The Starving Student Millett Hybrid
(click for a larger pic)
Webmaster's Note: the Starving Student Millett Hybrid, designed by Pete Millett and later modified for PCB use by Dsavitsk, has been so popular that it actually depleted the market of 19J6 tubes. Unfortunatley, the 19J6 tube was somewhat unique in its 19V heater voltage, which is integral to the heater-biased-MOSFET design of the Starving Student. Nevertheless, with some modifications such as those shown above, other tubes may be used in the circuit.
Head-Fi user "logistics" was the first to come up with a build using 12SR7 tubes. Later, Head-Fi user "the_equalizer" came up with a modification to use the ubiquitous 12AU7 or its variants. Below is a detailed description from "the_equalizer" on how he made this modification. It can be used as a guide in finding other tubes as well. Perhaps it can help continue the Starving Student legacy for others, even though the original 19J6 tubes have become rare and expensive.
Ok, this is a short description of how I did it. However, as I said before, I'm not an electrical engineer and there are a couple of things I don't (yet) understand about the circuit. If you look at the schematic above you can clearly see that there are two parts to the circuit:
a) the tube voltage gain stage (increases the volume of the signal)
b) the MOSFET output current gain stage (increases the power of the signal)
For simplicity, I'll refer to just the right channel part numbers. Naturally the left channel works in exactly the same way.
Capacitor C2 isolates the DC bias voltages and currents from these two stages.
Bias voltages and currents for the tube stage are set by resistors R1 (the plate load resistor) and R5 (the cathode resistor). The way to establish the value of those resistors is by drawing a load line and calculating the values. You can read how it's done here. After reading that, you might want to read this post from Dsavitsk. Theoretically, any triode (or maybe even any pentode) would do, I guess.
Now the interesting part of making tubes work with the SSMH is the MOSFET output stage. There are two bias voltages and currents you want to provide to the MOSFET
a) The drain-source voltage and current
b) The gate-source voltage
As to the drain-source voltage and current, you can see that the MOSFET drain and source are in a series circuit formed by the power supply, the MOSFET and the tube heater. The current in that circuit will be the tube's heater current, a value you can find in the tube datasheets; whereas the voltage will be the power supply voltage less the tube heater voltage. So in the 19J6 version, the MOSFET is biased at 150mA (the 19J6 heater current) and 29 volts (48V - 19V). In order to keep things simple, you want to find a substitute tube that has a heater voltage close to 19V and the same heater current as the 19J6.
There are several reasons why you want to stick to the same heater current: First, the heatsinking requirements of the MOSFET will grow enourmously with increasing heater current. Second, the standard PSU used in the SSHM has a current capability of 0.35 A, which is just enough to power two tube filaments, each drawing 150 mA; if your substitute tube draws more heater current you'll need to scale the PSU (more $$) . Finally, and this is something I haven't checked in the IRF510/610 datasheet - you want to make sure that the MOSFET can withstand your proposed heater current.
The heater voltage requirements are somewhat less strict, but you want to stay around 19 V. Too high a heater voltage and the voltage across the MOSFET will be too low for it to drive your cans (it wont have enough room to move, enough voltage to swing); whereas too low a heater voltage and you'll have the same problem as a high bias current: you'd be dropping too much power on the MOSFET and you'll need BIG heatsinks.
So, for example, using a 6J6 (a 6 volt heater equivalent to the 19J6) is almost impossible: its heater voltage is too low as you'd be dropping 42 volts across the MOSFET (compared to 29 V in the 19J6 version) and its heater current is 450 mA (!!!!!). Nor can you easily use 5J6 (4.7 V@600 mA heater) or 6DJ8 (6.3 V @ 365 mA heater) tubes.
Yes, you could insert a power resistor in series with the MOSFET and the heater to dissipate the extra voltage but the resistor would have to be well ventilated (it'd need to dissipate some 6 to 8 watts) and that still leaves you with the heatsinking required for the MOSFET to operate at 29 volts @ 450 mA !!!
Last but not least, we just need to talk about the MOSFET gate-source bias voltage. In the SSMH this voltage is set by the voltage divider formed by R2 and R4, and the tube heater voltage. You can see that the MOSFET's source pin (the pin tied to the tube heater) voltage to ground is equal to the tube heater voltage. So in the 19J6 version the MOSFET source pin voltage is close to 19 V, as indicated by Pete Millett in the schematic.
Now, in order to "turn-on" the MOSFET, the gate needs to be a little bit more positive than the source. How much more positive is something I deduced from the schematic (* webmaster's note: Gate-to-Source Threshold Voltage = 2 to 4 V for the IRF510). If the PSU voltage is 48 V, then the voltage at the midpoint of the voltage divider formed by R2 and R4 must be half that voltage (since R2 and R4 have the same value) thus the voltage with respect to ground at the MOSFET gate is 24 volts; if the source is at around 19 V, then the gate-source voltage is around 5 volts. Resistor R3 has no influence in this due to the way MOSFETs operate: there's no current flow in the gate-source circuit, so no current flowing through R3 and no voltage drop across it.
So, I assumed that that voltage had to be preserved. I was aided in this assumption by Dsavitsk's hint that "only one resistor" had to be changed, so I deduced, it had to be one resistor in the gate biasing voltage divider. Then it was a matter of using Ohm's law to calculate the value for such resistor so that if the MOSFET source pin were at around 12 volts, the MOSFET gate pin sat at around 17, thus mantaining the 5 volt difference.
The math is simple:
Total resistance in the voltage divider
390 Kohms + 220 Kohms = 610 Kohms
Total current in the voltage divider
48V / 610 Kohms = .08 mA
Voltage at the junction
.08 mA * 220 Kohms = 17.31 V
You can also see that Logistic (* webmaster's note: Head-Fi user logistic's 12SR7 mod) got the same result with different resistor values. The possible implications of choosing different values are something that I don't completely understand. I think you'd want to keep R4 the same value or higher, so that there's no bass roll off in the insterstage RC coupling formed by R4 and C2 but this I get from my incomplete understanding of the circuit; I could (very likely) be wrong.
Another thing I don't yet understand: does the source-gate voltage have to be around 5 volts ? Can it be higher? How much higher? Could the 12AU7 version work with the same voltage divider as the 19J6 version (e.g. with the gate sitting at 24 volts with respect to ground) ? I can see that this voltage (along with the source-drain voltage) determines how much voltage the MOSFET can swing, but I haven't yet sat down to carefully read the datasheet and run some experiments with the amp
So, that's pretty much it... it didn't end up as brief as I thought it would be . I hope it wasn't too confusing and that I managed to answer your questions.