Vacuum Tubes Part
As things stand, we've got a common cathode amplification stage with a cathode follower output to lower the output impedance. In a not-really-practical sort of way, the first stage could — in theory — do the trick all by its lonesome, but the high output impedance could cause problems when faced with capacitance or inductance from interconnect cables. The cathode follower alleviates those concerns by lowering the output impedance. The circuit, however, inverts the signal and — though there are simple external fixes, such as flipping the speaker cables — it is rather inelegant to have a circuit that requires that you fiddle with the rest of your system in order to accommodate its eccentricities.
It would be simple enough to add another common cathode gain stage. That would re-invert the signal, restoring the original polarity and all would be right with the world. Indeed, that's what many line stages do. The problem is that you then need to get rid of the excess gain. To those who feel that negative feedback is a gift from the audio gods, the answer is obvious:
negative feedback to lower the gain. My point of view is that nothing is truly
free and I'd rather not pay the subtle price exacted by feedback unless it is
absolutely necessary. A more elegant solution is to employ a circuit known as a
A differential consists of two identical gain devices — in
this case 6SN7 tubes — run back-to-back. Compared to the common cathode gain
stage from last time, the tubes are the same, the load resistors are the same,
the input network is the same...in fact, everything is the same except the
biasing resistor. Where before there was a 1300 Ohm resistor between the cathode
and ground, we now find the two cathodes tied together and a 620 Ohm resistor to
ground. The fact that the two cathodes are coupled is what makes the circuit a
There are a couple of things to point out before moving on.
The first is that R4, the biasing resistor for the differential, has had its
value lowered to account for the fact that it is now supplying current for two
tubes instead of one. Assuming that we want to keep the bias voltage the same
(on the order of 4V), then twice the current through the bias resistor means
that we need to halve the value. Simple Ohm's Law stuff. If you want to get
picky, 1300 Ohm / 2 = 650 Ohm, and while there's a 649 Ohm value in the 1%
resistor table, you can't count on finding some of the more arcane resistor
values when you want them. It's a pretty good bet that you can
find some of the more standard values, like 620 Ohm (619 Ohm in 1% lingo — and
if you want to lie awake at night fretting over the 1 Ohm difference be my
guest, but I have other fish to fry). Feel free to experiment with slight
variations in the values. The 620 Ohm will increase the bias current and
reduce the bias voltage by a smidgen, but you need not worry about exceeding the
limitations of the tubes. They're still well within spec.
The other thing to look at is that connection between the two
tubes' cathodes. That one connection is everything; the heart of the
differential. It also provides an opportunity to talk about the other input of a
tube — the cathode.
Anyone who has ever spent time looking into how tubes work
knows that the signal goes in the grid, which then uses an electrostatic field
to modulate the flow of electrons from the cathode to the plate. But what if the
grid was held constant and the electron flow was modulated directly? Believe it
or not, this works, and works well. It opens a whole range of other possible
topologies, but the one we need at the moment is the differential.
We'll need to keep an eye on those cathodes, though. The cathode, when used as an input, has a low impedance. In fact, the cathode has a low impedance no matter what you're doing with it. That's why the cathode works so well as an output node in the cathode follower configuration. At the moment, we really only care that the impedance at the "new" tube's cathode matches the impedance at the "old" tube's cathode — which, of course, they do, since we're using the same node on the same type tubes. Life would get complicated rather quickly if we tried to use, say, a 12AX7 as one side of a differential and a 6DJ8 as the other side. The circuit wouldn't balance and there'd be no benefit to using a differential, although you might be able to pull off some other tricks if you worked at it hard enough. The entire point of our differential is that it's the same on both sides. By the way, this is one time when tube people have a clear, unarguable chance to be smug about their chosen gain devices. While it's not impossible to find matched solid state devices (e.g. 2SK389), they've always been comparatively rare and a lot of the best candidates are out of production nowadays. Bummer. But all dual triodes are matched and nearly all of the most popular small signal tubes for audio use are dual triodes. You get your choice of characteristics, whereas the solid state contingent has a very, very limited palette to draw from, poor things...
Okay, enough bragging. (Not!)
