Calculation based on tube amplifier output transformer

1. Impedance calculation
All audiophiles with basic knowledge know that the square of the ratio of the number of turns of the primary and secondary sides of the transformer coil is equal to the impedance ratio, that is, R1/R2=(n1/n2)2, but the copper resistance of the coil is often ignored. Let the copper resistance of the primary side be r1 and the copper resistance of the secondary side be r2. The transformer reflects the secondary side speaker impedance Rx back to the primary side with the equivalent impedance R by the turns ratio n, and connects it in series with the copper resistance. The total output impedance is Ro, then Ro=R+r1+Z2, where Z2 is the equivalent secondary side copper resistance reflected back to the primary side by the secondary side copper resistance through the transformation ratio n, which is equal to r2n2, and the above formula becomes Ro=R+r1+r2n2. For a transformer with a reasonable coil arrangement, that is, a transformer with equal current density in the primary and secondary coils, the primary side copper resistance r1 should be equal to the copper resistance Z2 reflected back to the primary side by the secondary side copper resistance through the voltage ratio n, that is, r1=Z2, so the total copper loss of the transformer is r1+Z2=2r1. In this way, the previous formula becomes R = R + 2r1 or R = Ro-2r1. Please remember this calculation formula, as you will often use it.

[Example 1] The output impedance of an audio output transformer is Ro = 5kΩ, r1 = 350Ω, and the secondary load is 8Ω. Find the turns ratio n.
n = (R/Rr) 1/2 = [(Ro-2r1)/Rr] 1/2 = [(5000-700)/8] 1/2 = 23.2
If the copper resistance is not considered, the result is n = 25, and the impedance of the transformer produced will not be 5000Ω, but 5700Ω, resulting in errors.

The magnetic induction intensity in the core of the output transformer is very low, much lower than that of the power transformer, and the iron loss is small, so the loss is mainly copper loss. The effective impedance R = n2R in the transformer, the ineffective impedance r1 + Z2 = 2r1, and the proportion of the effective impedance R in the total impedance Ro is the efficiency η of the transformer, so η = (Ro-2r1)/Ro. In Example 1, η=(5000-700)/5000=86%.

2. Wire diameter
First, the current density should be considered, which is generally not greater than 2.5A/mm2. 2A/mm2 is selected for precision. Secondly, the transformer efficiency is considered, that is, the DC resistance is set to an index not greater than a certain value. For example, for a transformer with Ro=5000Ω, if η=90%, 2r1=10%, then 2r1=500Ω, r1=250Ω. Usually the first item is subject to the second item. When calculating, first measure the average length of each turn, multiply it by the number of turns, and get the total length of the primary side line. Then check the resistance per meter of the specification line, multiply it by the total length, and get the primary side DC resistance. If it is unqualified, select another specification wire diameter. When the primary side line selection is determined, the standard for the secondary side line selection is as follows:
In a transformer with two coils on both sides, the most reasonable line is used only when the current density in the coils on both sides is equal. Assume that the primary wire diameter is d1, the secondary wire diameter is d2, and the turns ratio is n.
According to the transformer principle, n=U1/U2=I2/I1, and the current is inversely proportional to the voltage. The current density on the primary and secondary sides is equal, and the cross-sectional area S of the wire is proportional to the current I, so I2/I1 =S2/S1, the area ratio is the square of the diameter ratio, S2/S1= (d2/d1)2, and
n=U1/U2=I2/I1=S2/S1=(d2/d1)2
, so d2=d1·n1/2, please remember this formula, only when the wire diameter ratio is calculated according to this formula, the current density on the primary and secondary sides is equal, and the wire is the most reasonable.

[Example 2] The wire on the primary side of an output transformer is φ0.25mm, n=25:1.
The diameter of the wire used on the secondary side is d2=d1*n1/2=0.25×251/2 =1.25mm

3. Air gap calculation
A type A single-ended output transformer has a DC current passing through it. To avoid magnetic saturation of the core, the core is changed from opposite insertion to parallel insertion, and an air gap is left. The size of the air gap is crucial. If it is too small, the core is easily magnetically saturated, and if it is too large, the inductance is insufficient. The factor that determines the magnetic induction intensity in the core of the transformer is the magnetomotive force, also known as the magnetic field intensity, that is, H=I·n, in units of ampere-turns (A·T), n is the number of turns, and I is the current. Under the pressure of the magnetomotive force, magnetic induction intensity will be generated in the magnetic conductive material. Therefore, the larger the magnetomotive force H, the higher the magnetic induction intensity in the core. When it reaches a certain level, the magnetic permeability μ of the core drops rapidly, and the core is magnetically saturated. At this time, the air gap should be increased to control the magnetic induction intensity. For a transformer with an air gap, the air gap width δ=I·n·r, where r is the practical coefficient of different magnetic conductive materials. In the past, when metallurgical technology was backward, r=1.8×10-4(cm/A·T). However, for the high-quality silicon steel sheets commonly used now, r=1×10-4(cm/A·T), and the air gap width has nothing to do with the size of the core.

[Example 3] For a Class A single-ended output transformer, the primary coil n1=3000T, driven by 300B, the plate current Ia=80mA, calculate the air gap δ.
δ=I·n·r=0.08×3000×1×10-4=0.024cm=0.24mm

The data measured under different inductance test conditions are also different. Generally, there are the following types:
1 Initial inductance
The no-load inductance measured by the inductance meter is the initial inductance. This data is only useful for the frequency correction network without load, because its conditions are far from the working conditions of the output transformer, so it is not very useful.
2 AC inductance
The data measured by loading AC voltage on the primary side of the transformer is the AC inductance, but additional test conditions such as frequency and voltage must be added. The inductance measured by this method is useful for push-pull output transformers because its test conditions meet the actual working conditions of push-pull output transformers.
3 Inductance after loading DC working current
Suitable for air-gap output transformers and power supply filter chokes for Class A single-ended machines. This method is relatively complex to test, and generally about 70% of the measured AC inductance value can be used to estimate the inductance after loading DC working current.