[Amps] Tubes, transistors, and 'abuse

Manfred Mornhinweg manfred at ludens.cl
Tue Apr 25 12:53:01 EDT 2017


Steve,

> Can you explain what this design is trying to achieve with separate 
> transformers for the output (T1) and for supplying DC (T2) ?  Why not 
> the more traditional center-tapped transformer that acts as the output 
> transformer and the method of suppling DC ?

I can explain the basics of this matter, but why the designer used the 
separate bifiliar choke in this case is something only he can explain. 
It's a personal decision more than anything else.

The basic issue is this, explained step by step:

When a single transistor works in class A, fed by an RF choke, the choke 
will force an essentially fixed current (pure DC), while the transistor 
will conduct a DC plus a sine wave current. The only other thing 
connected to the drain and choke is the output circuit. It follows that 
the output circuit must conduct the difference current - and that's a 
pure sine wave.

Instead if this same arrangement operates in class B (or class AB, which 
for the purpose of this analysis works essentially the same way), the 
choke will force a DC, while the transistor will conduct roughly a half 
sine wave current, and stay completely off during the rest of the time. 
The output circuit still must conduct the difference current - and 
that's now a DC minus a half sine. There is always some sort of DC 
blocking, either in the form of a series capacitor or in the form of a 
transformer. So the final current flowing into the output circuit is 
this DC minus a half sine shape, but centered on the zero line in such a 
way that the current*time integral on each polarity is identical. It's a 
highly asymmetrical waveform, one side being almost a square shape, and 
the other side almost a half sine.

Needless to say, this waveform has an enormous harmonic content, and 
whatever load circuit is connected to the class B transistor must be 
able to handle this waveform.

When two transistors work in push-pull in class A, each of the two 
transistors works exactly like the single class A transistor of the 
first case. Only the two are in opposite phase. This allows simply 
connecting a load directly between the two drains, or connecting it via 
a transformer, and all will be fine. This class A push-pull amplifier 
doesn't need any special feed arrangement. It can be fed through two 
separate chokes, or through a transformer center tap, or through a 
bifiliar choke, and will work exactly the same.

Not so with a class B push-pull amplifier. If the two transistors are 
fed by independent chokes we get those two DC-minus-a-half-wave outputs. 
Connecting a load drain-to-drain does not work, because the two 
waveforms aren't complementary, as they are in the case of class A!! In 
class B, while one side is trying to push an approximate half wave sine 
shape current into the load, the other side is trying to pull 
essentially a constant current, courtesy of that side's choke. The two 
currents simply don't match.

The result of attempting to operate a class B stage in this wrong way is 
   that the difference current between the currents of each side has to 
go somewhere. But where can it go? At high frequencies it can go, at 
least partially, into the transistors' output capacitances, and into any 
additional capacitors installed between each drain and ground. But at 
the low frequency end of the amp's bandwidth, those capacitances are all 
negligible. This causes a huge common-mode voltage swing on the drains. 
During the signal's zero crossings, when both transistors are off, or 
conducting only a tiny current, but the two chokes are forcing the same 
current as all the rest of the time, both of the drain voltages soar to 
high peaks. These peaks can be high enough to destroy the transistors. 
If the transistors have good enough avalanche ratings, the spikes can be 
clamped by the transistors, causing huge additional power dissipation 
and a consequent reduction in efficiency, along with lots of IMD. So 
these spikes need to be controlled in some way, and the way some 
designers choose is to implement a strong negative feedback from each 
drain to its gate. By doing so, the transistors increase their 
conduction when the drain voltages rise, pushing them into a kind of 
dynamic class A operation. It tends to reduce the voltage spike problem 
to an acceptable level, at the cost of a maximum efficiency of roughly 
45% instead of the 65% that should be typical for class AB amps.

Now enter a balanced feed arrangement, or good drain-to-drain coupling, 
which is the same thing.

Let's take a push-pull amp in class B, without any feed chokes, but with 
an output transformer that does have a real, good center tap. The 
requirements for this transformer are that all three windings are 
perfectly coupled to each other: Both halves of the primary, and the 
secondary.

What happens now is that when transistor 1 is pulling its half-wave 
current, this current flows from the supply through side 1 of the 
primary, and induces a corresponding secondary current that goes to the 
load. It also induces a similar but opposed polarity voltage in the 
other side of the primary, which is applied to drain 2, but no current 
flows in side 2 of the primary, nor in drain 2. When the signal gets to 
the zero crossing, all current ceases. Both drains are at the supply 
voltage level. And as transistor 2 pulls its half sine current, the same 
thing repeats with opposed polarities. All is fine and works 
beautifully,  with good efficiency and low IMD. Negative feedback is not 
required in this circuit, but can of course be added to further improve 
linearity.

This circuit can also be analized in another way: The center-tapped 
primary tightly couples the two drains together. Any voltage excursion 
of one drain, away from the supply voltage, must necessarily be 
accompanied by an equal but opposite voltage excursion on the other 
drain. But the currents are independent: There is absolutely no problem 
with the drain current in one transistor being totally different from 
that in the other. The secondary, and thus the load, carries whatever 
current is needed to keep the transformer in balance.

