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Re: XFRMR-T equivalent, was: Tube Data
Hi Simon,
I had a look to my old notebooks...
>
> Original Poster: "Simon Winder" <swinder-at-microsoft-dot-com>
>
> Isnt there a T - equivalent circuit for a loosely coupled transformer? Its a
> while since I studied this. What are the transformed values of the three
> inductors, in terms of L1, L2 and k? Anyone know?
Please view in fixed font (i.e. Courier):
The original transformer representation
i1 R1 R2 i2
o-->--RRRRRR----- ----RRRRRR--->-o-----
¦ ¦ ¦
L L Z
U1 L1 L k L L2 U2 Z Zb
N1 L L N2 Z
¦ ¦ ¦
o---------------- ---------------o------
... is converted to the equivalent T-circuit
i1 R1 Ls1 Ls2' R2' i2'
o-->--RRRRRR----LLLLLL-----o------LLLLLL----RRRRRR-->--o
¦ ¦
V io ¦
¦ ¦
-----o----- ¦
¦ ¦ ¦
iofe V V iom ¦ U2'
¦ E ¦ ¦
U1 R L Z
Rfe R L Lo1 = k*L1 (=t*M) Z Zb'
R L Z
¦ ¦ ¦
-----o----- ¦
¦ ¦
o--------------------------o---------------------------o
by the following formulas (prime: ' ,designates converted
secondary values):
If N1,N2 are the number of primary- and secondary turns
respectively, then,
with transformation ratio t = N1/N2 ~ U1/U2 if Zb=infinite
(=unloaded)
M = k*L1/t Mutual inductance
Lo1 = k*L1 Main inductance, as seen from the primary
side
Ls1 = L1*(1-k) Primary stray inductance
Ls2'= t^2*L2*(1-k) Secondary stray inductance (converted)
R1 Primary resistance (copper-loss, no
conversion needed)
R2' = t^2*R2 Secondary resistance (copper-loss, converted)
Rfe Iron loss resistance (to be left for air
xfrmer's, else =E^2/Pfe)
Zb' = t^2*Z2 Load impedance (converted)
U1 Primary voltage (no conversion needed)
U2' = U2 / t Secondary voltage (converted: ~equal to
unloaded primary voltage)
i1 Primary current (no conversion needed)
i2' = i2 * t Secondary current (converted)
iom Magnetizing current
iofe Iron loss current (zero for air xfrmer's)
io = i1 - i2' Magnetizing current including iron loss
(vector subtraction!)
-> -> ->
The resistances R.. are those at operating frequency, meaning
they include skin- and proximity effects, in addition to the DC
resistance, which may be important in the case of HF. If you like
to sketch a vectorial representation of the transformer, the
direction of the currents, as shown in the diagrams, as well as
the resistive- and reactive parts of Zb (cos(phi)!) are to be
considered. The above schematic is of general validity, even if
developed mainly for the application to power transformers,
including an iron core.
>
> I'd quite like to do the measurements and calculations and work out what ZL
> my tubes are seeing.
Simplifications of the above schematic diagram can be applied to
certain situations of actual transformers, which may simplify
connecting the results of short-circuit and no-load measurements
to the calculation of the circuit elements. I'm currently
generating an Excel-spreadsheet, in order to facilitate this
task. When ready, I could transfer it to you by off-list mail.
>
> Simon
>
I'm well aware, the above sketch is not solving the impedance
matching problem of a VTTC. Only lumped paramers are taken into
consideration, and a lot of the problem remains in the Zb
characterization. It's only an answer to the narrowed down
question of Simon. BTW: what's the meaning of "PFN-calculation" -
I'm only a thumb european...!
Cheers, Kurt
> -----Original Message-----
> From: Tesla List [mailto:tesla-at-pupman-dot-com]
> Sent: Monday, February 28, 2000 4:41 PM
> To: tesla-at-pupman-dot-com
> Subject: Re: Tube Data
>
> Original Poster: "Megavolt Nick" <tesla-at-fieldfamily.prontoserve.co.uk>
>
> Hi Malcolm,
> my comments:
> > Original Poster: "Malcolm Watts" <malcolm.watts-at-wnp.ac.nz>
> >
> > Hi Nick, all,
> >
> > > Original Poster: "Megavolt Nick" <tesla-at-fieldfamily.prontoserve.co.uk>
> > >
> > > Hi All,
> > > One of the main characteristics of an electron emission device
> > > like a vacuum tube is that it will have a comparatively high output
> > > impedance. The tube I mention below can supply 400ma at 5kV -
> > >
> > > Z = V/I
> > >
> > > Z = 5000/0.4
> > >
> > > Z = 12500
> >
> > I am aware of that - I do design valve audio equipment. However, the
> > only part the primary of the transformer plays as far as its
> > inductance goes is to present a sufficiently high shunt impedance at
> > the lowest frequency of interest, the impedance match actually
> > being effected by considering the turns ratio, coupling constant
> > (nearly 1 in this situation so is generally disregarded) and the actual
> > load impedance.
> Yes
>
> >
> > > This means that to avoid damaging the tube the load connected to it must
> > > have an impeadance of at least 12.5 kiloOhms. Therefore it must use a
> > > primary coil with a large inductance to provide this impedance.
> >
> > What happens if you shunt this impedance with a considerably
> > lower load impedance?
> You have to bother to do all the maths ;-)
>
> >
> > > For a typical disruptive TC operating at 300kHz with a 0.1µF primary
> > > capacitor you would need about 28µH of primary inductance to tune, which
> > > has an impedance of only 5 Ohms; whereas a vacuum tube design would use a
> > > much larger primary to provide the large impedance, about 6.6mH would be
> > > needed.
> >
> > Comparing disruptive discharge and CW is an apples and oranges
> > situation isn't it? In one case, you are simply transferring a block of
> > energy - in the other, you are supplying energy on a continuous
> > basis.
> Yes - the primary impedance of the disruptive tc is really calculated as a
> PFN calculation rather than a simple impedance.
>
> >
> > > This arrangement would suit a magnifier system as the large primary
> > > needed could be wound co-axially with the secondary for most of its
length
> > > giving the very high K factor a magnifier needs.
> >
> > I am inclined to think that you are really looking at a 1/4 wave
> > transformer situation for a CW magnifier. Quantifying the load
> > impedance is the interesting bit. Things would get more interesting
> > for effecting a match if you include loose link coupling.
> > Again, does anyone bother working this out properly or is the
> > general rule to take a suck it and see approach?
>
> The reason that my reply was so quickly shot down by anyone who knows what
> they're doing (malcolm :-) is that to actually calculate the load impedance
> of a vttc requires rather a lot of maths I'd rather not do - you have to
> take a sec. backwards approach and calculate the charateristic impedance for
> the topload, then work out the sec. impedance based on the load the sec is
> seeing (the topload) and the sec.'s own charateristics, then work out the
> presented impedance through the pri. from this, factor in the pri
> characteristics and you're there. Definitely not before breakfast ;-)
>
> The real problem is that there are two large unknowns - the k factor and the
> topload impedance. These would require some fairly concerted effort to
> measure to any reliable accuracy.
> Fairly sharp calculus would be a good idea as well :-)
>
> Regards
> Nick Field
>
> >
> > Regards,
> > Malcolm