John Broskie's Guide to Tube Circuit Analysis & Design |
30 Nov 2010
Cathode-Coupled Amplifier (CCA) 2005/April/blog 42 The last entry is actually a good place to start, as it shows several ways to overcome the problem of dissimilar cathode-to-plate voltages in a CCA. This problem is the CCA's one big hassle to overcome. In fact, if the using CCA circuit didn't entail dealing with dissimilar cathode-to-plate voltages, I am sure that this topology would be much more popular, as it has much to offer, such as low input capacitance and no phase inversion and a high-impedance feedback port.
In the schematic above, on the left, we see two triodes that share a common cathode resistor and separate B+ voltages, but the same DC grid voltage (0V). If we wish to use a single B+ voltage, then we must alter the cathode-to-grid voltages to compensate for the dissimilar cathode-to-plate voltages, as shown in the schematic on the right. Actually, the unspoken assumption here is that the two triodes share the same current draw. In fact, this is a precondition worth retaining, as the identical idle current helps linearize the two tube's gain. If we are, however, willing to run different idle current draws, then we can use the same grid-bias voltage for both triodes. An inspection of a triode's plate curves shows how this can be done. Note the -2V grid line and where it intercepts the 100V and 150V plate voltages; with 100V on the plate of this triode, the conduction equals 3mA; with 150V, 7.6mA. Yet, both plate voltages rely on the same -2V of grid voltage. Applying these two operational points in one circuit is straightforward enough: just add the 3mA and 7.6mA together to find the correct common-cathode resistor value and use just the 3mA idle current amount for determining the plate resistor value. The downside to this technique is an increased distortion figure. Furthermore, the right triode will be expected to drive the external load resistance and capacitance, but its idle current is less than half the left triode's current conduction, unfortunately. (On the other hand, if the right triode is only expected to drive a cathode follower's grid, then the weaker current is much less of an issue.) If we wish to keep the equal idle currents and the single B+ voltage, then we will have to give the triodes differing grid-bias voltages. Once again, an inspection of the plate curves shows how this is done. This time we hold the plate current steady (for both triodes), which the single green horizontal line indicates. At 5.2mA, the green intersects the two plate voltages used in following cathode-coupled amplifier design example; namely, 150V and 300V. The grid voltage intercepts occur at -3V and -9.5V, as marked by the red grid-voltage lines. In other words, the left triode will see a 300V differential and will need to see a grid voltage of -9.5V to draw an idle current of 5.2mA, while the right triode will experience only a 150V differential and will need to see a grid voltage of only -3V to draw an idle current of 5.2mA. Okay, I know many are scratching their heads right now, as the right triode is seeing a +6.5V grid bias voltage, not -3V, as the graph's inspection implied. Remember, the cathode-coupled amplifier's triode share a common cathode connection, so the common cathode voltage becomes our reference. Relative to the +9.5V cathode voltage, the left triode's grid is at -9.5V and the right triode's grid is at -3V, as +6.5V minus +9.5V equals -3V. Now that everyone is on the same page, the next question is: What happened to the negative power supply rail? A negative power supply could be used, but many tube fanciers just don't do negative power supplies and the relatively high common cathode voltage (+9.5V) means that, in a line-stage amplifier, more than enough input voltage headroom exists, as line-level input signals seldom exceed 2Vpk. Okay, but where do we get a 10.2mA constant-current source? We could build one out of discrete solid-state devices, such as a FET and a source resistor; or we could use a constant-current source IC, such as the LM334 or the new Linear Technology LT3092 constant-current source. The LT3092 is an extremely interesting constant-current source that only requires two resistors to set its idle current. Because the two resistors tie together at one end, the LT3092 is considered a two-terminal constant-current source, which can be floated or ground or B+ referenced. It can withstand up 40V and source up to 200mA. I haven't actually had any hands on experience with the LT3092 yet, but I am eager to try some tests on it. On the other hand, because the common cathode voltage is so high, we can get away with using