Part 1: How a tube works
A tube has 3 electrodes (hence triode, pentodes are for
another day) – the anode, also called the plate, the cathode, and the grid,
also called the control grid. When there is a positive voltage applied to the
anode, current flows from the cathode to the anode (Why? Click here http://www.tpub.com/content/neets/14178/css/14178_13.htm).
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| Figure 1: A Tube’s electrodes |
It is conventional to think of current flowing from anode to
cathode, but it actually goes the other way. That is, electrons move from
cathode to anode. Without the grid
present, current flow is basically unrestricted (i.e., the tube would be a
diode). However, by applying a negative voltage to the grid, the amount of
current flowing across the tube can be controlled.
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| Figure 2: Current flow |
As the grid’s voltage is made more negative, less current
flows. If it is less negative, more current flows. At some point, if the grid
is negative enough, then no current will flow.
This is useful because as an alternating signal voltage (such as from
music) is applied to the grid, the current flow across the tube will increase
and decrease with the signal.
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| Figure 3: Input and output signals |
Finally, if a load of some sort, such as a resistor, is
placed on the plate of the tube, it acts as an I to V (current to voltage)
converter. In essence, since V=IR, and since the resistor’s resistance is
constant, I and V are proportional. As
current (I) increases, so must voltage dropped across the resistor. As more current flows across the load, the
voltage generated is higher. As less
flows, the voltage is lower. Thus, an alternating signal on the grid turns into
an alternating signal on the plate. Moreover, since the current flow across the
tube is relatively large, compared to the input signal, and the swings in
current due to the changes in grid voltage are also large, there is an amplification
of the signal. Finally, note that since
decreasing the grid’s voltage results in decreasing current flow, and thus a
higher voltage on the output (less voltage drop across the resistor), the
signal is inverted.
Part 2: More on how a tube works
All triodes basically work as above. However, they all do so with different
parameters. There are three basic numbers that matter – mu, which is
amplification factor, gm, which is transconductance, and rp,
which is plate resistance. mu is basically the voltage gain of a tube, gm is
the current gain, and rp is the output resistance to these things. Thus, just as V=IR, mu=gm*rp. That is, there are tradeoffs.
For instance, if the amplification (mu) of the tube
is increased, the output impedance will be as well. This is why a 12AX7 with a mu of 100 has a
much higher output impedance (rp) than a 12AU7 with a mu of
20. Additionally, as gm is
increased, the output impedance is lowered.
Thus, a 12AU7 with a mu of 20 has a higher output impedance than
a high gm tube like the 6DJ8, even though the 6DJ8 has a higher mu.[1]
Finally, tube sections can be paralleled. Using two of a given triode doubles the gm
but leaves mu the same. That is,
voltage amplification is the same, but the amount of current gain is doubled. Thus, because V=IR, this reduces the plate
resistance by half.
Part 3: Putting the tube in a circuit
The load on a tube acts as the current to voltage
converter. Like with other circuits, the
higher this load is, the easier it is for the tube to drive. Additionally, the higher the load, the lower
the tube’s distortion will be. As a
general rule of thumb, a load should be at least 3x higher than the tube’s rp,
with higher being better.[2]
There are a number of ways to actually “load” a tube. Among these, in addition to a resistor, are a
transformer, a choke, and a constant current source (CCS). It is probably best to cover the ins and outs
of these various options via an example.
Resistor Loads
Imagine a hypothetical tube with a mu of 50 and an rp
of 1K when running with 150V on the plate and 20mA of current.[3] To appropriately load this tube with a
resistor (see Figure 3) of 5X rp would requite a 5K resistor load.[4] Notice that a 5K resistor, with 20mA across
it will drop 100V. Thus, the power
supply requirements for an appropriate resistor load of this tube is at least
250V. Moreover, increasing the load
increases linearity considerable.
However, to raise the load to 8K requires dropping 160V across the
resistor. This increases the necessary
power supply to 320V, and requires a resistor that can dissipate more than 3W
of heat.
Further complicating this is that the next stage that the
tubes “sees” is actually in parallel with the load. Thus, for instance, a tube loaded with an 8K
resistor driving a stage with a 50K input impedance, actually sees a load of
6.9K. In order to boost the actual load back to 8K requires a 9.5K loading
resistor, which increases the required power supply by another 30V to 350V and
increases the resistor load’s heat dissipation to closer to 4W.
Choke Loads
An alternative to the resistor load is a choke load. A choke is basically a transformer with a
single winding. Chokes have fairly low
DC resistance, but present a high impedance at AC. However, the impedance of a choke is
dependant upon frequency, and as such an appropriate choke must be selected to
work at the frequencies that one cares about.
The effective impedance of a choke is 2·Π·F·L where F is the frequency
and L is the choke’s inductance.
