I realize iron powder cores aren't or shouldn't be used for
transformers at LF. However they are popular for inductors in
filters, Class E amplifiers etc.

I was recently working with a Class E design using iron powder
cores. They worked fine in a ~175W version of the amplifier but I
don't know where I was with regard to saturation. What I did find is
that I had to go to a relatively large core to get enough turns (in
a single layer) with large enough wire to handle the current without
excessive heating of the wire.

Paul



On 05/04/2018 12:43 PM, Andy Talbot wrote:
> If you stack cores, you just add the core areas together.   There
> may be a bit of leakage inductance- it may be noticeable, it may not.
> Tape, gapes between them all add to leakage inductance, but doubt
> it'll be all that much.
> 
> Iron powder is an interesting situation as it has gaps inherent and
> built in.  The same equation applies but our inductance and ur are
> so much less that probably you'll need far more turns to get a
> workable inductance than you'd ever need to get saturation down.   
> Bsat for iron is higher, but different rules apply and to prevent
> losses you probably have to keep it way down.
> Shouldn't be using iron dust for transformers simply because you
> need so many turns to get L high enough to be insignificant, but
> some people do
> 
> Andy
> www.g4jnt.com <http://www.g4jnt.com>
> 
> 
> On 4 May 2018 at 12:29, N1BUG <[email protected]
> <mailto:[email protected]>> wrote:
> 
>     Thanks Andy.
> 
>     I've added a page on this in my handwritten notebook.
> 
>     I do have some related questions which go beyond what Chris
>     asked about.
> 
>     What about stacked cores? If we stack three cores, each having a
>     cross sectional area of 1.58 cm^2, does it become 4.74 cm^2? Do the
>     cores have to be in physical contact with each other? What if each
>     is first wrapped with insulating tape, then they are stacked?
> 
>     Finally, this dealt with ferrite cores. What about iron powder core
>     saturation?
> 
>     Paul
> 
> 
> 
>     On 05/03/2018 12:05 PM, Andy Talbot wrote:
>     > The peak of the fundamental sine component in a square wave is
>     4/pi
>     > times the square wave amplitude (yes, it is bigger).   The RMS of
>     > the sine is Peak / SQRT(2) .  Take the two together and the RMS of
>     > your sine is therefore 0.9 * Peak of the square wave
>     >
>     > So if you had a Vdd of 50V you would have 45V rms across each half
>     > of the winding, or 90V rms across the two halves.  Which answers
>     > your other question
>     >
>     > Actually, thinking about what I've just said, you are putting the
>     > square wave though the transformer, not a  sine.  So in that case
>     > you have a bit more leeway in that it can go to the same peak
>     value
>     > as the sine would have given you.
>     >
>     > Unless I'm building SMPSUs, I don't put square waves though
>     > transformers in transmitters - teh RF is always filtered
>     > beforehand.   There is an equivalent equation for square waves and
>     > used in SMPSU design,  it is   V.t = N.A.B    Now V is input
>     > voltage,  t is the on-time , and N.A.B as before
>     >
>     > And yes, your area is correct.  1.58cm^2 = 0.000158m^2
>     >
>     > Andy
>     > www.g4jnt.com <http://www.g4jnt.com> <http://www.g4jnt.com>
>     >
>     >
>     > On 3 May 2018 at 16:36, N1BUG <[email protected]
>     <mailto:[email protected]>
>     > <mailto:[email protected] <mailto:[email protected]>>> wrote:
>     >
>     >     I am still not sure whether I am using that equation
>     correctly.
>     >
>     >     Chris, if you will forgive me for tagging along on your
>     post, I'm
>     >     trying to get this equation sorted in my head. Put no faith in
>     >     anything I say with regard to this! ;-)
>     >
>     >     Andy, could you please check me out on this?
>     >
>     >     Let's use the case Chris asks about as an example.
>     >
>     >     Suppose I want to calculate the maximum RMS voltage to
>     stay out of
>     >     saturation.
>     >
>     >     The manufacturer of the core states its Ae (equivalent cross
>     >     sectional area) is 1.58 cm^2. I believe that works out to
>     >     0.000158 m^2.
>     >
>     >     F = 137000
>     >
>     >     I do not know if I should be using 4 turns or 8 turns for the
>     >     primary. It's 4+4. I will assume 8.
>     >
>     >     Vrms = 4.44 * 137000 * 8 * 0.000158 * 0.1
>     >
>     >     Vrms = ~77V
>     >
>     >     Does that look right?
>     >
>     >     As for practical application I have no idea what this
>     means. The
>     >     equation uses Vrms but we're talking about something
>     closer to a
>     >     square wave than sine wave. But then it's not really a
>     square wave
>     >     either...
>     >
>     >     Paul