a 913-ohm common cathode resistor, as 9.5V / 10.4mA = 913 ohms, which will make the solid-state-fearing types happy. The last unanswered question is: How do we establish the +6.5V bias voltage for the right triode's grid? Ah, finally we arrive at the meaty bits. We could use a two-resistor voltage divider that spans from ground to the B+ connection. While this would work well enough, this arrangement is not optimal. A better setup is to terminate the voltage divider's top resistor into the right triode's plate, rather than the B+ connection. Why is this arrangement better? If nothing else, we will not have to deal with power-supply noise leaking into the cathode-coupled amplifier's output, as the Aikido cathode follower's output should be dead quiet. Moreover, in this setup, the voltage-divider resistors constitute a DC feedback mechanism that works to keep the correct bias voltage needed to establish a plate voltage equal to half the B+ voltage on the right triode's plate. Why is this important? The cathode-coupled amplifier, however, needs a bit more help to make a first-rate line-stage amplifier, as its output impedance is too high and PSRR is weak (almost nonexistent). Adding an Aikido cathode follower to the CCA input stage makes a complete, high-quality line-stage amplifier that offers low distortion, low output impedance, wide bandwidth, no phase inversion, and a stellar PSRR figure. The Aikido cathode follower DC couples with the CCA stage, and the two-resistor voltage divider defined by resistors R5 & R6 provides the DC feedback loop to keep the Aikido cathode follower output centered at half the B+ voltage. The second voltage divider resistor string, R11 & R22, injects the required sampling of power-supply noise into the Aikido cathode follower's bottom triode's grid to null the power-supply noise at the Aikido cathode follower's output. The diode D11 is a safety device that protects the Aikido cathode follower top triode at start up, when the tubes are cold and not conducting; once the tubes are hot, the diode falls out of the circuit, as it will no longer be forward biased. The LM334 constant-current source is optional, as resistor R3 can be used by itself (the LT3092 cannot be used on the CCA PCB). On the other hand, the LM334 constant-current source does reduce the distortion substantially (-20dB improvement), but it does make the CCA a more sensitive to changes in the B+ voltage and will require more circuitry tweaking to obtain as low a noise figure as the common-cathode resistor achieves, such as lowering reisitor R9's value a tad relative to R8, say 270 ohms to 300 ohms. On the other hand, the current source does produce more voltage gain from the CCA, about +3dB more with a 12AU7. Finding the right values for resistors R5 & R6 is easy enough. As we all know, a triode's amplification factor is simply a measure of the relative effectiveness of the grid over the plate in controlling the current flow through the triode. Thus, a triode with a mu of 10 holds a grid that is 10 times more efficient than its plate. So an increase in plate voltage of 10 volts can be countered by a decrease of 1V on the grid. In other words, we can maintain a fixed idle current in spite of a 10V increase in plate voltage by moving the grid 1V more negative. If we inspect the plate curves from the previous examples, we will find the triode's mu equal to about 23.1. If we then divide the 150V difference in plate voltages by 23.1, we get 6.49V, which is close to the 6.5V difference in grid voltages (9.5V and 3V) that we gleaned from plate curves. Thus, we need a voltage that yield a 1/mu voltage division; so R6 should be (mu – 1) times bigger than R5. This wonderful trick works because the voltage from ground to the Aikido cathode follower's out is equal to the voltage across the plate resistor, R4, which is also the difference in plate voltages. Finding R11's value takes a bit more work, as the old Aikido formula does not apply here. If a constant-current source is used to load the common cathodes, then the cathode-coupled amplifier offers almost zero power supply rejection, so all of the power-supply noise must be fed to the Aikido cathode follower's bottom triode's grid. Thus, if the LM334 CCS is used in place of the simple common-cathode resistor, R3, then the CCA's plate resistor's value must equal B+/Iccs and R11 must be replaced by a jumper wire and R8's value must be slightly greater than R9's value to null the power-supply noise from the CCA's output. How much greater? It depends on the triode's mu and rp. The following formula establishes the correct ratio between resistors R8 & R9, when a constant-current source is used.
