak flux in each direction during a cycle. The
lower the number of turns, the faster the flux has to change to
produce the higher volts per turn, so the farther it ends up swinging
by the end of each half cycle. If we could come up with a low loss
transformer core that could handle twice the peak flux without
saturation, we could halve the number of turns in all windings for the
same integral volt seconds per half cycle. Integral volt seconds is
proportional to peak flux.
Because the low inductance of the primary will allow large current to
pass, saturating the core very quickly. If the frequency were high
enough that the high volts per turn did not have to be supported for
so long, you might get away with it. High frequency transformers tend
to have low numbers of turns compared to low frequency transformers.
It shifts the losses from core (because of lower hysterisis (which are
driven by peak flux swing) and eddy current (which are driven by volts
per turn) losses, while increasing the copper I*R losses (adding turns
A big transformer has a lot of core cross sectional area, so it can
carry a lot of total flux through the windings. Big rate of change of
flux means big volts per turn, means low number of turns. So there is
room for fat wire to get low loss at high current.

In general, the bigger the transformer, the smaller the waste compared
to the power carried through the transformer. This is not something
you get for free, but is a design necessity. You have to under
utilize the flux capability of a very big transformer core, compared
to a very small one, because the big one has a lower surface to core
volume ratio. The heat is produced in all the volume, but has only
the surface area through which to escape. Same for the copper
volume. So you have to design a big transformer to carry slightly
less power per pound of iron and copper than a small one. Stressing
that pound less produces less waste heat per watt carried in the big
one because that is the only way to keep the temperature the same.

So a wall wart transformer may be only need to be 80% efficient at
rated power to keep from self destructing, while a substation
transformer may have to be designed to be 95% efficient at full load
to have long term reliability (and this while being cooled with
circulating oil). It is all about temperature rise.

7. ### John LarkinGuest

Yes; John P explains it pretty well below.

Another way to look at it: assuming the wire resistance is pretty low,
disappear the core and apply 120 VAC. The current will be huge and
fuses will blow.

Now reinstall the core. The current goes down because there is a 'back
emf' effectively in series with the winding that cancels out most of
the applied 120 AC. This voltage is generated by the AC magnetic flux
within the core passing through the turns. Where does this flux come
from? From the applied AC! Current in the primary magnetizes the core,
the AC magnetic field induces voltage into the primary turns, and that
voltage semi-magically cancels out most of the applied 120V input,
leaving only enough residual current to magnetize the core!

More primary turns need less AC magnetic flux to induce the almost-120
volts back EMF.

Turns out that - first order approx - flux density is proportional to
volts-per-turn.

John

8. ### Rich GriseGuest

....
Strictly speaking, that's not faith, because it's observed. I think
the fancy-schmancy term is "empirical." I had a sort of "flash" of inspiration here - when you stack up 2
turns in the primary, each only gets half of the applied voltage,
which is then transformed over to the other wire. But the current
is applied in parallel, as it were, since it goes by twice.

At least it sounds pretty cool to me, and is fun to visualize
with some good bud. %-}

Cheers!
Rich

9. ### Rich GriseGuest

Actually, I rather like Robert Monsen's answer. At least it makes
sense to me, if I try to see your POV.

Cheers!
Rich

10. ### Guest

| Another way to look at it: assuming the wire resistance is pretty low,
| disappear the core and apply 120 VAC. The current will be huge and
| fuses will blow.

But there will be some inductance, just not very much? Could you wind
enough turns over air to draw no more than 10 amps at 120v 60Hz? Surely
I would think so (but it would be a lot of turns).

| Now reinstall the core. The current goes down because there is a 'back
| emf' effectively in series with the winding that cancels out most of
| the applied 120 AC. This voltage is generated by the AC magnetic flux
| within the core passing through the turns. Where does this flux come
| from? From the applied AC! Current in the primary magnetizes the core,
| the AC magnetic field induces voltage into the primary turns, and that
| voltage semi-magically cancels out most of the applied 120V input,
| leaving only enough residual current to magnetize the core!

Somewhat like a reflection of power that would not only be induced into
the primary, but also into any winding. So double the windings and you
get double the induced voltage back, given a particular flux?

| More primary turns need less AC magnetic flux to induce the almost-120
| volts back EMF.
|
| Turns out that - first order approx - flux density is proportional to
| volts-per-turn.
|

Actually, that would make sense to me. The secondary would be taking
power out of the system.

11. ### Guest

| wrote:
| (snip)
|> This can obviously
|> be seen by turning the transformer around and applying power to what
|> was the secondary, and getting more voltage out on the primary when
|> the primary has more turns.
|>
|> What I want to know is how it accomplishes this. I would think that
|> if I add more turns on the primary, I'm going to get more magnetic flux
|> in the core. Or is this simply not the case?
|
| The flux in the core is proportional to sum of the amp turns of all
| two windings (total amps going around the core). If we deal with an
| open circuit secondary, so we don't have to deal with any amp turns in
| the secondary (and just look at the voltage) then all the amps come
| from the primary. The primary RMS current is the applied voltage
| divided by the impedance of the primary, which is proportional to the
| primary inductance, which varies with the square of the primary
| turns. So doubling the primary turns, multiplies the primary
| inductance by 4, lowering the primary current to 1/4, lowering the
| peak flux in each direction to (1/4)*2=1/2 (amp turns, remember?).
| The volts per turn is proportional to the rate of change of the flux.
| Have you seen the V=L*(di/dt) differential definition of inductance?
| Keeping in mind that flux is proportional to amp turns, voltage must
| also be proportional to df/dt (f being flux). So doubling the primary
| turns also lowers the df/dt to (1/4)*2=1/2 also. So the volts per
| turn goes to 1/2, and we are back to another way of seeing that
| doubling the number of primary turns halves the volts per turn for all
| windings without assuming it.

