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
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