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).
|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.
|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.
|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.
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.
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.
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. To appropriately load this tube with a resistor (see Figure 3) of 5X rp would requite a 5K resistor load. 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.
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.
|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.
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.
|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.
|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.
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.
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.
 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.
 Higher being better in terms of distortion. However, the tradeoff is lower power.
 This is pretty close to a 6c45pi or a triode strapped E810F.
 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.
 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.
 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.