OPT Design

Output Transformer Design Basics

Contents

Tube Output Transformer DesignEdit

The goal of this book is to show the world how an output transformer or OPT may be designed and what to think of.

TheoryEdit

The fundamental consideration is that the core must not go into saturation at any voltage or frequency. That means that the core must widthstand:

 

Due to DC allmost always being present (especially true in single-end designs) we must also consider the magnetic intensity H:

 

where lm is the mean magnetic length around the core.

It can furthermore be shown that the primary inductance need emerges from the fact that

 

where RL is the reflected load resistance and rp is the plate resistance of the tube(s).

This yields the equation

 

where ZL is the loudspeaker nominal impedance and n the turn-ratio of the transformer.

It can furthermore be shown that maximum output power occurs when each tube is loaded by

 

Knowing this the minimum inductance L may be calculated.

The small signal model do however also put restrictions on HF parameters. If the OPT is carefully wound (in sectors and not bifilary as well as at least one layer of transformer tape inbetween each layer) the main HF problem will be a so called leakage inductance. The equation for calculating this is:

 

or

 

This parameter is however very hard to control. But experience has showed that one important thing is that the secondary winding must cover the whole (or both) primaries. Another idea is that the winding ends must not be folded. This last part might mainly eliminate HF resonances.

The inductance of a toroidal transformer can be expressed as:

 

where

 

where lg is the length of the air-gap and lm is the mean magnetic length.

If lg is zero, this simplifies to

 

where

 

is the relative permeability of the iron and

 

is the permeability of vacuum.

A Practical ExampleEdit

It can be shown that a single-pole low-pass filter roll-off yields an only -0,5dB impact on the frequency three times lower than fh. This means that if we want -0,5dB at highest audible frequency (20kHz) we would need a fh of 60kHz.

Because

 

this means a leakage inductance of less than

 

if we want to use KT66 tubes in push-pull (PP) where rp=2rp(KT66)=2500 Ohm. Observe that RL in PP-designs is 4 times the load on each tube. Optimum plat-to-plate load is thus 10k Ohms in our case.

The same thing is valid for single-pole high-pass filters, thus fl for -0.5dB@10Hz is 3,3Hz. The inductance therefor needs to be greater than

 

If we want to use the suggested core dimensions we get

 

and

 

Because P1+P2 needs to widthstand 230V@15Hz and a common maximum flux density for transformer irons is around

 

the number of turns can now be calculated using the first equation. This gives

 

Putting this into the equation for the primary inductance we get

 

Because we want to use the OPT in Class A push-pull configuration we need not take too much consideration of the DC that will flow due to tube aging and loudspeaker impedance variation with frequency. But it is recommended that the OPT should widthstand at least 10mADC thru both primaries. In SE-configuration a so called air-gap would be needed but this cannot (easily) be realised in toroidal transformers which makes us dependent on the width of the BH-loop of the used iron.

The magnetic intensity for a DC current of 10mA is for our transformer:

 

and here we want to have at least L/3=30H left.

Equation DerivationEdit

In this paragraph the derivation of the above used equations will be explained.

Standard OPT UsageEdit

 
Standard OPT Usage

This picture shows how an OPT is used in push-pull (PP) configuration. The below theory is however also valid for single-end (SE) configurations.


OPT Small Signal ModelEdit

 
OPT Small Signal Model

This picture shows the small signal model of the OPT.

Fig.1 shows the trivial OPT connection i.e driven by the generator G thru two plate resistances rp (because of PP).

Fig.2 shows what happens at low frequencies where the the OPT works as an ordinary transformer. Reflected impedance is therefor

 

where n is the turn ratio of the OPT and ZL is the loudspeaker impedence.

Fig.3 shows what happens at high frequencies where the leakage inductance LL is dominant over the interlayer capacitance (due to special winding techniques described earlier). The above expression still holds though.


Transformer BasicsEdit

Consider an ideal transformer without iron or copper losses. The output power will then be equal to the input power.

If you transform a high voltage to a low voltage you will be then able to extract a higher current at the secondary than you are putting in on the primary.

