Maker Pro
Maker Pro

Basic transformer question - number of turns depends on frequency?(Not turns ration, number of turns

A

Al Borowski

Jan 1, 1970
0
Hi,

For a university assignment, I have to design and construct a switchmode
regulator (flyback type). It has to drop 10V to 5V, using a certian
chip. I can choose the frequency to be between 20-100kHz.

Before I dive into the equations, could someone please tell me if the
number of turns required on the transformer depends on the switching
frequency? For instance, if a higher frequency means less turns, then
I'll go for it. If the number of turns is independant of frequency then
I'll use 100kHz for minimum ripple.

thanks alot,

Al
 
I

Ian Stirling

Jan 1, 1970
0
Al Borowski said:
Hi,

For a university assignment, I have to design and construct a switchmode
regulator (flyback type). It has to drop 10V to 5V, using a certian
chip. I can choose the frequency to be between 20-100kHz.

Before I dive into the equations, could someone please tell me if the
number of turns required on the transformer depends on the switching
frequency? For instance, if a higher frequency means less turns, then
I'll go for it. If the number of turns is independant of frequency then
I'll use 100kHz for minimum ripple.

Oh dear.
You really need to read your books/notes a bit more.

Saturation in transformers is a bad thing.
If the magnetic flux in the transformer goes over a limit, it
stops behaving as a transformer.
Look up saturation, and "leakage inductance"

All else being equal, higher frequencies mean lower peak fluxes, as
the energy transfer happens more often, so there is lowere energy to
be stored in each transfer.

A lower inductance is usually used at higher frequencies, so there are
fewer turns.

Look up how inductance varies as number of turns.

If all else fails, read the datasheet.
 
A

Al Borowski

Jan 1, 1970
0
Thanks for the reply.

Ian said:
Oh dear.
You really need to read your books/notes a bit more.

As I said, I haven't done *any* reading on this yet :) I just wanted a
quick answer before I 'hit the books'. Luckily I have plenty of time
before this is due.
Saturation in transformers is a bad thing.
If the magnetic flux in the transformer goes over a limit, it
stops behaving as a transformer.
Look up saturation, and "leakage inductance"

I'm OK with these - I think I understand the theory. Starting tomorrow
I'll get stuck into the assignment, but I first thought I'd ask here -
because I knew someone could tell me the answer right away :)
All else being equal, higher frequencies mean lower peak fluxes, as
the energy transfer happens more often, so there is lowere energy to
be stored in each transfer.

I thought so, which explains why the core size can be smaller, right?
A lower inductance is usually used at higher frequencies, so there are
fewer turns.

Thanks! That makes sense.

Look up how inductance varies as number of turns.

If all else fails, read the datasheet.

No worries, I plan to do so.

cheers,

Al
 
I

Ian Stirling

Jan 1, 1970
0
Al Borowski said:
Thanks for the reply.
No worries, I plan to do so.

Post in case of any problems...

There seem to be a lot of people who post saying "help me, I've
got 12 hours to complete the assignment and haven't read any books,
or done any work, can you please do all of the work for me".

Good luck.
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Ian Stirling
..net>) about 'Basic transformer question - number of turns depends on
frequency? (Not turns ration, number of turns)', on Fri, 9 Apr 2004:
Post in case of any problems...

There seem to be a lot of people who post saying "help me, I've
got 12 hours to complete the assignment and haven't read any books,
or done any work, can you please do all of the work for me".

Good luck.

Indeed. Al is a student, and an Australian student to boot. Yet he gave
a polite reply to a rather stern post, and he's got plenty of time for
his project. This man will go far!
 
G

Genome

Jan 1, 1970
0
message | Hi,
|
| For a university assignment, I have to design and construct a
switchmode
| regulator (flyback type). It has to drop 10V to 5V, using a certian
| chip. I can choose the frequency to be between 20-100kHz.
|
| Before I dive into the equations, could someone please tell me if the
| number of turns required on the transformer depends on the switching
| frequency? For instance, if a higher frequency means less turns, then
| I'll go for it. If the number of turns is independant of frequency
then
| I'll use 100kHz for minimum ripple.
|
| thanks alot,
|
| Al
|

There is no 'right' answer to your question.

