06 November 2016

More of the Cathode-Coupled Amplifier
Will the cathode-coupled-amplifier well ever run dry? I doubt it. Here is a simple example. The following cathode-coupled amplifier solves two problems: poor PSRR and the dissimilar cathode-to-grid voltages.

The left triode idles at 5mA, while the right triode idles at 10mA. The left triode sees a cathode-to-plate voltage of 241V, while the right triode sees a cathode-to-plate voltage of 136V. Both cathodes are tied together and sit at +9V. Thus, the left triode sees a cathode-to-grid voltage of -9V, while the right triode sees a cathode-to-grid voltage of -2V. The constant-current source draws 15mA.

What happens if the B+ voltage falls to 200Vdc? The left triode will then idle at about 6mA, while the right triode will idle at about 9mA. Interestingly, the distortion goes down, as the two triodes get closer to the same idle current flow.

Remember that the cathode-coupled-amplifier circuit offers two features: low input capacitance and lower distortion. The lower input capacitance results from the input triode's plate being fixed, so no Miller-effect capacitance results. The lower distortion results from the two triodes running in current anti-phase, so the output triode tends to unbend what the input triode bent.

A triode more easily turns on than to turn off; thus, the cathode follower's positive output swing is easier to make than its negative swing. With the grounded-grid amplifier, the opposite is true: the negative output swings are easier to produce than the positive swings. When the two circuits cascade, to a large extent the two opposite bends undo each other.

Returning to the circuit, what happens if the B+ voltage rises to 300Vdc? The left triode will see its idle current fall to about 3.5mA, while the right triode's idle will rise to about 11.5mA. Because the disparity between triodes has increased, so has the distortion. One workaround would be the following variation, which uses a PNP transistor to auto bias the grounded-grid triode so that it continues to conduct 10mA, in spite of changes in B+ voltage.

The 75k and the 10.1k resistors define a two-resistor voltage divider, which allows a portion of B+ ripple to appear at the grounded-grid triode's grid. Is that a good thing? Indeed, yes, as this creates a vastly improved PSRR. The way it works is easy enough to understand. As the power-supply noise pulls up in voltage, the grounded-grid triode increases it current conduction, so its output remains fixed at 149.7V; as the ripple drops in voltage, the triode eases up it current flow, so the plate voltage remains at 149.7V, which means that we have achieved a deep power-supply-noise null at the output.

The MJE5852G, which is a 400V PNP transistor in a TO-220 package, whose base-to-emitter voltage monitors the current draw through the 54.9-ohm resistor. If triode draws too much current, the transistor conducts more, which increases the output triode's grid voltage, which in turn cause its conduction to increase, thereby forcing a decrease in the input triode's current flow, as both cathodes share the single current path to ground through the constant-current source. Speaking of which, the 15mA flow exceeds the ubiquitous LM334's current limit. The workaround is to use either two LM334s in parallel or to use another device. One possibility is the famous LM317.

The LM317 makes a surprisingly good constant-current source, as Walt Jung pointed out in audioXpress.

Moreover they are neither large nor expensive, as you can buy two LM317L regulators in the TO-92 package for less than $1.

Ultra-Low Input Capacitance Cathode-Coupled Amplifier
Before moving on to the ultra-low-capacitance version of the cathode-coupled amplifier, I will detail how I came up with this variation. If you inspect the schematic below, you will see an extra capacitor, marked C. Why is it there?

Ideally, we want to see no power-supply noise on the plate of the input triode, which is why the large-valued electrolytic capacitor is there. But no matter how large we make this capacitor, some ripple must leak through. There are times when brute force isn't enough. Then it hit me: if we add an extra capacitor that terminated in the negative power-supply rail and whose impedance at 120Hz was the same as the plate resistor (ra) value, the two equal impedances would result in a tie, which would null the power-supply noise on the plate of the input triode. We still need the large-valued electrolytic capacitor, however, to shunt away the audio signal currents.

How do we find the right value for this added capacitor? The formula is easy enough:

       C = 159155 / 120Hz / Ra

where C is in µF. Why 120Hz? In countries with 60Hz wall-voltage frequency, the ripple will be greatest at 120Hz. (In countries that use 50Hz, we would use 100Hz.) For example, if Ra equals 15k, then a 0.0884µF capacitor is needed, with 0.091µF being close enough. This trick, of course, will only work with bipolar power supplies. In addition, the positive and negative power-supply rails must hold equal but out of phase ripple. What if no negative power-supply is available, as it isn't in the circuit below? Then we use the Aikido cathode follower to scrub away the power-supply noise.

Well, as I pondered this situation, I realized that I could lose the large-valued electrolytic capacitor by adding one resistor.

By selecting the right value for the added resistor, we archive an audio-signal null at the input triode's plate, as the output signal is out of phase with the signal the input triode would like to develop at its plate. The more I thought about, I came to the conclusion that I would like to see some audio signal at this plate; in fact, I would like to see the same amount that appears at the input triode's grid and in the same phase. Why? Since the two tied cathodes will partially follow the signal present on the grid, so why not have the plate do the same, which will result in lower input capacitance.