So, when you want to build a push-pull class B (or AB) amplifier, the 
use of an output transformer having a center-tapped primary is a perfect 
solution - and this is the way push-pull amplifiers have always been 
presented in books, since roughly a century ago! The catch is that the 
coupling between the two sides of the primary needs to be excellent, 
very close to perfect. And in RF work this is hard to achieve.

Most transistor amplifiers operate at low voltage and high current, 
leading to drain load impedances in the order of only a few ohm, or even 
less than one ohm! This requires the total leakage inductance of the 
primary, plus all stray inductance in the 
drains-transformer-bypass-sources circuit, to have a negligible 
reactance compared to that very low drain load impedance. And this 
negligible reactance - say, 0.1 ohm or less - needs to be maintained not 
only up to the max operating frequency, but has to cover all of the 
stronger harmonics as well! That's because the individual drain currents 
are half sines, and these have a very strong harmonic content.

Now, an inductive reactance of 0.1 ohm at, say, 200MHz is an inductance 
of 0.08 nanohenries. And a piece of straight wire 1 inch long (see, I'm 
making concessions to inch lovers!) already has an inductance 200 times 
higher than that!!! Do you now realize how difficult it is to make a low 
impedance RF output transformer that satisfies the leakage inductance 
requirement? People try to do it by using tubing and flat, wide straps 
instead of wire, but it's all futile. It can't be done, period.

Add to this a fact that I have been stressing for years, on my website 
and elsewhere: The most typical output transformer for such amplifiers 
has a single-turn primary winding, looped through the two holes of a 
binocular core. Some designers, even the famous Helge Granberg, have 
postulated the midpoint of this one-turn primary as a valid center tap. 
It is not!!!!!  The two half turns have no coupling whatsoever between 
them.

Many amplifiers use transmission line transformers in the output 
circuit. These too lack a center tap.

The designers who have realized the fact that all practical output 
transformers for these low impedance amplifiers lack a usable center tap 
have come up with an attempt at a fix: The use of an external, 
additional, bifiliar choke. Basically this choke is intended to provide 
the center tap that the output transformer lacks. It can in principle do 
so by having an even number of turns. It's usually wound with two wires 
in a bifiliar fashion, so that automatically the total number of turns 
will be even, and there will be a real center tap which can be bypassed 
to the transistor sources, and where the supply voltage can be applied.

The theoretical behaviour is now the following: When transistor 1 
conducts its half sine wave, half of this current flows through side 1 
of the bifiliar choke and into the drain 1. The other half flows through 
side 2 of the choke and through the output transformer primary, then 
joining the first half of the current to flow into drain 1. Drain 2 is 
off, of course. The voltages on both drains are symmetrically 
complementary: The drop in drain 1's voltage goes along with a rise in 
drain 2's voltage. And during the next half cycle of course the same 
happens with opposite polarities. All is fine, and the behaviour is the 
exact same as the one we would get when feeding the amplifier through a 
real, well coupled center tap of the output transformer.

This circuit can be seen as the two transistors plus the bifiliar choke 
and its bypass caps combining to form a full sine wave source. The ouput 
transformer is no longer exposed to half sines of any sort, and thus 
doesn't need to handle any significant harmonics. This makes the design 
of the output transformer far easier and more forgiving.

Alas, life is never easy, and a bifiliar choke that achieves the 
required ultra low leakage inductance is just as hard to make as a 
center-tapped output transformer that does!!! We have just moved the 
problem to another place. We haven't fixed the problem. So the bifiliar 
choke is no real solution, and that's probably why so many modern radios 
don't have it, while the older ones did. The result is that either way 
solid state, low voltage, high current, push-pull linear amplifiers all 
suffer from very poor efficiency and high IMD.

I have been pounding this problem of proper push-pull feeding and 
coupling over the last few years, and I have to admit that while my 
understanding of it has improved a lot, I have found no solution.

If only the IMD is a concern, then a solution is to run class A. As 
explained near the beginning of this post, class A push-pull amplifiers 
don't depend on coupling between the drains. But the efficiency is very 
low, typically 35%.

It's possible to achieve better results by using higher drain load 
impedances, ideally in a range of perhaps 50 to 100 ohm. That requires a 
high supply voltage, relative to the output power. 12V should be used at 
most for a few watts, 50 volts at most for 100 watts, and so on. But as 
the voltage rises, so does the product of ferrite cross sectional area * 
turns number required to keep the flux density in the ferrite at an 
acceptable level. And this requires longer wires in the windings, 
reducing the maximum frequency that can be handled before running into 
phasing trouble. We would need better ferrites, that can handle higher 
flux density without excessive loss. Also there aren't many high voltage 
RF transistors, and their capacitances are such that at increasing 
voltages they also can only handle ever lower frequencies. This makes a 
kilowatt class 30MHz linear amplifier with good IMD and an efficiency 
close to the theoretical value a very challenging proposition.

The little amplifier that motivated this post, having a two-turn 
primary, should in fact be able to work with the power applied to the 
transformer primary's center tap, and given that transformer's 
construction I would expect better performance than with the bifiliar 
choke the designer used. Also that little amplifier has a reasonably 
high drain load impedance, 11 ohm per side. That should allow pretty 
good performace on the lower bands, and maybe 50% efficiency on 10m, 
when feeding it through a center tap and optimizing the board layout for 
low stray inductance in the output circuit.

Manfred

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