So if

 

then

 

and if

 

it follows that

 

and finally

 

or

 

Small Signal Model EvaluationEdit

While using Norton and Thevenin circuit theory in Fig.2 we get:

 

In Fig.3 we may however just realize the fact that

 

because this is where the reactance of LL becomes dominant.

Standard Filter CharacteristicsEdit

Imagine a single-pole High-Pass filter. Then you might have a capacitor in series with a resistor to ground. The Laplace transfer function then yields

 

or simplified

 

Putting

 

we get

 

Putting s=jw we get

 

and the amplitude of the transfer function gets

 

or

 

where f is the frequency.

Putting

 

we get

 

Field MagneticsEdit

In this paragraph we will show the world the electromagnetic fundamentals.

Maxwell's EquationsEdit

The normal form of Maxwell's Equations is

 

 

 

 

The first equation, Gauss's Law, describes how electrical fields are caused by electrical charges.

The second equation states that there are no "magnetic charges", or so called magnetic monopoles.

The third equation, Faraday's Law, describes how electrical fields are created due to magnetic field variations.

The fourth equation, Ampere's Law (with Maxwell's correction), describes how magnetic fields are created from electrical field variations.

A List of the Used QuantetiesEdit

E : Electric Field Intensity [V/m]

D : Electric Flux Density [As/m^2]

H : Magnetic Field Intensity [A/m]

B : Magnetic Flux Density [Vs/m^2]

Jf : Free Current Density [A/m^2]

Integral FormEdit

 

 

 

 

Boundary LimitsEdit

Going from medium 1 to medium 2 Maxwell's Equations gives

 

 

 

 

where

  is the surface charge density och Kf the free surface current intensity between the mediums.

Faraday's LawEdit

Consider Faraday's Law where we have from Maxwell's Equations:

 

If we define:

 =emf induced in the curvature C [Volt]

and

 =magnetic flux thru the surface S [Vs or Weber]

then we get:

  [Volt]

If we use several turns N of wire we get

  [Volt]

And if the magnetic flux flows thru an iron where

 

the magnetic flux will stay in the iron only, yielding a secondary voltage proportional to the turn ratio n.

Magnetic Flux Density in an Iron CoreEdit

From Faraday's Law we have

 

And due to no variations in the surface S, we have

 

Using the equation

 

we then get

 

A sinusoidal magnetic flux density yields

 

and thus

 

which maximum occurs when

 

Thus

 

or

 

where A has been substituted for S

And if the voltage is sinusoidal

 

we get

 

Magnetic Field Intensity in an Iron CoreEdit

From Maxwell's Equations we get

 

because we are considering DC only and a homogenous surface.

So if we are using a toroid, then

 

and if

 

then

 

Effective Permeability due to Air-GapEdit

Applying Ampere's law we once again get

 

Here we also have

 

But in the core we will have

 

and in the air-gap

 

This gives

 

This may be rewritten as

 

or

 

where

 

which gives

 

or

 

Toroidal Core InductanceEdit

Consider cylindrical coordinates. Then we get

 

 

 

Since the path encircles a total current NI, we have

 

Knowing the relationship

 

it is easy to relate to the earlier equations, thus

 

and

 

which equals

 

Using that the flux linkage is

 

and that the small signal inductance is independent of the current, we get

 

C-Core InductanceEdit

This is not so easy to calculate but we can do some approximations if the mean magnetic length could be defined by

 

where c is the shortest leg (at the center of the iron) and d the longest dito leg.

Approximating this to a circular toroidal shape, we get

 

Adding half of the thickness of the iron to this we get b, substracting half of the thickness we get a. Then we might reuse

 

This should be quite valid due to the magnetic flux staying in the iron because of

 

Optimizing Tube Amp LoadEdit

 
Maximum Available Power for Triodes

If we consider

 

 

and realize that

 

 

output power may be written

 

and derivated by Ia this gives

 

with maxima for

 

thus

 

From the image we can see that

 

where Ua disappears so that

 

or

 

thus

 

A more simple way to prove this is by inspection of image:

 

This finally proves that optimum load for a triode is double it's internal resistance. It should however be pointed out that plate voltage should be the limiting factor, for higher voltages where plate dissipation comes into the picture, Ra must be higher.

It is also interesting to note that the efficiency is only 25% in this case, we can prove this by putting:

 

with

 

thus