calculating transformer saturation

"Jamie Morken"

You seem to be designing an inductor as you state the max current as 10 amps.

Know the difference?

....... Phil

Cheers

Nope, thats right. You have implicitly taken Ae into account by using Bsat = Phi_sat/Ae.

What are you trying to make - a choke, or a transformer? the calculation method is slightly different (although all the maths is the same).

BTW, 370mT might work at 20C, but at 100C the choke will well and truly saturate - Bsat DECREASES with increasing temperature. I inherited a UPS design once that suffered from this particular problem.....

It sounds like you want a choke, and you want to bang 10A thru it. Given that its a planar core, its probably an output choke for a forward converter, or maybe a buck choke.

what you have discovered is that ferrites dont make very good high-current chokes, unless you stick in an air gap.

(unless, of course, 1T and about 10uH is OK, along with saturating)

Initially, ignore the permeability, and use the maximum MMF,

Hsat = NI/le

for 35G material (3F3), Hsat = 50 - 70 A/m

(its a gradual curve, which means you get to pick "saturated")

say Hsat = 70A/m = N*10A/0.1m so N = 0.7T

Wow, this is a lot lower than your 1.3T. BECAUSE mu_i falls off with increasing H......

ignoring the practical difficulties associated with winding 0.7T (and the reduced permeability), we end up with L = 4.9uH and E = 245uJ

OK, so now what?

Well, when you add an air gap, the BH curve shears over - more H before it saturates.

originally, you had Ho = (NoIo)/le

when you add an air gap, this becomes:

H = Ho*[1+(mu_i*Ae*lgap)/(Agap*le)]

and if Agap = Ae (it does if we ignore fringing flux) this gives:

H = Ho*[1+(mu_i*lgap)/le]

So, if Io stays the same, we can write:

H = N*Io/le = (No*Io/le)*[1+(mu_i*lgap)/le]

i.e.

N = No*[1+(mu_i*lgap)/le]

so now you can pick some arbitrary number of turns N, and work out how big a gap you *NEED* to prevent the core saturating

(this is a terrible way to do it, but its a tutorial....)

say you want 4T and mu_i = 2200. We already calculated No = 0.7T so:

lg = (4/0.7 - 1)*0.099/2200 = 0.2mm

Alas by bunging in an air gap, the effective permebaility is reduced. For an air gap bigger than about 0.1mm, its a reasonable approximation to say ALL of the energy is stored in the air gap itself. So we can calculate the inductance as:

Lgapped = mu_0*N^2*Ae/lg

Ae = 10.2mm x 50.3mm = 500mm^2

Lgapped = 1.257e-6*16*500e-6/0.2e-3 = 50uH

and E = 2.5mJ

So although the effective permeability went down, the total inductance went up, and there is now 10x more energy stored in the choke c.f. the

There is enough information here for you to derive a closed-form solution for lg & N, given I and the desired L. I'll leave that as an exercise for you ;)

alternatively, whack up a quick spreadsheet, MathCad etc, and bang some numbers in until you are happy with the result. Dont make your gap too small, there are finite manufacturing tolerances - the guy I use gives

When designing a transformer, the procedure is different. We can just use the transformer equation, V = N*dPhi/dt

Phi = B*Ae

V = N*dB*Ae/dt

for square wave, dB = Bsat, dt = Ton so:

Vin*Ton/(N*Ae) = Bsat

which can be solved for N (use min Vin, Max Ton)

Then calculate L = N^2*Al

If you need a precise value of L (eg flyback), this result is probably too high, so introduce a gap to reduce L to the required amount. If its too low, the core is too small.

Note that the transformer equation is often written V = 4.44*N*B*Ae*f

this is confusing, because it assumes sinusoidal operation:

B = Bpk*sin(wt)

and dB/dt = w*Bpk*cos(wt) which has the maximum value w*Bpk

and w = 2*pi*f

giving Vpk = 2*pi*f*N*Bpk

then just to confuse you, they use V = Vrms = Vpk/sqrt(2)

Vrms = sqrt(2)*pi*N*Bpk*f = 4.44*N*Bpk*f

Voila!

