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Audio Transformer Substitutes

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Joe Sousa
Joe Sousa
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23.Nov.09 08:53

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Hello fellow radiophiles:

Exact replacements for audio output transformers in old tube radios are often hard to find. Some of the problem is that there is some competition on the market from the audiophile world.

But there is a general purpose multi-Impedance transformer type that has not been "discovered", and is easy to find at modest prices and in new production.

I am talking about the so-called "70V Line Matching Transformer". These transformers are marketed for public address use, as you might find in large stores and other public spaces.

The engineering concept behind these transformers is that the power amplifier puts out several watts of power at high impedance, and each of the remote speakers, which could be located hundreds of feet away, has it's own step-down impedance matching transformer.

The reason for this arrangement is low loss transmission of power over long distances with wires that have significant resistance. This is the same reason why Nikkola Tesla transmitted power from Niagara Falls hydro-electric power station over long distances with high voltage transmission lines over a century ago. This concept became universally used.

For example, consider an 8_Ohm speaker driven by 100feet(30m) of #26 (0.4mm diameter) wire. The resistance in 200feet of this wire is 8_Ohms. This means that half the power would be lost in the wire resistance. The problem gets worse with large installations drivine several speakers, where the net parallel speaker impdedance could drop to less than 1_Ohm.

If the same 100 foot speaker wire is used to drive a 1kOhm load, the 8_Ohms of wire resistance are negligible.

Application to tube radios

These transformers come with several primary and secondary taps, so that the output power delivered from the public address amplifier to the speaker can be adjusted without the loss of a rheostat . The power levels that are marked in the primary winding assume an 8 Ohm or 4 Ohm load to the appropriate secondary tap.

 

I added voltage, or turns, ratio markings to the impedance markings on the box schematic diagram:

The primary impedance is usually not stated in the service data, but tube data sheets often contain this information for individual tubes:

These two examples for a 6L6 and a 1C5, require primary impedances of 2.5K and 8K, respectively. The 2k and 8k taps of the example transformer would work for these two cases. If the requirement is for a 3k tap, you can experiment with the 2k tap and 4k tap for the sound you like better. Perhaps you will not notice a significant difference between the two taps. Be sure to insulate any unused tap wires, which must remain unconnected.

Low frequency response

The measured secondary inductance at the 8 Ohm tap is 4.5mH. The inductance at the various taps is directly proportional to the tap impedance. This means that the primary inductance at the 8kOhm tap is 4.5H.

The inductance establishes the low frequency cutoff point from F=R/(L*2*pi) as 8_Ohm/(2*pi*4.5mH)=282Hz, if the primary transformer is driven by the high output impedance of a pentode or a bipolar transistor. By high output plate impedance, it is meant that this impedance is greater than the impedance seen at the primary, which can be taken to the the stated impedance on the box, if under ideal 8 Ohm or 4 Ohm load conditions.

A lower output plate impedance driving the transformer, such as that at the plate of a power triode will extend the low frequency cutoff. For example, if a triode with an 8kOhm plate impedance drives the 8K tap, the low frequency cutoff is extended to 141Hz. This happens because the 8k of the triode combines in parallel with the 8k load impedance for an effective resistance term R=4k in the general formula F=R/(L*2*pi) for the low frequency cutoff point.

Negative feedback from the speaker terminal to the the input of the amplifier also extends the low frequency cutoff.

In general, good results can be obtained for transformers of the same or smaller physical size, as long as the correct primary and secondary taps are used.

The Physical size establishes the maximum amount of magnetic flux that the transformer can store. The highest flux requirements are at low frequencies, so the size of transformer affects low frequency power handling ability and distortion.

Push-Pull substitutions

The primary taps are arranged in 1.4X step turns ratio or voltage increments. This means that every two steps the turns ratio is 2X. This is ideal for a Push-Pull substitution. A few examples of Push-Pull combinations for various output tube impedances.

Transformer tap in Ohms

Voltage or turns ratio to output

2k in p-p 1k in p-p 500 in p-p
Push-Pull Configurations for various push-pull output impedances
8000 32x Plate A    
4000 22x   Plate A  
2000 16x Center Tap   Plate A
1000 11x   Center Tap  
500 8x     Center Tap
common 0 Plate B Plate B Plate B

This table shows three possible impedance configurations for push-pull. The push-pull impedance in the table headers is the plate load impedance that each plate will see, assuming that the other plate has the high impedance of a pentode.

The 2kp-p example would be good for a 6L6 push-pull configuration. The last example column is more likely to be useful in a p-p transistor amplifier substitution.


