Output Transformer Lumped-Element Model

Many characteristics of a transformer are distributed. There is capacitance between turns of a winding, for example, that is evenly distributed along the winding. A transformer model can be substantially simplified if the distributed capacitance is lumped into a single, ideal capacitor. Here is a lumped-element model of a real-world output transformer.¹

real-world transformer lumped-element model

C_P = primary shunt and distributed capacitance
L_P = primary leakage inductance
R_PW = primary winding resistance
R_CL = resistance as a result of core (eddy current and hysteresis) losses
L_M = primary winding inductance
C_PS = primary-to-secondary inter-winding capacitance
L_S = secondary leakage inductance
R_S = secondary winding resistance
C_S = secondary shunt and distributed capacitance
R_L = load resistance across the secondary
N₁ = number of turns in the primary winding
N₂ = number of turns in the secondary winding

At the core of the model is an ideal transformer formed by N₁ primary windings and N₂ secondary windings.

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Midrange Transformer Losses

At midrange frequencies the large magnetizing inductance and the capacitances are open circuits and the small winding inductances are short circuits. This gives us just the pure resistances shown here.

real-world transformer lumped-element model for midrange frequencies

To simply the notation, let's define n = N₁/N₂ to be the winding ratio, the number of turns in the primary divided by the number of turns in the secondary. The equivalent resistance at the ideal transformer primary, based on the secondary resistances, is

where R_P is equal to R_L translated from the secondary to the primary by an ideal transformer. This gives us, from the perspective of the primary circuit, this equivalent AC circuit, where V_P = nV_out and R₁ = n²R_S.

midrange-frequency equivalent primary circuit

Notice that there are no inductances or capacitances in this circuit, only pure resistances. The output is in-phase with the input. If there are no transformer losses, then R_PW = 0, R₁ = 0, and R_CL is infinite, corresponding to an open circuit. This means V_P = V_in and the transformer output is

which is the relationship for an ideal transformer. For the lossy circuit, on the other hand, we get an attenuated output of

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For audio transformers operating over midrange frequencies, the voltage losses due to winding resistance are more important than core losses, and current is relatively unaffected.² This gives us this approximate circuit.

midrange-frequency equivalent primary circuit for audio transformers

The output voltage is approximately

Manufacturers describe midrange losses in terms of efficiency and insertion loss. Transformer efficiency is defined as the output power P_out divided by the total power P_total, which includes losses. For the approximate audio circuit, we get a transformer efficiency of

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Consider, for example, the Amplimo BV Type 3A524 output transformer. According to its data sheet, R_P = 3.545kΩ, R_PW = 72.9Ω, R_S = 0.423Ω, and n = 21.05. This gives us an estimated efficiency of 93 percent, which is excellent.

The insertion loss is the power loss induced into the circuit by the addition of the transformer. It is equal to the output power relative to the input power, measured in dB.³ For the 3A524, for example, we get

Low-Frequency Transformer Response

At low frequencies, the primary winding inductance L_M is a significant factor, so the circuit looks more like this.

transformer equivalent circuit for low frequencies

The inductance L_M is very high, usually on the order of several hundred henries. Its reactance 2πfL_M decreases when the frequency decreases, however, so eventually it begins to short out the resistances in parallel with it, attenuating the output signal. The voltage driving an inductor leads its current by 90 degrees, which creates the shift in phase angle shown here.⁴

typical frequency response of an output transformer

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High-Frequency Transformer Response

At high frequencies, especially when the transformer is driven by pentodes or beam power tetrodes, things become more complicated. The midrange frequency circuit needs to be modified to include the total leakage inductance reflected in the primary and the equivalent winding shunt capacitance C_eq that are shown here, where L₁ = L_P + n²L_S and R₂ = R_PW + n²R_S.

equivalent transformer circuit at high frequencies

The resistance due to core losses R_CL comes from eddy currents and hysteresis and is about as large as the plate resistance of a pentode or beam power tetrode. The inductance and capacitance in the circuit cause increasing attenuation with increasing frequency. The voltage in a capacitor lags the current through it by 90 degrees, so when combined with the effects of the leakage inductance we get the opposite phase angle effect that we observed at low frequencies.

References

¹F. Langford-Smith, ed., Radiotron Designer's Handbook, 4th ed., (Harrison: RCA, 1953), p. 204.

²F. Langford-Smith, ed., Radiotron Designer's Handbook, 4th ed., (Harrison: RCA, 1953), p. 206.

³Edward C. Jordan, ed., Reference Data for Engineers: Radio, Electronics, Computer, and Communications, 7th ed., (Indianapolis: Howard W. Sams, 1985), pp. 13-14.

⁴Reference Data for Radio Engineers, 3rd ed., (New York: Federal Telephone and Radio Company, 1949), p. 198.