1970s Design Indulgence

The "right-hand rule" (as taught in technical colleges when I was a youth), helped us remember the direction of the magnetic field surrounding a wire.

It is almost gyroscopic, in that, the magnetic field opposes any change in current, to a degree.

Now, what property do you think it had? Correct, if you said inductance!

Audio beginners might be in awe to learn that a straight piece of wire has inductance. Still, in the absence of engineering education, it is understandable, but now you know. Whatever you do, please don't ask me how much inductance it has.

It is good to understand that conductors have inductance when current flows, which happens when the product (or system) containing the conductors is switched-on.

When modeling in SPICE, the simulator assumes all conductors have zero resistance and impedance (impedance being the alternating current version of resistance). In a valve set, where the current was quite low, the inductance wasn't very significant. The wiring to common / 0V / ground could often be daisy-chained - but whenever there is a large current, the inductance is substantial too. In a solid-state amplifier, the current at specific parts of the circuit is large enough to make the inductance do a lot of signal damage.

It is not only straight wires which possess inductance, but printed circuit traces too. The moment you curve them or they need to take a corner; the inductance becomes larger.

The reason for explaining this helps in the understanding of grounding.

Apparently, there was a time when every piece of successful hi-fi equipment ever made was designed by one man. To paraphrase Bowie, "He made everyone, and he's been everywhere." Everybody else seems to have been unnecessary. Slight exaggeration? I'm not sure.

He was so big, he makes Doug Self look amateurish, and as for Bob Cordell?

Perhaps that is why so little is written about the real physics of amplifiers - the essential stuff I'm finding (or not finding), which keeps an amplifier's pecker up, no matter how long it's left on.

I suppose if one man knows everything and it's his road to riches, he isn't going to let any secrets out. Who is he? Find out yourself! I'm just me, and I prefer to keep things that way.

Perhaps writers like JLH (no, it isn't him) didn't feel confident enough to explain parasitics and negative impedance, so simply chucked in EF2 driver base stoppers of any value between 300 ohms and 3 kilohms, because it looked good? I only read two of his books on amplifier design, and I'm still looking for an explanation.

Looking back, I've never done a complimentary EF2 before. Correction, I did one for a commercial headphone amplifier and gave up. It did nothing but oscillate, and there was no cure. In the event, there wasn't really any need for such a high beta to drive headphones.

Some might think it silly to take on a vintage transistor design and use modern transistors, but I think I explained that back on page 1. The design has developed and become audibly better along the way and has measured better on occasion.

The problem is still one of going horribly bright after some days switched on permanently. Everything I know, and that (most) others know has been tried, but I seem to keep returning to the possibility of parasitic oscillation.

Parasitic oscillation is triggered by a stimulus, and once established, stays like that unless switched off and given time, such as 24 hours, after which it sounds good again. It can be imagined as the amplifier equivalent of tinnitus.

An EF2 amplifier with a stabilised supply doesn't seem to suffer "tinnitus," probably due to "stiffness of supply," and is worse of all using a toroidal transformer unregulated supply. I read somewhere that a full bridge HT supply leads to greater instability, which is a bit of a shame because this single rail design has to use one.

At this moment in time, I am convinced it is output stage parasitics causing the problem, and I'm sure there is a cure. The difficulty is that although two white papers have been written on the subject, they say very little about practical solutions, although one would appear to do.

I am encouraged to tread this path by looking up old designs of commercial circuits, where most can be seen trying to overcome something in the HF region. The work of Otalla also points in the approximate direction. It's as if many had an inkling, but few could tie it down.

And as for the man who worked for them all, I'm following where he went.

You can look at a transistor datasheet until you're blue in the face, it will never tell you exactly what will happen.

https://www.onsemi.com/pub/Collateral/MJE243-D.PDF

Try finding a.c. current gain? It doesn't say. It only shows DC current gain. However, it can be taken as a.c. current gain at low frequencies.

The problem is that the current we are using is off the left-hand side of the graphs. The actual DC current flowing in the emitter resistor of the driver transistor is 4mA, and the chart stops at 40mA.

To get a better idea of current gain, we need to look at the tables, where we see minimum current gain - also known as H parameter hFE. It shows 40 or 15, and as 40 is at a lower current than 15, and our actual current is much smaller, we assume it's going to be 40.

We can then see what the transition frequency is (Ft), and that's 40MHz. At 40MHz, it has no current gain because it transitions to an hFE (or beta) of one.

Divide 40MHz by the hFE, which is conveniently 40, and we see it has current gain up to 1MHz. Great.

But then we connect its emitter to the base of our power transistor, and everything changes. We've joined two transistors in a Darlington configuration, and they're going to inter-react, but at least we might have some control over the outcome.

We now cast our eyes over the output transistor datasheet: https://www.onsemi.com/pub/Collateral/MJL21193-D.PDF.

Looking at Ft and minimum hFE, we see 4MHz and 25, which by simple maths tells us we can have full power up to 160kHz.