A note about output impedance requirements for audio amplifiers

The first consideration for the load impedance of output amplifier is the impedance that can deliver the highest level of undistorted power.

This is a very different requirement than matching the high output impedance of a pentode or transistor to an equally high load impedance.

The maximum power that the amplififer can deliver can be first determined from the the curve family of the tube or transistor, such that the load impedance can be represented by a load line that connects the point of lowest voltage and highest current with the point of highest voltage and lowest current.

The following curve family has a 4kOhm load line that crosses pentode curves of a substantially higher impedance. The more horizontal the curve, the higher the impedance.

The two blue dots mark load line limits over which the curve spacing is even enough for low distortion. The red dot marks the plate supply voltage of 105V and 43mA of quiescent current.

 

Regards,

-Joe

 

 

This article was edited 23.Nov.09 09:31 by Joe Sousa .

Bernhard Nagel
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23.Nov.09 11:54

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Dear Joe,

for single ended applications as most of the tube radios and audio amplifiers are built, the output transformer needs an air gap to prevent it from DC saturation caused by the plate current. Most probably this Speco T-7010 transformer doesn't have one, because the originally purpose was for public adress usage where no air gap is required.

For push-pull applications an air gap may be obsolete due to the compensating effect of the DC magnetization. For a more linear frequency response and less distortion factor, a small air gap will often be used anyway.

Regards, Bernhard

This article was edited 23.Nov.09 11:55 by Bernhard Nagel .

Dietmar Rudolph † 6.1.22
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23.Nov.09 11:55

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Line Matching Transformers are a good idea in principle.

One problem remains, however. Single ended (not push pull) output transformers have an air gap in the iron core. So a magnetic saturation of the iron due to the anode DC current can be avoided.

Line Matching Transformers usually don't have such a gap, and the core will be magnetized by the DC current giving lower impedance.

Choosing a type with greater power is therefore recommended.

If the core is not glued, probably disassembling is possible and assembling again in one direction and add a small gap could be a solution.

Regards,

Dietmar

This article was edited 23.Nov.09 12:03 by Dietmar Rudolph † 6.1.22 .

Joe Sousa
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24.Nov.09 19:16

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Gentlemen, thank you for your nearly simultaneous contributions. This is exactly the most cherished kind of Forum feedback I look for. It is the kind of feedback I get when discussing a topic with my engineering colleagues.


Utility of a Gap

I must admit that I knew about gaps in chokes, but I had not seen them in the few low end audio transformers I have disassembled. I will have to look more closely through my audio transformer stock to find the gaps.

Your correction made me think about how a gap could be designed for an audio transformer, or for a choke.

The utility of the gap is that it drops most of the magnetic potential across it, such that much less total flux is needed to store the same magnetic energy.

A good design goal for the gap could be that the total DC flux is dropped by about the same amount as the crest factor of music. In a recent post, Prof Rudolph mentioned this crest factor as 10dB, which is to say that the average rms content is about one third of the peak voltage and peak current.


Gap size estimation

So we need to increase the total magnetic resistance in the magnetic path by 3X, but we still need the inductance that we are going to loose with the addition of the air gap.

Fortunatelly, the inductance is proportional to the square of the number of turns, while the flux is directly proportional to the number of turns. This means that we can regain the inductance by increasing the number of turns by the square root of 3, which is 1.7, or add 70% more turns to get the inductance back.

L=N*Flux/Current

"N" is the number of turns, "Flux" is the magnetic flux usually denoted by the letter Phi, and "i" is the current.

The magnetic flux is also a function of the number of turns "N", so "N" appears twice in the product shown in the equation, and explains the dependency of "L" on the square of the number of turns.

If we want to add an air gap, but retain the same inductance, then the magnetic path resistance needs to increase by the square of the net reduction in DC flux that we want. In the example we are working with, this means a magnetic path resistance increase of 9X instead of 3x.

If the relative permeability of the core (ur) is 1000, then the magnetic path resistance will be increased 9x if the air gap measures 8*Path/1000. A rough estimate of the magnetic "Path" length for the Speco transformer would be about 4 inches (10cm). We need a gap that is 32mils (0.8mm) long.

Keep in mind that if using the original EI lamination shape, three gaps will form if all the I pieces are grouped over the E pieces. One gap is at the center, the other two are in parallel, on the outer legs, and can be thought of as parallel gap. The net gap should then be about half of the 32mils. The area of each gap could also be taken into account to adjust the gap size.

Note that the core permeability was only a guess on my part. I have not measured it. But this illustrates the calculation method.

The number of turns would also have to go up by 3x to retain the original inductance, while achieving a 3x increase in the saturation range of the transformer.

