AUDIO OUTPUT TRANSFORMERS

An audio transformer can be modeled as below. Image courtesy of Igor Popovich

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ri = anode resistance of output tube
Cp = capacitance of the primary winding
Rp = resistance of primary windings
LLP = leakage inductance of the primary winding
Rc = core power losses, due to eddy currents and hysteresis
Lp = primary inductance
LLSTR2 = leakage inductance of secondary reflected to the primary side. TR = turns ratio
RsTR2 = secondary resistance reflected to the primary side
Cs/TR2 = parasitic capaticance of secondary reflected to the primary side

The parallel combination of Rc and Lp is the transformer itself. An ideal transformer doesn't present any impedance to the input voltage. The transformer is merely a device to transform the secondary resistances (by a factor of 1/T2). Now, this is a very complicated model to analyze, so it will be analyzed for the Low frequency model, midfrequency model, and HF model.

MID-FREQUENCY MODEL

At mid-frequencies (100 - 5000Hz), all inductances and capacitances can be pretty much ignored, so you're left with just a simple voltage divider:
Vo = RLTR2 / (ri + Rp + RsTR2 + RLTR2).

LOW FREQUENCY MODEL

At low frequencies (200Hz to 5kHz) the inductances and capacitances can be ignored except for Lp. So you're left with the following simplified model:

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where R1 = rI + Rp and R2 = (Rs + RL)TR2

Now the frequency where the output voltage drops by 1/√2 relative to the midband gain is the -3dB frequency fL. This frequency is determined by the following:
fL = [R1R2TR2 / (R1 + R2TR2)]/2πLp, or more simply
fL = (R1 ∥ R2TR2) / 2πLp       ∥ means "parallel with"
Or, even more simply:
fL = RPAR / 2πLp,    where RPAR = R1∥R2TR2    

Remembering that R2TR2 = primary impedance of XFMR, also known as ZP:
RPAR = R1 ∥ ZP.     Now, rI >> Rp, so Rp can be ignored (it's usually less than 100). Thus:
RPAR = rI ∥ ZP

So, finally;

You want fL as low as possible, so you want LP above a certain minimum. Rearranging the above, gives:
LPMIN = RPAR / 2πfL

LPMIN is the minimum value of L needed to get the lower -3dB frequency fL to the desired level to allow for adequate bass.
Looking at the above, a lower RPAR means a lower (and easier to make) LPMIN. This means that low impedance tubes (those with low rI) are more desirable for the beginning choke maker.

HIGH FREQUENCY MODEL

For reasons I don't fully get, the shunt capacitances of the transformer model can be ignored, so you are left with the following model.

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This is a simple low-pass LR filter and the upper -3dB frequency relative to the midband gain is the frequency at which the reactance of the total leakage inductance = resistance of the 3 resistors in series. In other words:
2πfULL = RSER,      where LL = LLP + LLSTR2 and    RSER = rI + R1 + R2TR2. So...

fU = RSER/2πLL
LLMAX = RSER/2πfU

UNIVERSAL FREQUENCY AND PHASE RESPONSE CURVES

It turns out that if you look at values of RPAR and RSER, RSER is usually about (and that's a very big about) 5X larger than RPAR. So, looking at equations above, it's easy to see that:
fU / fL = 5Lp / LL

All the above can be summed up in a graph below, found in Igor Popovich's book "Transformers for Tube amplifiers." Note that for both the high-frequency and the low-frequency halves, the gain has dropped to .707 (1/√2) when ωL = R

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The equation at the very bottom right is derived as follows:
LP = NP2μRμ0A/ℓMP    (see the Transformer Theory page)
LL = μ0NP2VW/A,     where VW = volume of the windings.
So, fU / fL = 5LP / LL = 5μRA2 / (ℓMPVW)

This equation shows that increasing cross-sectional area A of the laminations seems to increase the bandwidth, but this is offset by the increase in ℓMP and VW, so the only real way of increasing bandwidth is to increase permeability by using high quality iron.

Looking at the equations for fU and fL, the following is seen:
On the low frequency side, you want to minimize rI and maximize LP.
Unfortunately on the high frequency side you want to maximize rI. This favors pentodes and tetrodes. LP doesn't really matter, but you want to minimize LL.

LEAKAGE INDUCTANCE

Leakage inductance LL is due to magnetic flux between primary and secondary windings that is not coupled through the magnetic core. If LL = 0, then all of the flux is shared by the windings. A sad fact of life is that as f increases LL increases.

Without going through a derivation (which I don't know anyway), the following expression is given:
LL = u0V/CL2Np2,    where V =