In an 'ideal' world higher frequency would equate to an overall smaller
inductor with, perhaps, fewer turns.

Unfortunately

At low frequencies.Core losses are less so you are saturation limited,
0.3T. Skin and proximity effects are less so you can use larger wire
sizes. At high frequencies. Core losses are greater so you are loss
limited, <0.2T Skin and proximity effects are greater so you have to use
wire ropes.

Discontinuous operation gives you high peak and RMS currents with higher
copper losses implying thicker wire and more winding area. Continuous
operation has lower RMS currents and losses but needs more inductance,
more turns, and suffers from the right half plane zero.

These are just a few of the design decisions you will have to make and
justify. There are probably more and at the end of the day you have to
get out your pen and paper and run through some iterations.

I hope you have some good sums.

DNA
 
J

Jan Panteltje

Jan 1, 1970
0
Hi,

For a university assignment, I have to design and construct a switchmode
regulator (flyback type). It has to drop 10V to 5V, using a certian
chip. I can choose the frequency to be between 20-100kHz.

Before I dive into the equations, could someone please tell me if the
number of turns required on the transformer depends on the switching
frequency? For instance, if a higher frequency means less turns, then
I'll go for it. If the number of turns is independant of frequency then
I'll use 100kHz for minimum ripple.
Yes, the higher the freq, the less turns you need.
This because there is less inductance needed for a low (zero load)
current in the transformer.
You can calculate that from the switch time and L with i = t / L
Beware of saturation.
 
R

Robert Baer

Jan 1, 1970
0
Al said:
Hi,

For a university assignment, I have to design and construct a switchmode
regulator (flyback type). It has to drop 10V to 5V, using a certian
chip. I can choose the frequency to be between 20-100kHz.

Before I dive into the equations, could someone please tell me if the
number of turns required on the transformer depends on the switching
frequency? For instance, if a higher frequency means less turns, then
I'll go for it. If the number of turns is independant of frequency then
I'll use 100kHz for minimum ripple.

thanks alot,

Al

Turns per volt depends on the frequency, the core material, and the
power level (ie: the size of the core).
 
A

Al Borowski

Jan 1, 1970
0
Thanks to everyone who replied... clearly I need to bone up on
electromag theory :)

cheers,

Al
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Al Borowski <aj.borowski@erasethis
..student.qut.edu.au> wrote (in <[email protected]
..net.au>) about 'Basic transformer question - number of turns depends on
frequency? (Not turns ration, number of turns)', on Sat, 10 Apr 2004:
Thanks to everyone who replied... clearly I need to bone up on
electromag theory :)
You may find this easily-memorable equation useful:

E = BAWN

E = voltage (volt)
B = induction ('flux density') (tesla)
A = area of cross-section of core (m^2)
W [omega] = angular frequency (rad) = 2 [pi]f
f = frequency (hertz)
N = number of turns.
 
R

R.Legg

Jan 1, 1970
0
You may find this easily-memorable equation useful:
E = BAWN

E = voltage (volt)
B = induction ('flux density') (tesla)
A = area of cross-section of core (m^2)
W [omega] = angular frequency (rad) = 2 [pi]f
f = frequency (hertz)
N = number of turns.

......if you can figure out whether it's peak to peak, peak, average,
or RMS for the flux quantity.

For voltage, rms or instantaneous values are assumed, unless stated
otherwise. This is not so for flux density.

RL
 
J

John Woodgate

Jan 1, 1970
0
<[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sat, 10 Apr 2004:
You may find this easily-memorable equation useful:

E = BAWN

E = voltage (volt)
B = induction ('flux density') (tesla)
A = area of cross-section of core (m^2)
W [omega] = angular frequency (rad) = 2 [pi]f
f = frequency (hertz)
N = number of turns.

.....if you can figure out whether it's peak to peak, peak, average,
or RMS for the flux quantity.

