A vacuum tube hi fi primer
Part 2: Direct coupling
This is meant to be a continuation of the concepts discussed in the original Vacuum Tube Hi Fi Primer. This time we will be taking the knowledge we gained from that howto, and using it to design a slightly more challenging amplifier. Again, the result will be an inexpensive amplifier you can build. The goal is not only to be able to build vacuum tube hi fi amplifier, but to fully understand how it works, and how the design was conceived - and ultimately how to modify it or design a different amplifier using these same concepts.
Conventional RC coupled amplifiers are very easy to design, and can sound very good. Direct coupling is a clever way to eliminate the coupling capacitor. Capacitors are a necessary evil in amplifier design, and many believe that eliminating as many capacitors as possible is a good thing. Like Einstein said, "make everything as simple as possible, but no simpler" :) I chose direct coupling as the topic for this howto because I learned so much designing my first direct coupled amplifier that once I had figured it out, I felt like a real designer. Direct coupling relies on concepts that require a slightly more thorough understanding of how tubes operate, and how to work our way through several interrelated calculations.
So let's get to it.
The purpose of a coupling capacitor is to block the voltage present on the plate of the driver tube to keep it away from the grid of the output tube. If this cap were not present, the grid of the output tube, which is referenced to ground, would be at the potential of the driver tube's plate - a couple hundred volts. This at the least would ruin the bias point of the tube, and at worst cause the tube to conduct as much current as it can get, destroying the tube and possibly a couple resistors and maybe even the rectifier. Not good. So clearly we have to do more than just remove this cap to create a direct coupled amplifier :)
What if the plate of the driver tube was at exactly the same voltage as the grid of the output tube? If that were true, the coupling cap would have nothing to do, and could be eliminated. That is the trick. We must find a way to raise the grid of the output tube to a voltage equal to the voltage at the plate of the driver.
In a typical RC coupled amplifier, the grid of the output tube is referenced to ground by the grid resistor. Since no current flows to or from the grid during idle, there is no voltage developed across the grid resistor and the grid is therefore at 0V. To bias the tube, the grid also must be somewhat negative in relation to the cathode. So if the grid is at 0V, we must raise the cathode up off of ground by the bias voltage, so that the grid will be at a negative voltage relative to the more-positive cathode. So we add a cathode resistor with the bias current flowing throught it to develop a voltage drop across the resistor.
Confused? Let's look at what the voltages are between each of the parts of the tube. Let's take as an example a 1626 output tube at an operating point of 200V plate voltage, -18V bias voltage, and 25mA plate current. To make things even more explicit, lets call the plae voltage the plate-cathode voltage, because there is a 200V difference between the plate and the cathode, not between the plate and ground. Let's also call the bias voltage the grid-cathode voltage, because it is really the voltage difference between the grid and the cathode. The tube doesn't care where it is in relation to ground or to any external voltage reference, all it cares about are the voltage potentials between it's plate, grid, and cathode.
The voltages that affect the tube are like a fly trapped in your car as you speed down the highway. If the fly flies from the back of the car to the front of the car, all it knows is that it just flew 4 feet at maybe 2 miles per hour. Viewed from outside the car, it may have traveled 127 feet at 62 miles per hour, but the fly doesn't care. The car could have been anywhere travelling at any speed. And don't worry, calculating how many feet the fly travelled is probably harder than designing this amplifier :)
First I suppose we should decide what tubes we want to use. I want somethng that you will be able to afford, and something relatively common. But it must also sound good. One tube that meets these criteria is the 1626 indirectly heated triode. This tube was made famous by Bob Danelak's Darling amplifier and the many people who built one, or some version of it. We will be continuing the tradition by building a DC Darling.
Now we must find a suitable input tube. I have seen many many tubes used as input/driver tubes on Darling amplifiers. I will use the 6sn7/12sn7/6j5 tube. The 6sn7 is a 6V heater, dual triode. The 12sn7 is the 12V heater version. The 6j5 is basically a single-triode 6sn7. The only criteria we must meet for choosing a driver tube is that it must have enough gain to drive the output tube to full power. The typical operating point for the 1626 shows it biased at somewhere around 18-20V. An easy rule of thiumb is to assume that we will have a 1V input sensitivity. That way we can expect something slightly less that 1X the amplification factor of the tube. The 6sn7 has an amplification factor of around 20, so we are all set.
As with a conventional RC coupled amplifier, or any tube sircuit for that matter, let's start with the load lines. We will begin by determining an operating point for the driver/input tube. Most small signal tubes like the 6sn7 can be operated just about anywhere because their plate curves are quite parallel along the majority of the area underneath the maximum plate dissipation. We want an operating point with a relatively low plate voltage, and relatively high plate current. High plate current decreases the plate resistance and keeps the driver tube from running out of current to feed to the output tube when the output tube is clipping heavily and drawing plate current. It also raises the load line well above the 0mA axis where it is least linear. Let's use 100V plate voltage, and -1.2V bias for a corresponding 8mA of plate current. The 1.2V bias will let us use a battery instead of a bypassed cathode resistor. This eliminates a capacitor (which is almost always a good thing) and allows us to experiment with one more form of bias.
The operating point we will use for the 1626 will be 175V plate voltage, -18V bias, and 25mA plate current. This is a bit more current and a bit less voltage than some amplifiers use, but it should be very linear, and I don't see the point of trying to squeeze every last drop of power out of a 1W amplifier.
So now we have the driver tube's plate sitting there at 100V. In an RC coupled circuit, we would simply sonnect it to the grid of the output tube via a series capacitor and a grid resistor to ground, which would basically tie the grid of the output tube to ground, and let us cathode bias it just like the driver stage. We aren;t going to do that.
If we are going to tie the grid of the 1626 output tube directly to the plate of the driver tube, then we must configure the output tube so that the grid is properly biased when it is at 100V. Bias means simply that the grid is 18V less positive than the cathode. If the grid is at 100V, then all we have to do is put the cathode at 118V, and the tube will be proplerly biased. I am making this bold, because this is the "trick" to direct coupling. That is really all you have to know. So, with 25mA flowing through the 1626, we can calculate the proper cathode resistor to put the cathode at 118V. R = V/I = 118V/.025A = 4720 ohms. Let's just say 4.7k.
Next, let's look at what B+ power supply voltage we will need. If the cathode of the 1626 is at 118V and we are biasing it at 175V plate voltage (remember plate voltage is really the plate-cathode voltage) then we need the plate to be at 293V. Since the output transformer will have some low but nonzero DCR, and will therefore drop a couple volts, let's plan on a 300V B+ power supply.
Now that we know what voltage our power supply voltage is at, we can calculate the plate resistor for the 6sn7. The plate of the tube will be at 100V, and the B+ is at 300V, so we need the resistor to drop 200V. Plate current is 8mA. R = V/I = 200V/0.008A = 25k ohms. 25k ohms should be sifficiently large to adequately load the 6sn7.
So there is our design. Now all we have to do is design a power supply, find parts that will work, and build this thing.
Stay tuned for the rest of the design, lots of diagrams, and eventually photos of the complteted amplifier...
