PG2DTA: Choosing a Power Tube and Transformer

In my first post of the Practical Guide to Designing Tube Amplifiers (PG2DTA) I defined 10 steps for designing an amplifier. This post will go over Steps 1 and 2:

1. Choose a power tube
   
2. Determine the output transformer to use
   

While I'm sure there are a number of technical reasons that determine why an output tube is suitable for a particular application, I'm not going to attempt to describe them. These days, people aren't building tube amplifiers for technical reasons, they are building them because they like tube amplifiers. The tubes that are suitable for audio output are pretty well known so I'm not going to deviate from the list (tube depot, boi audio works, the tube store).

For the purposes of the guide I am going to use a triode-connected 6V6. Why? Because I like 6V6 tubes in guitar amps and the 6V6 is reasonable inexpensive. In addition, there is a local amplifier competition that I am entering so this is a good exercise.

The simplest way of determining which transformer to use is to look up what other people are using (Hammond, VT4C). However, you will notice that there is not a single value to use for a particular tube. Recommended values for the 6V6 range from 3,500 Ohms to 8,000 Ohms. This value is known as the load and will be used to draw the loadline in step 3 of the PG2DTA. For now, let's try to figure out where these numbers came from.

Generally speaking, a reasonable load for a triode is twice the Anode Resistance or Plate Resistance:

RL = 2 * ra

So, all we need to do is look up the Plate Resistance for the tube and we are done, right? Wrong! Unfortunately the Plate Resistance value changes depending upon the operating point of the tube. This is where things start to get more technical in the books and I get lost, but its really quite simple.

First thing to do is find the data sheet for the tube you want to use. A great source for finding tube data sheets is Duncan Amp's Tube Data Sheet Locator.

Find the maximum plate voltage (Va-max) and the maximum plate dissipation (Pa-max) on the tube datasheet. Unfortunately, there doesn't seem to be a consistent naming convention for these values. Some use the term Anode, some use Plate so you will have to become familiar with all of the terms. I am using General Electric's 6V6 data sheet for this example.

Find the graph that plots the plate current against the plate voltage labeled average plate characteristics. For the 6V6 tube, find those that are for triode connection. Draw a vertical line representing the maximum plate voltage (Va-max). For the 6V6 this will be 315 Volts.

Determine points on the graph representing the maximum plate dissipation (Pa-max). This requires some math, but its simple math. Using Ohm's law P = I * V we can calculate the current required for 9 watts at a number of different voltages with a quick rewrite of the formula. In other words, I = P / V where P = 9 and V = voltages we choose. For example:

9 / 60 = 0.150 or 150mA
9 / 100 = 0.090 or 90mA
9 / 120 = 0.075 or 75mA
9 / 200 = 0.045 or 45mA
9 / 250 = 0.036 or 36mA
9 / 300 = 0.030 or 30mA
9 / 320 = 0.028 or 28mA
9 / 340 = 0.026 or 26mA

You will end up with a graph that looks something like this:


To determine the Plate Resistance, find a grid curve that is closest to the intersection of Va-max and Pa-max and draw a line tangent to the curve. A tangent is the straight line that just touches a curve at a specific point on the curve. Without having the formula that determines the curve, I think the only way to determine the tangent is to do your best drawing it. Here is what I came up with:

Now that we have the Plate Resistance line plotted, calculate the slope of this line by using another of Ohm's laws V = I * R. Because the graphs are not exactly high-resolution, the easiest thing to do is look where the line intersects the grid. In this case, I=0, V=240 and I = 30 and V = 320. Thefore:

I = 30 (30 = 30 - 0)
V = 80 (320 - 240)
R = V / I
R = 80 / 0.030
R = 2,666

As I said above, optimal load for a triode is twice the Plate Resistance:

RL = 2 * ra
5,332 = 2 * 2,666

So, a tranformer with a primary of around 5K is appropriate for a triode-connected 6V6. This is consistent with many designs so I think that we calculated the plate resistance correctly.