(personally I hate formulae with magic numbers)

HTH

Cheers Terry

Ok thanks, I thought Bsat was a property of the ferrite material, that didn't depend on size, but I see now that in the 35g.pdf they specify:

"All measurements are made on ring core T 22 14 07."

I still think that Bsat is a property of the ferrite material, or else a very large ferrite core would have to have a very large Bsat, I guess this is the case, like a large core could have a Bsat of

It is a 3.6kW planar transformer used in push pull fullbridge input with max +-500VDC squarewave input, designed to output +-200VDC

Ya I noticed this in the datasheet, I will have to find the Bsat for the core I am using, if its a bigger core the Bsat may be

Thank you for all of the information! :)

I remember hearing it is not common for a transformer designed for fullbridge operation to be gapped, but I am starting to think that my transformer was airgapped, as the inductances are a lot lower than would be expected for the cores Al of 10000n. I will have to ask for more data next time, instead of doing detective work.. :)

cheers, Jamie

Bsat is indeed a property of the material, and is related to Hsat by B=uH [u being u_0*u_i(H,T)]

my previous statement is a bit confusing. Bsat and Hsat are properties of the core material alone. Core dimensions come into it when you start pushing amps thru a choke (suddenly you need le to work out N given Hsat & I) or volt-seconds for a transformer (you need Ae to work out N given Bsat & V & Ton)

in that case, ignore the choke design stuff.

nope, material property alone. 300mT is a good choice.

no worries. Use it when you design the output choke.

Nope, I just didnt talk about forward converter transformers. I will now.

There are basically two different types of converter topologies using transformers - flyback & forward.

A flyback converter transformer is really a coupled inductor. During Ton, energy is stored in the primary winding of the coupled inductor; during Toff it is delivered to the secondary winding(s). You end up working out Ecore = Pout/Fsmps = 0.5*Lp*Ip^2, and using Vin = Lp*Ip/Ton, choose Fsmps, Ton and hence Lp & Ip. The core then gets gapped to:

a) not saturate when it holds Ip

b) have the correct amount of inductance, so the load power can be delivered.

A forward converter transformer is just that. The magnetising inductance is NOT used to store energy that gets delivered to the load - Lp is in fact a parasitic component, and should be made as large as possible.

When you say 10A, you are referring to the reflected load current. Due to the magic of transformer action, as long as you suck it out of the secondary AT THE SAME TIME AS YOU PUMP IT INTO THE PRIMARY, you can stick as many amps as you like thru a forward converter transformer and it wont saturate. Of course the winding may melt, which is bad...

so it turns out higher power converters use larger cores SO THEY HAVE ENOUGH WINDING AREA TO GET THE REQUISITE AMOUNT OF COPPER IN. Sorry to shout, but thats whats really going on.

This is how you design your forward converter transformer:

Vin = Vin_min

Ton = Ton_max

(assuming some form of closed-loop control so duty cycle = K/Vin)

B = Bsat = 300mT

Np_min = Vin*Ton/(Bsat*Ae)

Then, calculate Ns:

Vout = Vin*(Ns/Np)*D

now be careful which D you use here. For a push-pull (or full-bridge) use D = Ton/(Tsmps/2). This is becuase a full-bridge/push-pull converter does this:

Ton1---Toff1--Ton2--Toff2--Ton1..... ^^^^^^^^^^ ^^^^^^^^ one side the other side

Typically Ton1 = Ton2 and Toff1 = Toff2

D = Ton1/(Ton1+Toff1)

and D is typically pretty close to 1.

Now Ns has to be integer, and Np cant go down (picking 300mT for Bsat gives you a bit of wiggle room), so fix Ns then re-calculate Np, thence Bsat and D.

Now calculate Lmag = Np^2*Al, and Imag = Vin*Ton/Lmag

Thanks Terry! :)

cheers, Jamie

Terry Given wrote: