Tube CAD Journal Page 9

What we need to know is the plate voltage at idle, the plate's voltage swings at maximum output, the idle bias voltage, and the grid's voltage swings at full output. One possible circuit arrangement taken from SE Amp CAD (click to download PDF) is the following: a Bartolucci 24 output transformer with a 4k reflected impedance as the plate load, a 60 mA idle current, a -87 volt bias voltage, a 494 volt B+ voltage, a plate voltage of 488 volts, a cathode-to-plate voltage of 400 volts (the output stage is cathode biased, a grid voltage swing of at least 80 volts, and a output of 7 watts. If we were given just this limited amount of information, we could deduce several missing pieces of data. The most important is that the plate must swing 240 volt swings to develop 7 RMS watts into a 4k load. SE Amp CAD's simulation of plate voltage swings (and grid swings) were not used because its simulation calculates the 300B's actual raw distorted voltage swings that contain a large helping of 2nd harmonic distortion, which we hope to reduce. How big a reduction can we expect? The formula for distortion reduction is as follows:

D´ = distortion / (1 + bGain),

where gain equals the open-loop gain and b equals the feedback ratio, but as our hope is to realizes close to 100% feedback, so we can round up to 1. The last missing piece of information is the value of the Rfb resistor. It can be found by first taking the gain + 1 and multiplying this sum (3 + 1) against the 300B's 15 pF of grid-to-plate capacitance; second, we chose the acceptable limit to high frequency response. In this example lets pick 30 kHz. Reordering the last formula gives us the value of for resistor Rfb:

Rfb = 1 / (2piFC)

In this example, Rfb = 88.5k, which we round down to 80k. Understand, this value was almost arbitrarily picked and may not ultimately work, but is a convenient starting point, as 3 mA against it equals the 240 volts of plate swing needed.

Now that we have the design specifications for the output stage, we move back to the driver stage. Because our output stage consists of an I-to-V converter, the driver stage must not drag down the output stage by having too low an output impedance. Ideally, what we need is a V-to-I converter, which is convenient for us tube fanciers, as tubes are basically V-to-I converters. In a tube amplifier, the grid presents an extremely high input impedance and it acts upon voltage signals by varying the current through the tube in response.

In other words, it converts input voltage into output current, which when this current flows through an impedance or resistance, develops an output voltage. Since the V-to-I conversion is inherent in the tube's functioning, all that remains is to guarantee a high output impedance. Using a pentode or a triode with a large value cathode resistor has already been mentioned and both solutions work well. My preference (this week) is for the triode, if for no other reason that it does not require a screen power supply; next week, I may find the pentode intrinsically high output impedance and wider range of operation more compelling.

Given that the 300B's plate must swing 240 volts, we must find positive grid swing needed to achieve that plate swing. The SE Amp CAD report shows +80 volts are needed. Given that grid must see at least 80 volts of peak swing and that the series resistor equals 80k, we determine that the peak current swing into the output stage must equal 4 mA. (Before reading too far ahead, stop and see if you can figure out how this value was derived.) The 300B's plate must swing ±240 volts to realize 7 watts into the output transformer's primary and its grid must see at the same time see at least ±80 volts, as the two voltage swings work in anti-phase. In other words, the 80 volt grid swing must be added to the 240 volt plate swing. The total, 320 volts, is then divided by the 80k resistor: 320 / 80k = 4 mA. This is the peak current swing, not the peak-to-peak current swing, which would be 0 to 8 mA. If the inductive load is used, the peak-to-peak current swing would be -4 mA to +4 mA, the idle current through Rfb is 0 mA.

I do not want output lose any readers, so I will try to keep the math simple here. Now 4 mA is also the value that 4 volt across 1k equals or 40 volts across 10k equals...and so on. In the first case, a 1 volt input voltage results in a 1 mA output current; in the second case, 1 volt results in .1 mA. So in keeping to our pledge of simple math, let's opt for the first case and accept a conversion ratio of 1 mA per volt. This means that we will need at least 4 volts of input voltage swing.

Which tube to use? Several tubes can be biased with a 1k cathode resistor (the 12AX7 to the 6BX7). But few, if any tubes, will find an idle current equal to 4 mA at the desired plate voltage (168 volts). There is one workaround for those cases where idle current against the cathode resistor is greater than the required bias voltage.

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