With a common-cathode resistor and 6CG7s, in contrast, there is a 17% reduction in power-supply noise at the cathode-coupled amplifier's output. I used the following values in my own CCA setup (all tubes are 6CG7s): If a 12AU7 is used as the input tube, use a 931-ohm common-cathode resistor and set R11 to 15k and R6 to 175k. Wait a minute! Don't you first need to know what the B+ voltage is? No. That's the amazing thing about this topology—these values work perfectly with a B+ voltage that spans 100V to 400V. On the other hand, if a CCS is used in place of the common-cathode resistor, then the CCA's plate resistor's value must equal B+/Iccs. While testing my CCA prototype, I was thrilled to see the output stage always center to within 1V of the half-the-B+ target, as spun the variac's AC output voltage up and down. The PSRR figure, even without the LM334 CCS, was stellar and distortion quite low, about 0.1% with 1Vpk at 1kHz into a 47k load. I wondered if the DC feedback loop was not also contributing AC feedback, so I tried the line-stage amplifier with resistor R5 bypassed, so the small AC signal present would be shorted to ground. The results were interesting; with the bypass capacitor in place, the distortion almost imperceptibly budged upwards. in other words, very little negative feedback obtains. This makes sense, as the line-stage amplifier's open loop gain would have to be greater than the input triode's mu for negative feedback to take hold. With the 6CG7s, the AC gain is only 5.2 (or +14dB).
Relatively Low-Voltage CCA Line Amplifier
See Blog 62 for more information. So before proceeding, WARNING DO NOT ATTEMPT THIS AT HOME, UNLESS YOU REALLY KNOW WHAT YOU ARE DOING. I tried the 6DJ8 in the CCA, with a B+ voltage of only 150V. (I started with a 120Vac power transformer, which rectified up to about 165Vdc, which was then dropped to 150Vdc through resistors R12 and R17.) Nothing exploded or arced. Furthermore, the sound was surprisingly fine and the gain was close to +20dB. The lower B+ voltage allowed me to use much larger-valued capacitors in the high-voltage power supply (220µF/200V rather than 47µF/450V for C7 & C8 and 270µF/200V rather than 150µF/400V for C5), which greatly reduced the ripple the Aikido cathode follower had to wisk away. In other words, it was dead quiet. I used the following resistor values:
LM334 constant-current source
The LM334 requires only one resistor, Rset, to establish its idle current. The following graph shows the turn-on voltages and bias currents set by various Rset resistor values. Note that 10mA, the optimal value for most CCA setups, does not receive a resistor value, but we can readily see that 6.8 ohms is the correct value. (The LM334 datasheet goes into much more detail, but for tube work, 6.8 ohms is close enough.) Although 10mA is the LM334's maximum current flow, the device sees less than 10V with most tubes, such as the 6CG7 and 12AU7, so the device's dissipation is usually less than 100mW, well below its 400mW limit. Nonetheless, it is a good idea to attach a small heatsink to the IC, as it better ensures an accurate idle current.
The CCA Power Supplies THe CCA PCB can accept either a center-tapped or non-center-tapped high-voltage power transformer. (In my own CCA setup, I used a 240Vac non-center-tapped Hammond power transformer.) The CCA PCB holds the heater power supply as well and it is regulated. As can be seen, the power supply can accept either full-wave bridge rectifier circuit or a full-wave voltage doubler rectifier configuration. When used as a full-wave bridge rectifier circuit, the two power supply filtering capacitors are placed in parallel by orienting their positive leads to where the heatsink sits; and the secondary attaches to the two encircled AC pads. Configured as a voltage doubler, these capacitors placed in series by being rotated 90 degrees clockwise, so the positive leads point to the center-tap (CT) pad at the bottom of the PCB; the transformer secondary attaches to both the single AC pad in between capacitors C13 and C14 and AC pad that feeds rectifier D10 and D8; and D7, D9, C9, C11 are left off the PCB. If used as a full-wave center-tap circuit, the two heater capacitors, C13 & C14, are placed in parallel by orienting their positive leads to where the heatsink sits; and the secondary attaches to the two encircled AC pads while the secondary center-tap attaches to the CT pad. I used a 12Vac/3A transformer, as I like being able to switch the heater power supply on independently from the B+ power supply. By the way, the CCA can easily be built upside down, wherein the tube sockets are soldered to the top of the PCB, but everything else attaches to the PCB's bottom. Why? This allows the tubes to protrude through holes in the top of the chassis. The usual roadblock is the heater voltage regulator, but on the CCA PCB, I placed redundant solder pads on the bottom for the LD1085. The heater regulator uses the LD1085 low-dropout, adjustable, three-pin, 3A voltage regulator. The regulator can be set to an output voltage between 6V to 25V, but the assumption is that a 12Vdc output voltage will be used for the heaters, so that 6.3V heater tubes (like the 6FQ7 and 6DJ8) or 12.6V tubes (like the 12AU7 or 12BH7) can be used. Both voltage types can be used exclusively, or simultaneously; for example a 6GC7 for the input tube and an ECC99 for the output tube. Thus, if the input tubes (V1 and V2) are 6CG7s and the output tubes (V3 and V4) are 12BH7s and the heater regulator output voltage is 12Vdc, then use jumpers J2, J4, and J6. More RMAF Details Imagine a wife or a girlfriend being dragged into a car part store by her significant other. Her expectations of having a good time are low, so she is startled see an impossibly handsome man standing behind the counter. "Maybe I should shop here more often," she says to herself. Upon approaching the counter, she is disappointed to see that the gorgeous face was only printed on a life-sized poster cutout of a famous race car driver and that the actual living fellow behind the counter was only okay looking, with a bad haircut and in need of a shave. Well, this is how I felt, for as pretty as the picture the solid-state amplifier painted was it never felt real. And even when the tube power amplifier rendered a tarnished sonic picture, zits and pockmarks included, the sound still maintained a believability about it. In a nutshell, this is why tubes still matter.