When a signal enters the grid of the first tube of a
differential it modulates the flow of current in that tube in the normal way. In
consequence, the current at the cathode of that tube ebbs and flows, depending
on the needs of the moment. That constitutes a signal, just as though we were
using the first tube as a cathode follower. That signal then enters the second
tube via its cathode and is voltage-amplified, passing out through the plate
just as you'd expect in a common cathode amplifier. Incidentally, the second
tube, with the signal entering through the cathode and the grid held constant,
is operating in what is known as common grid mode. Yes, you can drive the second
tube's grid, but we'll get to balanced signal inputs later.
Now, watch closely. I promised that this circuit would cure
the out-of-phase problem that results from using a single common cathode
amplifier stage. Here's how it works:
1) The signal enters the grid of the first
tube. For the sake of argument, let's say that it's swinging positive, which
we'll designate with a +1.
2) That signal then presents itself at the
plate (inverted, hence -1) and at the cathode (+1). We're done with the signal
at the plate for the moment, but we'll keep following the signal at the cathode.
3) The signal from the first tube's cathode
then makes its way to the second tube's cathode. The interesting thing about a
signal that enters a tube through its cathode is that it appears at the plate
with the same phase — in this case +1, thus giving us the option of an inverted
output signal at the plate of V1 or a non-inverted output signal at the plate of
V2...and the neat thing is that both of these signals are available
simultaneously. You don't have to choose just one.
So, looking at the plates for the differential, we have both phases available.
At the first tube's plate, we have a -1 and at the second tube's plate we have a
+1. Cool. So we're ready to go, right?
The signal at the second tube's plate is weak. Really weak.
What went wrong?
Strictly speaking, there's nothing wrong. The circuit is functioning precisely
as designed. To understand why, let's take a closer look at the bias resistor
for the differential.
As I've mentioned, the cathode is a low impedance node and, as
it happens, that 620Ω bias resistor is pretty low also. Suppose you are an
electron that's just left the first tube's cathode and you're presented with the
option of going either of two ways, both having similar, fairly low impedances.
If you think of the impedance as the "cost" of choosing one path or the
other, then you'd be just as likely to take one fork in the road as the other.
The result being that although some of the signal goes into the second tube's
cathode — which is what we want to happen — some goes down the 620 Ohm bias
resistor to ground and is lost.
Rats. And just when things were starting to get interesting.
Don't despair. All is not lost. We can raise the toll on the
bias resistor path. The classic way is to use a much larger resistor. Something
like 10 or 20k would do a pretty good job, but if you do a little math you'll
soon see that we've run into trouble. If we assume that we want to keep the same
bias current (and don't forget that we've got two tubes to bias), then Ohm's Law
tells us that a 10k resistor will develop a 60 or 70V drop instead of the measly
4V we had before. And 20k? Twice as much...roughly half of our rail voltage will
disappear into the bias resistor before we even get started. Whew!
That's going to make a mess of things.
Let's put on our thinking caps and see if we can come up with
a clever way to resolve the problem. One way is to add a negative rail of
sufficient voltage and attach the bias resistor to that negative rail instead of
ground. That's like standing a tall pole in a deep hole, so that only the very
tip shows above ground. As a matter of fact, if you look at enough older tube
schematics, you'll find that that's exactly what some people did and it works
like a charm. If you're a tube purist and you want to follow that route, then do
so with my blessing.
Another possibility is to use capacitors to decouple the input grids from the
outside world, add resistive dividers to pin the grids to +60V or so, then
increase the rail voltage to compensate. Hmmm... well... okay, with enough extra
parts and fiddling this can be made to work, but it's unnecessarily complicated.
But what if we could squeeze a really, really high impedance bias circuit into
that little 4V space that we've already got? Turns out it can be done. Although
it rankles those who believe that tubes and solid state devices should be kept
separate, a solid state current source will do the job, and depending on the
particulars, it can sport an impedance all the way up into the MOhm range...
which, in practical terms for a circuit like this, is as close to
infinite impedance as you could ask for. Very effective for balancing the
outputs from the two sides of our differential.
There are more current source options than you can shake a stick at and people tend to develop preferences for one circuit or another. We'll look at a few simple, yet effective possibilities next time.