For voltage, rms or instantaneous values are assumed, unless stated
otherwise. This is not so for flux density.

It is so for ALL quantities (that can have p-p, peak, average and r.m.s
values - you don't normally get peak-to-peak d.c. resistance!). Why
should there be an exception for B?If you understand SI units, it is obvious that E in that equation is
r.m.s. if B is r.m.s.
 
R

R.Legg

Jan 1, 1970
0
John Woodgate said:
<[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sat, 10 Apr 2004:
You may find this easily-memorable equation useful:

E = BAWN

E = voltage (volt)
B = induction ('flux density') (tesla)
A = area of cross-section of core (m^2)
W [omega] = angular frequency (rad) = 2 [pi]f
f = frequency (hertz)
N = number of turns.

.....if you can figure out whether it's peak to peak, peak, average,
or RMS for the flux quantity.

For voltage, rms or instantaneous values are assumed, unless stated
otherwise. This is not so for flux density.

It is so for ALL quantities (that can have p-p, peak, average and r.m.s
values - you don't normally get peak-to-peak d.c. resistance!). Why
should there be an exception for B?If you understand SI units, it is obvious that E in that equation is
r.m.s. if B is r.m.s.

It is B that is not obvious.

Of actual interest is Bpk, as this will tell you if the circuit is
linear (nonsaturating), and will also allow determination of core
loss, per typical published mfr data that refers to the peak value.

Bpk is also the product of V.t - the result of an average voltage, not
an rms one. At least, that is my understanding.

RL
 
J

John Woodgate

Jan 1, 1970
0
<[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sun, 11 Apr 2004:
It is B that is not obvious.

Of actual interest is Bpk, as this will tell you if the circuit is
linear (nonsaturating), and will also allow determination of core
loss, per typical published mfr data that refers to the peak value.

But if you put Bpeak in the equation, you lose the easily remembered
form and you have to explain the 1/sqrt(2) factor. It's even worse of
you 'simplify' it by also including the numerical value of 2[pi]. You
get a factor of 4.44 to explain away.

Anyone who doesn't understand peak and r.m.s. values has no business to
be designing anything, let alone wound components.
 
G

Genome

Jan 1, 1970
0
| <[email protected]>) about 'Basic
| transformer question - number of turns depends on frequency? (Not
turns
| ration, number of turns)', on Sun, 11 Apr 2004:
| >It is B that is not obvious.
| >
| >Of actual interest is Bpk, as this will tell you if the circuit is
| >linear (nonsaturating), and will also allow determination of core
| >loss, per typical published mfr data that refers to the peak value.
|
| But if you put Bpeak in the equation, you lose the easily remembered
| form and you have to explain the 1/sqrt(2) factor. It's even worse of
| you 'simplify' it by also including the numerical value of 2[pi]. You
| get a factor of 4.44 to explain away.
|
| Anyone who doesn't understand peak and r.m.s. values has no business
to
| be designing anything, let alone wound components.
| --
| Regards, John Woodgate, OOO - Own Opinions Only.
| The good news is that nothing is compulsory.
| The bad news is that everything is prohibited.
| http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

The 'easily' remembered form for those designing switch mode power
supplies, or otherwise, is;

N = VinTon/B.Ae

It's just a statement of Faraday's (apostrophe included) law in other
terms that B is proportional to the integral of the applied volt
seconds.

Anything else is just pants

DNA
 
G

Genome

Jan 1, 1970
0
|
| (in
| | <[email protected]>) about 'Basic
| | transformer question - number of turns depends on frequency? (Not
| turns
| | ration, number of turns)', on Sun, 11 Apr 2004:
| | >It is B that is not obvious.
| | >
| | >Of actual interest is Bpk, as this will tell you if the circuit is
| | >linear (nonsaturating), and will also allow determination of core
| | >loss, per typical published mfr data that refers to the peak value.
| |
| | But if you put Bpeak in the equation, you lose the easily remembered
| | form and you have to explain the 1/sqrt(2) factor. It's even worse
of
| | you 'simplify' it by also including the numerical value of 2[pi].
You
| | get a factor of 4.44 to explain away.
| |
| | Anyone who doesn't understand peak and r.m.s. values has no business
| to
| | be designing anything, let alone wound components.
| | --
| | Regards, John Woodgate, OOO - Own Opinions Only.
| | The good news is that nothing is compulsory.
| | The bad news is that everything is prohibited.
| | http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
|
| The 'easily' remembered form for those designing switch mode power
| supplies, or otherwise, is;
|
| N = VinTon/B.Ae
|
| It's just a statement of Faraday's (apostrophe included) law in other
| terms that B is proportional to the integral of the applied volt
| seconds.
|
| Anything else is just pants
|
| DNA
|
|