The load is 8 ohms, and the combined hFE's are 1,000, and so the load on the voltage amplifier stage is (8 x 1,000) 8 kilohms. At 160kHz and above, the load will be more substantial, at an expected rate of twice the load per doubling of frequency, so will be four kilohms at 320kHz. This continues until the linearity of the voltage amplifier becomes profoundly distorted.

However, most of the time, we will be listening to a few hundred milliwatts, and the voltage amplifier has an easy job. But it doesn't change the break frequency of the output stage, which remains at 160kHz.

At 160kHz, the load on the VAS must be reactive - what else can it be? The phase displacement at the turnover of a single-pole filter is 45 degrees. The VAS must be swinging its signal into a capacitor of some sort.

That capacitor, or real capacitance, exists because of the power transistor and is transferred back to the VAS through the beta of the driver transistor. Something isn't the way we expect.

If we could increment slowly at 160kHz, and read-off measurements as we go, we might be able to construct a curve of what is happening. Unfortunately, we cannot increment anything slowly at 160,000 times a second.

This is where a considerable knowledge of transistor physics would come in handy, but things become complicated for the average working engineer.

An intuitive description might help, but for the fact that in all the amplifier engineering books I have read, nothing of the sort exists.

However, between the mathematical symbols and calculations demonstrating the prowess - and possibly superiority - of one author to another (for who else could understand them?), there might be a slip where intuition can gain a grasp.

A clue is the "foldback" drawings, which seem to fall out from the mathematical jargon, and the foldback here can mimic the letter S.

This gives us the impression that as the input rises, the output rises, but then stops rising, but as the input is still rising, it appears that the output has reversed. It then "switches back" to rising again.

The non-linearity is caused by multiple factors combining. Considering there is an inter-reaction between betas and transition frequencies of the cascaded emitter followers, this should hardly be surprising.

The moment where the input rises but the output stops rising - and this can be a current rather than a voltage - and this can be out of phase relatively - might suggest that the impedance (the resistance at some frequency) has gone negative.

We have created an oscillator at some yet as unknown frequency, which reacts to a stimulus, which, although the signal does not contain such a high-frequency stimulus, its harmonics most likely do!

To prevent such harmonics (distortion), we feedback a portion of the problem after the event. All the negative feedback can do is worsen this situation.

Somehow, we must get in there and tackle the cause.

In search of negative resistance

By now, we know that negative resistance occurs at the base of the transistor. What transistor will that be? Intuitively we can say, the input transistor of the Darlington pair, or EF2, but why? Intuition says that, because it connects to the voltage amplifier load, it will affect the voltage amplifier load.

The second transistor can also exhibit negative resistance but inflicts it on the first transistor's emitter, but that doesn't contribute gain as such. The gain is below unity, and although the second transistor's negative resistance can translate it into gain, the first transistor's emitter is a better buffer than the VAS collector.

The effect of negative resistance from the base of a driver is obviously going to change the voltage amplifier's gain characteristics, and intuitively, it might be a more significant influence on the resulting amplitude. In other words, it might make it more noticeable.

Rather than having to work out the actual negative resistance of the driver, can we ignore it? The reason for asking is that we can't calculate the negative resistance, because, to some degree, it's load (loudspeaker load) dependent, which seems to be a good excuse.

Chessman and Sokel suggest the negative resistance lies somewhere between 0 and minus 500 ohms. The reason for it going negative is because the transistor has parasitic capacitances, which are unavoidable. It can be seen as a negative resistance in series with the base (on the way into the base). It isn't a full-time occurrence - it happens when the stimulus causes a match where the frequency causes the reactive components to result in a negative resistance.

It is doubtful that at high volume, the ephemeral duration will register with us. Still, I want to make this amplifier as listenable as a valve amplifier at the low volumes some of us are used to, in a domestic setting (and so did JLH).

A possible 1-volt aberration is more significant, in ratio to a sub-1 watt output.

By now, I'm hoping that this will register with the few readers I have left, following this thread. You must have noticed sometimes that bass can improve by turning up the volume?

An output stage, having to deal with ephemeral distortion when swinging a small signal, might emphasize the distortion. The distortion because of negative resistance is obviously at high frequencies because that is where the reactance comes into play.

We have established that, on occasion, an imaginary negative resistor inserts itself into the base lead of each driver transistor. This turns the dynamics upside down, where soft becomes loud, and loud becomes soft. It is such a transient event that an audio analyzer cannot capture it as distortion, but it is actually a distortion.

Now for the crazy bit. I cannot give it the eloquence that Marcus Graham Scroggie, "Cathode Ray," did in Wireless World, but I will try.

In our mind's eye, we see the imaginary negative resistor in series with the driver base, and we decide to place a real resistor (which is obviously positive because it obeys ohms law), in series with it. We make it such value which, when added to the negative resistance, the result is >0, then the negative resistor will cease to exist!

What we are left with is a real resistor, having a positive value. That value cannot change, because by making it larger than the negative resistor, the imaginary parts of the equation we decided not to examine, cancel.

So, where does all this leave us? It leaves us still guessing the actual value of resistance we need to choose, and what if we did? I will try to examine that in my next post.