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| Figure 4: Choke loaded tube |
Back to our tube example, if we want a 9.5K load (to give us
an actual 8K load when working into a stage with a 50K input impedance), and if
we want good performance down to 20Hz, we will need a choke with about 75H of
inductance at 20mA. One can purchase
such a thing, for instance, from Electra-Print audio, for about $100 for a
pair. This is not necessarily
prohibitive, but it is certainly more than a resistor.
There are two additional benefits of using a choke. The
first is that the DC resistance of a choke is relatively low, about 1500 ohms
for the above choke. Thus, at 20mA the
choke only drops about 30V. This relaxes
the power supply requirements considerably as to have 150V on the tube’s plate,
the power supply in our example only needs to be 180V, which is much easier
than 350V.
At first blush this might, or at least should, seem
odd. Half of the point of the resistor
load, after all, is that as current oscillates between 0mA and 40mA (centered
around the operating point of 20mA) the 8K resistor load swings the output,
theoretically, from 0V to 320V. If the
power supply is reduced to 180V with the choke load, how is this possible? It turns out that chokes (and transformers)
are energy storage devices, and that a choke can “swing” above the power supply
by releasing this stored up energy.
Thus, even with the lower power supply, the choke can still swing just
as much as the resistor load with the higher power supply.[5]
The second benefit to using a choke load is that chokes can
present a very high impedance at higher frequencies. The choke in the example needed to be 75H in
order that the load at 20Hz would be 8K.
However, since the impedance presented by a choke is frequency
dependant, at 1KHz this same choke presents a 470,000K load. At 2KHz it is close to 1M. Thus, for all practical purposes, above a
certain frequency the choke can be considered an infinite load with the next
stage providing the actual load.
Finally, there are also disadvantages to using chokes, aside
from the expense. The first is that,
while better than with a resistor, PSRR is not great. The 75H choke from the example only presents
an impedance of 28K at 60Hz, or 56K at 120Hz, which can let power supply ripple
through. Additionally, chokes can couple
magnetically with power supply transformers and chokes which can actually
inject noise into a circuit unless one is very careful with layout. Third, chokes interact with capacitors and
parafeed transformers in sometime unpredictable ways leading to resonances in
the circuit. Complicated models can help
with minimizing the effect these have, but in the end a circuit must be built and
tweaked with a scope to get the circuit to operate properly. Finally, because the load at low frequencies
is much lower than the load at high frequencies, distortion at lower
frequencies is higher. Fortuitously, low
frequency distortion is less audible, so this is not necessarily a huge issue.
Constant Current Source (CCS) Loads
The final loading option covered is the so-called Constant
Current Source, or CCS, load (transformer loads will have to wait for another
day). A CCS is, as the name implies, a
circuit that provides a constant current.
They can be as simple as a single transistor, or extremely
complicated. While providing almost no
DC resistance, a CCS provides an extremely high AC impedance. Indeed, designs exist with AC impedance
measured in the giga-ohms range (1,000,000,000 ohms) or higher. Moreover, this AC impedance is more or less
constant over the entire audio range.
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| Figure 5: CCS loaded tube |
To return to the example, even the simplest CCS will provide
an AC impedance in the mega-ohm range from DC to well over 20KHz. As such, the following stage effectively
provides the entire load. The tube
working in to a 50K load, effectively sees a 50K load.
Another benefit of a CCS is that it provides extremely high
PSRR. This is an artifact of the fact
that a CCS is a constant current device – ripple is effectively nullified by
the constant nature. Indeed, power
supply ripple reduction of 120dB is routine, and 150dB is possible.
The part about a CCS that seems to be the most confusing is
the relationship to the power supply. In
short, the CCS will drop whatever volts are necessary to keep the tube at it’s
operating point. Return again to the
example where the tube’s plate sits at 150V when passing 20mA. With a 20mA CCS, the tube will operate at
that point regardless of the power supply’s voltage. If the power supply is 160V, the CCS will
drop 10V across it in order to keep the tube’s plate at 150V. Similarly, if the power supply is 300V, the CCS
will drop 150V.
The high impedance, the lack of importance of the absolute
correctness of the power supply voltage, and the generally low cost are all
advantages of a CCS over a choke. The
disadvantage is that unlike a choke, a CCS cannot swing higher than the power
supply. Thus, as in the case of a
resistor, the power supply must be high enough to accommodate the amplifier’s
swing. In order for the example tube to
swing 200V peak to peak, the CCS needs 100V of headroom. This makes CCS loading less appropriate on a
power output stage, but near ideal for a lower power driver stage.