So to build a higher KVA transformer at a given voltage means fewer
turns because you have to have more current ... and a correspondingly
larger core? That would explain why these big KVA transformers have
so few turns.

|> I'm sure that if I build a transformer with a nice solid
|> iron core and wrap 2 turns on the primary with some nice thick 0 gauge
|> wire, and wrap 4 turns on the secondary, and plug it in, I'm going to
|> get a blown fuse or tripped breaker, instead of double the voltage out
|> on the secondary (well, maybe it will be for an instant).
|
| Because the low inductance of the primary will allow large current to
| pass, saturating the core very quickly. If the frequency were high
| enough that the high volts per turn did not have to be supported for
| so long, you might get away with it. High frequency transformers tend
| to have low numbers of turns compared to low frequency transformers.
|
|> Suppose I have a given transformer with room to add more windings to
|> both primary and secondary, and I add an equal proportion to both so
|> the ratio stays the same. What does that do to the capacity of the
|> transformer?
|
| It shifts the losses from core (because of lower hysterisis (which are
| driven by peak flux swing) and eddy current (which are driven by volts
| per turn) losses, while increasing the copper I*R losses (adding turns

My original presumption of more turns = more combined flux was just all
wrong. The flux is proportional to the current and more inductance means
less current (as well as more copper resistance as a small side effect).

|> I've seen the internal windings on a 1000 KVA transformer, and it just
|> didn't look like much. They were thick, but not that many. It just
|> seemed like it would "nearly short out" whatever power was applied to
|> either primary or secondary. Or is it the case that this sizing is
|> also going to translate into the waste factor of a transformer (e.g.
|> the waste in a 1000 KVA transformer with a 120/240 volt primary would
|> be more than a normal home 15/20 amp circuit would take)?
|
| A big transformer has a lot of core cross sectional area, so it can
| carry a lot of total flux through the windings. Big rate of change of
| flux means big volts per turn, means low number of turns. So there is
| room for fat wire to get low loss at high current.
|
| In general, the bigger the transformer, the smaller the waste compared
| to the power carried through the transformer. This is not something
| you get for free, but is a design necessity. You have to under
| utilize the flux capability of a very big transformer core, compared
| to a very small one, because the big one has a lower surface to core
| volume ratio. The heat is produced in all the volume, but has only
| the surface area through which to escape. Same for the copper
| volume. So you have to design a big transformer to carry slightly
| less power per pound of iron and copper than a small one. Stressing
| that pound less produces less waste heat per watt carried in the big
| one because that is the only way to keep the temperature the same.

That volume vs. surface issue sure shows up in a lot of places | So a wall wart transformer may be only need to be 80% efficient at
| rated power to keep from self destructing, while a substation
| transformer may have to be designed to be 95% efficient at full load
| to have long term reliability (and this while being cooled with
| circulating oil). It is all about temperature rise.

Which is a critical rating on both oil and dry distribution transformers.

12. ### Guest

|
| | ...
|> When answering this, please try to avoid falling back to turns ratio
|> itself, since I am accepting that already, though on faith (because it
|> can be observed to work).
|
| Strictly speaking, that's not faith, because it's observed. I think
| the fancy-schmancy term is "empirical." OK, sounds good to me.

|> I'm trying to understand _why_ it works.
|
| I had a sort of "flash" of inspiration here - when you stack up 2
| turns in the primary, each only gets half of the applied voltage,
| which is then transformed over to the other wire. But the current
| is applied in parallel, as it were, since it goes by twice.

And there being 2 sets of turns in the primary, twice the inductance
and half the current?

| At least it sounds pretty cool to me, and is fun to visualize
| with some good bud. %-}

It's starting to resemble radio transmission antennas. If I can connect it
all back to that, I think I'll have it (I do know more about that).

13. ### John LarkinGuest

Yup, a lot of turns. The permeability of air is something on the order
of 1/10,000 or so that of a transformer core, so the 'magnetizing
current' (just the primary current into the primary inductance) is
correspondingly higher if you remove the core. In most practical
situations, an air-core coil will have more copper-wire resistance
than inductive reactance at 50/60 Hz.

A single turn of #0 wire, about 2 miles in diameter, should be close.

John

14. ### Robert MonsenGuest

Right, with fixed V, N and dPhi/dt are inversely proportional.

V = N.dPhi/dt

Thanks for pointing that out.

Regards
Bob Monsen

15. ### PeterGuest

Great thread. I learned a lot.

---

16. ### Don KellyGuest

------------
John, I am late jumping into this but one point that is generally forgotten
or ignored by many is that, on the basis of e=N(d(flux)/dt) , for AC the
relation Eave =4(frequency)*(Maximum flux) * Number of turns (For a
sinusoid change the Bougerre factor to 4.44 ). Trade offs between flux,
frequency and turns, as you indicate, then occur. The core material
doesn't enter this. The core affects only the excitation current needed to
get the required flux.

By the way, I would not be at all interested in a substation transformer
with only a 95% efficiency at full load. I would want what is readily
obtainable, an efficiency in the order of 99.5% at full load. I have tested
3KVA air cooled 120/240V transformers which had an efficiency over 95% from
1/4 load to 1-1/4 times full load rating, peaking about 98%.   