The Aikido cathode follower can easily drive the 13k resistor and it provides a fine PSRR and low output impedance. In SPICE simulations, the results were exemplary. Of course, many different tubes could be used, with suitable changes in part values. One combination I would like to try would be a 6SU7 (a 6SL7 with matched triodes) as the cathode-coupled tube and a 6SN7 as the Aikido cathode follower tube. One problem with high-mu triodes, such as the 12AT7, 12AX7, 12AY7, 5751, 6072, is that the cathode voltage will prove too low to use most solid-state constant current sources, such as the LM334. The workaround would be to use a low-voltage negative power-supply rail, say 5Vdc or 6.3Vdc or 12Vdc, which would give the LM334 enough voltage headroom and could be used to power the cathode-coupled triode's heater element. If you are a bit more ambitious, the following high-voltage shunt regulator could be used to make use of the clean -12V provided by a linear regulator.

Reading from left to right, we see the raw DC high voltage enter the circuit. The TIP50 NPN transistor then measures the voltage at its base. If this voltage is too high, the transistor will increase its conduction until the voltage drops to the transistor's nominal base-to-emitter voltage, about 0.6 volts. The three diodes are essential, as they protect the TIP50 from seeing excessive base voltage.

Balanced Cathode-Coupled Amplifier
The feature of balanced audio is common-mode rejection ratio (CMRR). The higher, the better, as we only want to pass differences and ignore what is shared between the two input signal phases. Extraneous noise tends to be equal and in phase, which means that a high CMRR will largely ignore it, passing the desired audio signal. Having just said that, can you see what is wrong with the following balanced cathode-coupled amplifier circuit?

Everything seems to look good, as the shared 15k plate resistor will see a constant current draw, as the two audio-signal currents flowing through are out of phase with each other. What is wrong is that the circuit offers no CMRR, as an in phase signal presented at both inputs will result in equally large output signal, which will be in phase with each other. Of course, the next balanced piece of equipment, such as a push-pull amplifier may offer a fine CMRR and clean up the mess that it received. But think how much better things would be if it had less of a mess to contend with. The workaround is to add a third constant-current source.

Now things get interesting. Imagine that we present both the inverting and non-inverting inputs with +1Vdc. What will happen? The output triode plates will not change in DC voltage and the input triodes will not alter their idle current! How is that possible? The coupled plates have only one path to the B+ voltage, which is through the added constant-current source. If both input see +1Vdc, the input triode plates will fall by mu times 1V, or 20V, so the plates will sit at 130Vdc, rather than 150Vdc. Conversely, if both input see -1Vdc, the input triode plates will rise by mu times 1V, or 20V, so the plates will sit at 170Vdc. Since the output triode plates will not change in DC voltage, we have achieved an excellent CMRR.

By the way, this technique will not work with pentodes or FETs or transistors—only with triodes. Why? Plate resistance and, by extension, amplification factor. Triodes exhibit a relatively low plate resistance, which means that a change in plate voltage results in a surprisingly large change in current conduction. This is not true of FETs, as once their drains move beyond the FET's triode region, the FET is indifferent to changes in drain voltage, which is why they make such fine constant-current sources.

The next problem we must overcome is poor PSRR. My solution is to use two Aikido cathode followers.

Click on schematic to see enlargement

The added constant-current source must withstand high voltages, which means that an LM334 or LM317 won't cut it. Here is one possible circuit.

The MJE5852G PNP transistor can withstand a 400V collector-to-emitter voltage. Here is the complete circuit.

Click on schematic to see enlargement

Note that each side gets a negative feedback loop. The gain is set to about +10dB, not much in other words. Why? Most balanced signal sources run hot from my experience, so little gain is needed.

Four tubes are needed per channel. Yes, this is a lot of tubes, but it is also a lot of performance. In my last post, I mentioned that I needed to start using three icons, so that I could label schematics as being either textbook examples or modifications of the standard circuit or an altogether new circuit. Well, I could add a few more icons, for example, one for performance and another for expense. The above circuit, wrapped in luxurious metal work, powered by a regulated power supply, would garner rave reviews from the audio press and sport a $5,000 price tag.

For many newbies, all tube circuits are equal. They aren't. My issue with sonic ratings is that I hate all the hype in high-end audio and I do not want to throw any tube circuits away—even the SRPP. Thus, I rather not proclaim too much enthusiasm or too much scorn. Moreover, I know that many believe that they do not need to waste their time with anything less than the best. Really? In many applications, silver if often better than gold. In spite of the popular notion that you are only as exalted as is your consumption, the truth is that you must understand why the less-than-the-best is less, before you can understand why the best is the best. As Plato wisely pointed out:

for a doctor to heal, he must first know what kills.

In other words, I won't be using the above icons. Well, at least not the sonic bells.

Another approach to building a balanced cathode-coupled amplifier is the following.

The assumption here is that the signal source, say an external DAC, offers an XLR balanced output, which means that the signal source was designed to drive a 600 ohm load. The 1:1 signal transformer is a 600-ohm : 600-ohm type. The word "semi" appears in the circuit title because this circuit only half counts as a cathode-coupled amplifier. The two cathodes are sort of tied together as they share the same cathode resistor. On the other hand, the input impedance is very low, whereas a true cathode-coupled amplifier offers a hugely high input impedance. Resistor Rk's value wasn't specified, as it would depend on the DCR of the transformer's secondary winding.

The tubes are 6DJ8 types, but other high-gm tubes, such as the 5687 and ECC99, could be used. The input impedance presented at each cathode is equal to

    Zin = (Ra + rp) / (mu + 1)

where Ra is the plate resistor; rp, the triode's plate resistance; and mu, the triode's amplification factor.

Aikido mojo is in play with the two-resistor voltage divider that terminates into the B+ connection. All in all, an interesting circuit.

//JRB

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Next Time
A good question. I believe it's time to return to single-ended power amplifiers.

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//JRB