With a Name Like Schiit, You have to Sound Good I just do not understand how Schiit can make a such an amazing headphone amplifier, an amplifier that is made in America (not China), that uses such high-quality parts (Alps, Nichicon, Wima, Dale, and Neutrik), that uses internal power transformers and not a switcher wallwart, that comes with a 5-year warranty, that sports four tubes and does not use a solid-state output buffer, and that sells for only $349. Amazing stuff. Schiit also makes a solid-state headphone amplifier, the Asgard (with a name like Asgard...), that sells for only $249. From the Schiit website:
If I didn't already own about ten headphone amplifiers, I would certainly buy a Valhalla from Schiit. My only wish is that they paint their boxes flat-black enamel paint. Here's why:
Taller feet would also prove effective in keeping the little amplifier cooler.
At the Other $$$ Extreme So, the only thing I can conclude is that the nice people at Schiit must be making a killing. Of course, ours is a bad economy and I heard from many American firms that if it weren't for the export market, they would be out of business. I also encountered near universal scorn and derision for high-end cable manufacturers. I can understand the logic behind such animus, as it must be painful to see $20 worth of wire selling for more than your power amplifier or loudspeaker or turntable... These cable guys run a markup that would make a jewelry store owner blush. Sure, cables can prove important, but they should never cost more than the rest of your system. Or am I getting this wrong? Would $20 speakers hooked up with $10,000 cables sound much better than $10,000 speakers hooked up with $20 cables?
New Books This handsome 600-page book covers aspects of solid-state power amplifier design, from the basics to the secrets only the masters know. The following table of contents gives you a good idea of much material is held between the covers:
As soon as I have finished reading Bob's book, I will give it a proper review; but if you have any interest in solid-state or hybrid power design—or audio design in general, as much of what Mr. Cordell writes about also applies to phono stages and electronic crossovers—then be sure to drop many hints to your loved ones that you would enjoy receiving this book for Christmas. Well written, thoroughly researched, beautifully illustrated, highly recommended. The second book is not a book, at least not in the typical sense, as it a bookzine. A magazine as thick as a book, but ad free—this is the answer to the question What is Linear Audio? Jan Didden, famous for his many articles in Audio Amateur and audioXpress, is the editor and publisher. He has brought out Volume 0 (yes, zero) and filled it with a who's who of audio design:
From the Linear Audio website:
Loudspeaker, solid-state, and tube design articles abound in this 169 page bookzine. And those 169 pages are ad free. Another way of looking at it that most magazines run about 50% content and 50% ads; thus, Linear Audio would have been 338 pages long had it been filled with ads. In addition, regular magazines engage in funny math to convince their advertisers that they are getting their money's worth, such as printing twice as many magazines as they have subscribers. In other words, the ads pay for magazine's printing. With Linear Audio, the reader pays for it all. Sure that is all very nice, but can't I just read about audio off the net and pay nothing? Indeed, but high-quality articles like these deserve to be printed, which would certainly cost more in paper and toner or ink than the cost of this beautifully put together bookzine. In short, also highly recommended.
Next Time
//JRB |
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