The winding resistance that additional turns causes is usually tolerated in an amplifier design, but represents too much loss in a power transformer. This explains why most power transformers run at the saturation limit when idling without a load. A load current cancels enough of the magnetic field, such that core saturation is eliminated. This explains why transformers often run quieter under load.


Gap to estimate core permeability

Another use for the gap is to estimate the permeability of the core. The idea is to introduce a known short gap that only drops the inductance by about half. The gap and inductance are related linearly. If it took 10mils to drop the inductance by half, then permeability is Path/10mils.

I suggest dropping the inductance by not much more than half to retain reasonable accuracy in the estimation. If the inductance is dropped by 1/10, then the estimation becomes more error prone.


Future Measurements

Now I should measure how much DC current it takes to saturate Speco transformer. Perhaps, as Prof Rudolph suggested, the Speco has a high enough saturation current to be useful as a replacement for small audio output transformers, like the kind found in AA5 radios.

I am also going to go through my collection of spare audio transformers and look for those that have gaps and compare them to those that don't.

Comments invited, in particular, relating to some of the engineering assumptions I proposed.

Regards,

-Joe

p.s.: I apologize to non-London residents for being unable to extend meaning the pun in the title of this post into different cultural contexts. Londoners, have a good chuckle!

Dietmar Rudolph † 6.1.22
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In a German article written by Hans Knoll the calculation for output transformers is presented.

In an appendix formula and examples are given. (Origin: Telefunken)

Regards,

Dietmar

Joe Sousa
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Dear Prof. Rudolph,

Thank you for the reference to Hans' article. Somehow, I had not caught that thread when it came out, or perhaps I looked at it superficially. Now I have a good reason to study it carefully.

I tried OCR on the first page with "Softi FreeOCR" in German mode, and got good results. So I will proceed with the rest. Then google-translate. The good ilustrations and formulae will help keep the translation reasonable.

One part of the core saturation considerations I will look for, is low frequency behaviour where the VoltXsecond product is highest. That is to say that a voltage remains applied during a half cycle, for the longest time, and has the highest chance of puting the core into saturation.

Another aspect of trasnformer behaviour I will look for is the magnetic Flux cancelling effect of load current, and the benefit that this has on saturation reduction.

Speaker self-resonance would tend to greatly reduce loading at resonance, and thus hasten output transfomer saturation at a low frequency. Negative feedback would tend to moderate this effect.

This is, indeed, an interesting topic.

Regards,

-Joe

Joe Sousa
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28.Nov.09 21:57

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A further clarification about Volt X seconds (Volts times seconds):

When a voltage is applied to an ideal inductor, the current and flux start at zero and increase linearly over time, until infinity.

We know that real inductors have limitations. One limitation is saturation at the core, which causes a drop in inductance and a much faster increase in current over time, until the maximum current is limited by the winding resistance.

Volt X seconds then becomes a measure of how close you are to saturating the inductor. Cored inductor and transformer vendors often specify how many Volt X seconds can be applied to an inductor until a pre-defined level of saturation occurs. This limits the maximum magnetic energy that can be stored in the inductor or transformer.

My power supply design coleagues at Linear Technology use VoltXseconds as a key figure of merit to select inductors for their designs.

High Pentode plate impedance

Most tube amplifiers have the high impedance of a pentode, or beam tetrode, driving the lower impedance of the primary of the output transformer. The plate impedance can be a factor of 10x higher (i.e. 40k) than the load impedance of the transformer primary (i.e. 4k).

We can think of the plate as a current source forcing current nearly instantaneously onto the transformer primary. This, in turn, means that the magnetic field created by this current is also formed nearly instantly, after the core material has had time to respond. The core material can change it's magnetization at rates faster than the audio range, so we will neglet any core magnetization delays.

In short: a current drive to the primary translates nearly instantaneously to a flux level in the core.

The concept of Volt X seconds does not apply in the case of current drive.

OK, so how do we apply the concept of Volt X seconds in a pentode plate circuit?

The answer is that the total impedance seen by the transformer includes the speaker load and, in this case, it is it's only significant load, if there is no negative feedback.

Now we can apply the concept of volt-seconds to the transformer primary impedance under output load.

The current that is forced by the plate first "sees" the reflected secondary impedance as a resistance (i.e. 4k), and a voltage appears nearly instantaneously at the primary, that is the current divided by the loaded transformer primary impedance. Then the internal current will increase, at first seemingly linearly, then exponentially, with an R/L time constant, where R is the primary impedance with the transformer under output load, and L is the primary unloaded inductance.