And a better one for those designing flyback converters is.....

N = L.Ipk/Bpk.Ae

These sums are incestuous.

DNA
 
L

legg

Jan 1, 1970
0
<[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sun, 11 Apr 2004:
It is B that is not obvious.

Of actual interest is Bpk, as this will tell you if the circuit is
linear (nonsaturating), and will also allow determination of core
loss, per typical published mfr data that refers to the peak value.

But if you put Bpeak in the equation, you lose the easily remembered
form and you have to explain the 1/sqrt(2) factor. It's even worse of
you 'simplify' it by also including the numerical value of 2[pi]. You
get a factor of 4.44 to explain away.

Anyone who doesn't understand peak and r.m.s. values has no business to
be designing anything, let alone wound components.

Which isn't made any easier if the limitations to the formula aren't
pointed out. If you remove the equals sign and just stick an
approximation in the formula, it would serve just as effectively as an
educational tool, without promising exact solutions.

Why not just

Bpk = vt/NA ?

Teslas
volts
seconds
turns
meters^2

This allows the end user to work out his own waveshape effects, based
on the learned or worked out relationships between peak, rms and
average of the waveform in question.

At the college level, you can throw in the integral symbol in the
appropriate position, to allow for flashy numerical solutions that
include the phase angle that your initial reference to radian
measurement inanely suggests.

Just because 2xpixf (W in BAWN) is easy to remember, doesn't mean you
should chuck it in everywhere, as a sign that 'this is where there's a
relationship but I can't be bothered to go into it now'.

RL
 
J

John Woodgate

Jan 1, 1970
0
(in <[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sun, 11 Apr 2004:
The 'easily' remembered form for those designing switch mode power
supplies, or otherwise, is;

N = VinTon/B.Ae

It's just a statement of Faraday's (apostrophe included) law in other
terms that B is proportional to the integral of the applied volt
seconds.

Anything else is just pants

So is your equation unless you define the variables. I know that YOU
know what they are, but a lot of people don't.
 
J

John Woodgate

Jan 1, 1970
0
(in <[email protected]>) about 'Basic
transformer question - number of turns depends on frequency? (Not turns
ration, number of turns)', on Sun, 11 Apr 2004:
Why not just

Bpk = vt/NA ?

Because it isn't as easily remembered.

You are clearly determined to have the last word, so have it.
 
G

Genome

Jan 1, 1970
0
| I read in sci.electronics.design that Genome <[email protected]>
wrote
| (in <[email protected]>) about 'Basic
| transformer question - number of turns depends on frequency? (Not
turns
| ration, number of turns)', on Sun, 11 Apr 2004:
| >The 'easily' remembered form for those designing switch mode power
| >supplies, or otherwise, is;
| >
| >N = VinTon/B.Ae
| >
| >It's just a statement of Faraday's (apostrophe included) law in other
| >terms that B is proportional to the integral of the applied volt
| >seconds.
| >
| >Anything else is just pants
|
| So is your equation unless you define the variables. I know that YOU
| know what they are, but a lot of people don't.
| --
| Regards, John Woodgate, OOO - Own Opinions Only.
| The good news is that nothing is compulsory.
| The bad news is that everything is prohibited.
| http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

Ooooops, flak fall out alert.......

Give me the address of your local drinking hole and the name of the
landlord and I'll post a fiver for you and your good one to have a pint.

DNA
 
Top