# calculating number of turns on a transformer

Discussion in 'Electronic Design' started by Jamie Morken, Oct 24, 2007.

1. ### Jamie MorkenGuest

Hi,

I have a flyback transformer design in ltspice with a 1mH primary coil,
I would like to see how many turns this would require on this transformer:

core size: E375

falco part#: 1831-331-002

cross section area Ac(cm^2) =0.840

magnetic path length (cm) = 6.94

WaAc(cm^4) = 0.856

core volume(cm^3) = 5.830

Is there a formula to calculate the required number of primary
turns to get about 1mH inductance from this data? Thanks for
any help.

cheers,
Jamie

2. ### Jamie MorkenGuest

I know the flyback configuration will decrease the inductance
too, I guess this makes it harder..

3. ### Tam/WB2TTGuest

They can boil all this information down into one constant, sometimes called
Al, given in uH/100 turns. You might want to see if they publesh that for

Tam

4. ### EeyoreGuest

I suggest you get a copy of Epcos's Ferrite Magnetic Designer program.

Graham

5. ### John LarkinGuest

Call them and ask what's the Al value. It's amazing that many
magnetics suppliers don't furnish this. They could also furnish a
number of other handy electrical and thermal values that are a
nuisance to calculate, like gauss per ampere-turn and temperature rise
per watt.

John

cheers,
Jamie

7. ### Jamie MorkenGuest

I guess also they might have Al values for different core gaps, or can
this be guesstimated too? I don't know much about using core gaps
for flybacks, is it even necessary to do if you have a core big enough
that it won't saturate without a gap?

cheers,
Jamie

8. ### leggGuest

You would not normally need AL values for a ferrite core used in a
flyback, as this is determined by the gap required to store the energy
needed.

L = uo x N^2 x Ae / Lg

L = inductance in henries

uo = permeability of free space = 4 x pi x 10E-7

N = turns count

Ae = cross-sectional area at the gap in meters^2

Lg - length of the gap im meters.

.....................................

You have to determine N based on peak core flux change, limited by
core loss or core saturation at the frequency and duty cycle you are
preparing or able to use. To do this you need more core material
information concerning it's power loss characteristics, whether the
material is ferrite or powdered material.

As powdered material has a distributed gap, it will not normally be
merchandised without reference to it's permeability or the part's AL
value

For ferrite parts, if the frequency is lowish or the part is very
small, the flux density will likely be determined by the saturation
limit.

In this case,

Nmin > V x t / ( Bsat x Ae )

Nmin = minimum turns

V = applied voltage in volts

t = period of applied voltage in seconds

Bsat = saturation flux density in Teslas

Ae = minimum cross-sectional area of the ferrite material in meters^2

( Bsat of ferrite ~ 0.33T @ room temperature )
.........................................................

You seem already to have determined that 1mH of primary inductance is
desirable, by some method or other. Note that depending on the
operating frequency, a certain peak current will be expected in this
primary inductance in order to deliver the required output power.

This is determined by the rough formula

P = L x Ip^2 x f / 2

P = delivered power and all losses in Watts

L = primary inductance in Henries

Ip = peak primary current in Amps

f = pulse repetition rate in Hertz

cover flyback converters and flyback transformers pretty clearly.

RL

9. ### GenomeGuest

Blork.

So... In the limit as Lg tends to zero inductorance tends to infinity?

What is life like when you get dialed in as low quality?

DNA

10. ### leggGuest

When the ratio between the lengths of the gap path and the ferrite
path lenths approaches 10E-7, then the permeability of the ferrite
comes into effect.

A gapless ferrite core isn't much use in storing energy, which is what
is intended in a flyback circuit. Whether a 20mW circuit can be said
to have a topology is another thing. erp.

I thought I might get a respnse from Master Jamie, but I guess he's
just fooling around again.

Why didn't you offer a link to your own blither. I recall it being
modestly illuminatin'.

RL

11. ### Jamie MorkenGuest

I did some testing with an ETD29 ferrite core:

Ferroxcube ETD29 core and bobbins bought from Farnell

3C90 ferrite material

AL = 2350 +-25% (with no airgap)
Ue = 1850 (with no airgap)

(wound with 22 gauge wire, 25 turns from one end of bobbin to the other end)

25 turns no airgap gives 1.8mH

25 turns with 1 layer kapton tape airgap gives 0.446mH (0.002" airgap)

25 turns with 2 layer kapton tape airgap gives 0.260mH (0.004" airgap)

25 turns with 3 layer kapton tape airgap gives 0.192mH (0.006" airgap)

now how many turns do we need with these kapton tape airgaps to get 1mH
inductance?

fullcoil_turns =
sqrt((fullcoil_inductance*(coil_turns(turns)^2))/coil_turns(inductance))

1layer of kapton tape:
------------------------
fullcoil_turns = sqrt((1mH*(25^2))/0.446mH)
= 37 turns for 1mH with 1 layer kapton tape airgap

2layers of kapton tape:
------------------------
fullcoil_turns = sqrt((1mH*(25^2))/0.257mH)
= 49 turns for 1mH with 2 layers kapton tape airgap

3layers of kapton tape:
------------------------
fullcoil_turns = sqrt((1mH*(25^2))/0.192mH)
= 57 turns for 1mH with 3 layers kapton tape airgap

So how big should I make the airgap?

cheers,
Jamie

12. ### leggGuest

What is your

a - input voltage (at drop-out)
b - operating frequency
c - maximum duty cycle

Core saturation limiting is best handled by a current limiting control
topology.

For core-loss limited applications, determine the permissible core
loss; assuming 1degC surface rise in every square centimeter for every
milliwatt dissipated. (+/-20% )
.....

All of the milliwatts are generated by the total core volume, giving a
core loss density. (mW/cm^3 = Kw/m^3)
.....

This loss density will correspond to a peak flux density at a specific
operating frequency in the core material's published characteristics.
.....

Is 1mH capable of storing your power requirement ?

The peak primary current must be achievable for the same operating
conditions as the peak flux calculation.

V = L x di / dt

V = minimum input voltage

di = minimum required current peak

dt = maximum conduction period (at frequency and duty cycle limit)

L = maximum primary inductance.

RL