This is where saturation trouble will arise at low frequencies, if a half cycle takes long enough for the current to rise in the R/L time constant regime. A high DC standing current was the chief concern about using the 70V transformers for single ended use, because it brings the transformer much closer to saturation.

Negative feedback emulates a transformer load.

If the amplifier has significant negative feedback from the speaker output, or from the plate output, to the input, then the transformer voltage is regulated with negative feedback, and the concept of Volt X seconds can be applied again, even if there is no speaker load.

By significant negative feedback, it is meant that the output voltage follows the input voltage with the gain established by the feedback network. If the application of negative feedback to the amplifier lowered the net closed loop amplifier gain by 1/10, then the output follows the input within 10%, at the new closed loop gain.

This amount of feedback also lowers the open loop output impedance by a factor of 10. This means that a 40k plate impedance will look like a 4k impedance, which is comparable to the loaded transformer primary impedance.

I had not thougth about much about this topic until this thread. Your feedback stimulated a further appreciation of transformer behaviour. Please feel free to share any insights.

Regards,

-Joe

Joe Sousa
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03.Dec.09 04:51

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This post was originally located in the German language thread "Calculation of push-pull and single ended output transformers" by Hans Knoll, and was in response to his German language post "Matching of speaker impedance to output tubes".

Comments on administrative issues moved to the talk section. (JS 2009-12-13)

Regards,

-Joe


Hello Hans,

I have read your article about output transformers. I was familiar with some of the points, but I also learned new ideas. I will share with you a few lessons I learned in your article.

The Boucherot series R-C network

One idea was the use of the Boucherot series R-C network in the plate circuit to cancel the inductive increase in impedance above 1kHz, that is seen in typical full range speakers with a single driver.

Somehow, I had never appreciated this important point before.

Speakers are usually designed for a flat response when driven from a voltage source. The increase of impedance at high frequencies does not affect a voltage source, but it affects a pentode plate.

In a silly case of missing the obvious, while being concerned about second order effects, I thought that the highest concern about high frequency at pentode plate primary, was excessive voltage swing caused by the leakage inductance.

Another way to explain my myopia, is that I was not thinking of the pentode-transformer-speaker as an inextricably connected system.

If the leakage inductance is low, it's effect should start well beyond 1kHz, and the high impedance of the pentode plate is very good at forcing the appropriate high frequency current despite the leakage inductance. So the leakage inductance should have no effect on the frequency response at the transformer output of a pentode amplifier.

The high frequency voltage swing should be put into the context that the largest swings at the plate usually come from bass frequencies, so there is usually plenty of voltage swing left in the treble range.

Once the Boucherot series R-C network is applied at the primary, the second order effect of leakage inductance can cause loss of treble. The treble loss should start at the R-L time constant where R is the Boucherot resistor in series with the reflected resistive component of the primary impedance, and L is the leakage inductance.

One way to have the speaker impedance normalization benefit of the Boucherot series R-C network, while taking advantage of the high impedance of the pentode plate to overcome the leakage inductance, is to place the Boucherot R-C network in parallel with the speaker terminals. The new R-C impedance, is divided by the square of the transformer turns ratio.

Triode vs Pentode plate impedance

One of the interesting points to me, was your direct comparison of triode plate impedance to pentode plate impedance.

My first guess about the optimum load impedance for a triode was to match it to the plate impedance for maximum power output.

I knew that a higher load impedance would produce less distortion at the expense of delivered power, as the triode transfer characteristic would be closer to a current source load, which has the lowest distortion. But I did not realize that a somewhat increased load impedance was a common circuit realization. You pointed this out in your article amendment of February 2009, as Ra=3..4xRi.

I have occasionally thought that a Pentode could be made to respond like a triode, while retaining it's power capability and plate efficiency.

The simplest way that I can think of, to do this, is to apply negative feedback from the pentode plate to the pentode control grid with an AC-coupling capacitor.

I used this concept with the Trioderizer.

This resistive network would have an attenuation ratio that is the inverse of the "mu" of the desired triode response.

This circuit is not very different than applying negative feedback from the transformer secondary to the cathode of the pentode. The significant difference is that including the transformer in the feedback loop  "hides" the leakage inductance, and linearizes transformer response.

Part of what is understood as "triode sound', as you point out, is the high frequency roll-off from leakage inductance.

The reason to use the plate feedback is that the rest of the design can be conducted as if a triode were used.

The most important difference is the new closed loop pentode plate impedance would look like that of a triode. The simplest approximation of this impedance is that the closed loop plate impedance would be mu/gm, where mu is the inverse of the plate feedback attenuation, and gm is the transconductance of the pentode.