Part 4: On to Parafeed
The basic circuit topology discussed above is known as a
grounded cathode circuit. In order to make it a parafeed circuit, one need only
AC (capacitor) couple it to an output transformer. The transformer works to transform the high
impedance, high voltage, and low current signal at the tube’s plate to a low
impedance, low voltage, high current signal at the output.
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| Figure 6: Parafeed Circuit |
Because tubes have a relatively high output impedance, they
work well into relatively high loads.
Using a transformer allows one to connect a low impedance pair of
headphones and reflect this as a high load to the tube. Transformers are designated as having an
impedance ratio. For instance, an 8K to
32 ohm transformer would allow a 32 ohm load connected across the secondaries
to appear to the tube as an 8K load.
By stepping down the impedance, the transformer also reduces
voltage swing, but increases current.
The voltage step down is the square root of the impedance step
down. If a transformer has an impedance
ratio of 4:1, then it has a voltage step down ratio of 2:1. In the case of the 8K:32 transformer, this
represents an impedance ratio of 250:1, which is a voltage ratio of
approximately 16:1.
Returning to the example, since the hypothetical tube has an
output impedance of 1K ohms, using an 8K:32 parafeed transformer converts this
into an output impedance of approximately 4 ohms. In order to swing 1V into the headphones, a
16V swing is necessary on the plate.
Recalling that the mu (or voltage gain) of the tube is 50, this
means that an input at the grid of 0.32V is necessary. In other words, the amplifier has a voltage
gain of around 3 in to 32 ohm headphones.[6]
It is common to see a headphone amp specification that is
different for 300 ohm and 32 ohm transformers.
The reason for this is that the amplifier can swing a certain voltage on
the output, and generally can supply enough current to support the voltage. For instance, an amplifier than can swing 4Vrms
puts about 500mW into 32 ohm headphones, but only 50mW into 300 ohm
phones. A transformer coupled amp with
taps for the particular phones used, however, does not behave this way.
Returning to the example, assume that we want to power 300
ohm headphones. Instead of running them
off the 32 ohm tap, we’ll assume that the transformer also has a 300 ohm
tap. 8K:300 is an impedance ratio of
about 27. This means that the output
impedance into 300 ohm loads is about 37 ohms.
Additionally, the voltage step down ratio is around 5:1. Thus, the same 16 ohm plate signal will now
be just over 3V, and the overall voltage gain will be around 10.
Further, note that the wattage output is the same for both
sets of headphones. Since watts are
equal to V2/R, 3 volts into 300 ohms and 1 volt in to 32 ohms are
both just over 30mW of power. This can
also be seen at the plate, where 16V in to 8000 ohms is the same. That is, a transformer conserves wattage.
Conclusion
The topic of this article is specifically to do with a
single tube “spud” parafeed circuit for headphones. That is, it deals with circuits that need to
provide on the order of a quarter of a watt of output power (about 45VRMS
into 8K). As opposed to a 2 to 8 watt
speaker amp, this changes the calculation of what circuit is appropriate. For instance, in order to generate 5W,
assuming the same 8K ohm output transformer primary, the circuit needs to swing
close to 300V peak to peak. A tube that
can do this, like the 300B, would need around 425V across the tube itself (including
grid bias) at over 60mA. To use a CCS
would add another 175V or more to the necessary power supply, making the power
supply circuit impractical and, even in SET terms, wildly inefficient. Thus, a choke would be the appropriate load.
However, for a headphone amplifier than can produce on the
order 250mW of output, a CCS is an excellent choice. It is still wildly inefficient, but it is at
least bordering on reasonable. Moreover,
the performance increase of the CCS over the choke in terms of the higher, and
more consistent, load presented to the tube and the excellent power supply
isolation, make it a good choice.
[1] Numbers
given for mu, gm, and rp are always as a particular
operating point as they change depending upon the operating point of the tube. As current across a tube is increased, mu
and gm both tend to increase, while rp tends to decrease. The consistency, or lack thereof, for these
parameters, changes the distortion spectrum at various operating points and is,
in part, what causes different operating points to “sound” different.
[2] Higher
being better in terms of distortion.
However, the tradeoff is lower power.
[3] This is
pretty close to a 6c45pi or a triode strapped E810F.
[4] A tubes
inherent plate resistance is often abbreviated rp (sometimes ra
instead). The resistor on the plate is
often abbreviated Rp (or Ra), the difference being the
capitalization of the R.
Moreover, this convention is often not adhered to, and periodically one
must figure out whether rp refers to the load or the plate resistance by
context.
[5] Gapped
transformers in single feed single ended triode amps work the same way, and
this efficiency gain is a big part of the reason they make sense as output
devices for speaker amps.
[6] As you
might imagine, it is actually considerably more complicated than this
example. Transformers, tubes, and
circuits all have various inefficiencies that make the real output impedance
higher, and make actual signal swings and power lower.
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