The reason why the plate efficiency is not lost with this approach, is that the Screen voltage remains at full pentode bias, as usual. In a triode connected pentode, the efficiency is lost because the screen voltage drops when the output swings downward.

A related approach is the Ultra-Linear circuit, where a tap from the output transformer drives the screen grid. This approach looses some of the pentode plate efficiency because the highest plate conductance is reduced by the simultaneous drop of screen voltage during the most negative plate swings.

I think I have seen direct feedback from the plate to the control grid of pentode in some radio design, but I can't remember where. Global feedback is certainly more popular, with the many ills that it corrects.

Thanks for the lessons,

-Joe

This article was edited 14.Dec.09 05:29 by Joe Sousa .

Dietmar Rudolph † 6.1.22
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An example for such a direct feedback is the DKE (Deutscher Klein-Empfaenger) also called "Goebbels Schnauze" which is a very simple and cheap receiver from 1938.  It has no output transformer.

Regards,

Dietmar

This article was edited 03.Dec.09 14:52 by Dietmar Rudolph † 6.1.22 .

Joe Sousa
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Thank you very much for the reference. I had seen the DKE38 schematic before, and now I remember being intrigued by the 2Meg negative feedback from Pentode plate to grid as the classic Rf of an opamp circuit, with the internal ~30k triode plate impedance as the grounded Ri of an opamp circuit.

I have been reading though a lot of material that has been posted on the DKE38. There were interesting extensive articles written by you, Hans Knoll, and Wolfgang Holtman, along with shorter contributions that were also interesting.

600 Ohm "Entbrummer" resistor

One very interesting point was the dual purpose of the variable 600 Ohm "Entbrummer" resistor. It creates a negative DC drop that serves as the DC bias point for the Pentode control grid. It also injects the correct amount of AC hum which is in opposite phase with the hum present at the screen grid G2. The result is that the Hum component is reduced in the plate current. After proper adjustment of the 600_Ohm resistor, which was eventually replaced with a fixed 300 Ohm resistor, the pentode plate voltage has a hum component that matches as closely as possible in shape, phase and amplitude, the hum voltage component at the supply side of the speaker terminals.

The DKE has a very sophisticated low frequency design in Audio and hum reduction.

The 2Meg negative feedback resistor

The 2Meg resistor, like almost any other component in this radio, serves more than one purpose. It provides negative feedback from plate to grid of the pentode.

The closed loop gain from pentode grid to plate, as established by the ratio between the 2Meg resistor and the ~30k internal triode plate resistance is around 70. Without feedback, the intrinsic voltage gain of the pentode is around 70k*5mS=350. The pentode has a 5mS transconductance, and 70k internal plate resistance.

This negative feedback also lowers the output impedance of the pentode from ~70k open loop to something near 70/5mS=14k.  This reduced plate impedance makes the pentode plate behave like a triode, while retaining the high pentode efficiency that is provided by the fixed screen voltage.

We should expect a low pass filter behaviour from the ~14k closed loop plate impedance and the speaker inductance. The low pass filter behaviour is in the current through the speaker coil, not as a terminal voltage, which should rise with frequency.

The 2Meg negative feedback resistor also lowers the load impedance for the triode plate. This reduces the voltage swing at the triode plate, with the load current pushed across the 2Meg resistor.

The triode plate runs at a very high voltage for a grid leak detecting stage. This is made possible by the high intrinsic gain of the triode, which runs around mu=50. The high mu means that the plate voltage has little stealing effect on grid current, which is essential for efficient detection.

100k grid resistor and 30pF Miller feedback

I neglected the 30pF feedback capacitor and the 100k input resistor at the grid in these first rough calculations of circuit behaviour. But these should be considered for the overall audio gain of this stage. The 30pF behaves as a classic Miller feedback capacitance in a triode from the 1920's and will reflect the inductance of the speaker as negative resistance at the pentode grid input. This would be a good reason to add the 100k resistor and maintain stability. Perhaps these values were cleverly chosen for some added "secret" gain, that plays the reflected negative resistance at the grid, against the positive resistances driving it.

Another function of the 100k - 30pF combination is to serve as low pass filter to reject RF content in the detected audio signal.


It is amazing to see RC based designs of such sophistication in a low end receiver, when the carbon composition resistor had been around for hardly more than 10 years in 1938.

One important missing component from my estimates is the inductance of the speaker coil. I conveniently assumed the impedance of the coil to be high enough to be neglected in most of the calculations.

These are all back-of-the-envelope calculations that helped me appreciate the low frequency design of the DKE38, but a more rigorous analysis could, and perhaps should, be done.

Regards,

-Joe

 

  
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