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  • Writer's picturePerry

Tube Amplifiers Explained, Part 6: Load Lines and Operating Point

Updated: Mar 29, 2020

Part of a blog series Tube Amplifier Circuits Explained

The physical characteristics of each type of tube—materials, how far apart is the cathode, grid, anode, etc.—determine how the tube will operate and the effect of different voltages. Tube manufacturers provide datasheets that include a variety of information about these characteristics and limiting values. Search online and you’ll easily find these as PDFs—often looking like bad photocopies of documents from the 1950s—and of course there’s a long history of tubes from that era or earlier, so it’s no surprise!

If you haven’t studied these, they might seem confusing, but you are a brilliant learner and this is how you will be empowered to begin understanding or even designing your own circuits. Let’s use the 12AT7 tube and take a look at an example datasheet:

12AT7 Datasheet (example pages)

Go search for one of these online (e.g. “12AT7 datasheet”) and pull one up for reference. We won’t go into all of the details, but you’ll find some useful information and I always keep these handy especially for the pin diagrams to tell you which pin number is the anode, grid, etc., or to know how much current the heaters will draw, or what maximum voltages can be used.

One of the charts you will find is the plate characteristics. On the x-axis is voltage and on the y-axis is current. There are a series of curves that represent different voltages that you could possibly have on the grid, and this will tell you what current would correspond to a particular anode voltage. Using the chart below, if you had the grid at -2V and you put 200V on the anode, then the tube (valve) will allow about 6mA of current. You will notice these are curves, not straight lines, because the tube does not operate as a perfectly linear device. We will talk about this later, regarding distortion.

Before we look at an actual circuit and load line, let’s consider a few attributes of the valve that we can visualize on this chart. First, remembering that grid voltage controls the amount of current, let’s see what this ratio is. If we pick a place on the chart and hold anode voltage constant and measure the distance between two grid curves, we can see that a 1V change in grid voltage results in around 4-5mA of current change. This is the purple line on the chart below, and is referred to as transconductance (Gm), often measured in a funny unit called mhos (reverse spelling of ohm, conductance being the opposite of resistance!) and tube datasheets usually use micromhos (one millionth of a mho). So look on the 12AT7 datasheet and you’ll see transconductance of 4,000 – 5,500 (depending on operating conditions), which would match our chart estimate after converting units.

Now, consider another property called amplification factor (a ratio abbreviated with the Mu symbol: µ). Holding current constant, if we measure between two grid curves, we see that a 1V change in grid voltage will result in a change in anode voltage of 60V (blue line above). Aha! Here is the leverage that we have been looking to understand. We could swing 60 volts of anode voltage for every one volt on the grid. Powerful, yah? So the µ of the 12AT7 is 60 and you’ll see this on the datasheet. Some tubes have lower or higher amplification factors that could range from 20 to 100, for example. After we plot a load line and calculate the gain of the amplifier, you’ll see that we won’t expect to get this full factor of amplification in this application (a regulated current source is one technique that could be used).

Now let’s take our basic circuit and consider what happens when we have a particular anode load. Let’s say we had a load resistor that is 50k Ohms and we supplied 400V B+. If there was zero current flowing (tube totally in cutoff), what voltage is on the anode? Ohms law tells us that with no current, there would be no voltage drop across the load resistor, so the anode will have the full 400V. Let’s plot a point at zero mA and 400V representing this extreme situation. And imagine the opposite end of the spectrum—what if the valve was wide open so that maximum current flows and the entire 400V were to drop across that resistor, and there was zero volts on the anode. Ohms law in this case to solve for current:

I = V/R

I = 400 / 50,000

I = 0.008A or 8mA

So let’s plot another point at 8mA current and 0 volts on the anode. Ohms law is a linear relationship, so this is all we need to draw a load line between these two points.

So what we see here is the actual linear relationship between voltage and current for a 50kOhm load resistance and 400V B+ supply. Our tube must operate on this line somewhere. Where? That depends on what voltage we put on the grid. Look again at the grid curves and find the points that intersect with our load line. If we set the grid to -2V, then in this circuit we will have about 4.5mA of current and 175V on the anode. If we change the grid to -1V (less negative), then we will have 5.5mA of current and 130V. Notice that this is a change of 1V on the grid and a change of 45V on the anode—not the full 60 amplification factor because our load line isn’t horizontal, it has a slope. If you were to alter the supply voltage or the load resistance, you can control the position and slope of this line, which impacts gain (and distortion).

Now it’s time to visualize your audio signal on this load line. The input signal is an AC voltage that goes peak to peak from, let’s just say, +1V to -1V (depending on your source level). If we simply put this signal onto the grid, it would cause the grid to fluctuate both positive and negative. We don’t have any positive grid curves shown here. Why not? Because when the grid is positive it is attracting electrons just like the anode. This isn’t what we intend and will leak current out of the tube through the grid, which would then further impact the voltage potential between grid and cathode, etc. It’s a problem we will avoid by keeping our grid operating in a better space, more central on our load line.

Let’s pick a spot to represent the operating point, or the quiescent state when there is no audio signal impacting the grid, but there is enough room above and below this point for the audio signal to raise/lower the grid voltage. If we pick an operating point where the grid is -2V, this puts us roughly in the middle of the load line and there will be a steady-state current of 4.5mA, and 175V on the anode.

Now envision an AC audio signal on the grid that causes it to have a range between ‑1V and ‑3V and the anode voltage will fluctuate from around 130V to 220V. You can see a sine wave visualized on the horizontal axis. You should also recognize that we have current flowing continuously, it just varies in how much, based on the input signal and grid voltage. This is a Class A type of amplifier, conducting current across the entire input signal, not going into cutoff at any point. You can see that Class A consumes power continuously regardless of input signal amplitude and is therefore less efficient than other options, but it has excellent characteristics for high fidelity amplification and simplicity in our design. A single-ended tube amplifier uses one tube to produce an output, as opposed to other options such as a push-pull tube amplifier that would use two tubes amplifying portions of the signal.

One more consideration regarding load line and operating point. The tube will have maximum rated limits of operation. In the case of the 12AT7, the datasheet tells us that the anode cannot be higher than 300V in steady operation. So our operating point must be left of that point. And the tube is rated for maximum power dissipation of 2.5 watts. The red dotted line on the previous chart shows where current and voltage would exceed 2.5W of power dissipation on the anode, so our load line and operating point needs to be below that line.

Now let’s figure out how to get our grid to be at our desired voltage for this operating point. One option is to use a negative DC power supply and adjust it so that it is 2V lower than the 0V potential of the cathode. Some amplifiers use this technique and it’s referred to as fixed bias. It requires calibration to ensure the right relative voltage between grid and cathode to achieve a target operating bias.

Ok, consider the new circuit shown. We inserted a cathode resistor Rk. What value should we use? If we want to elevate the cathode to +2V so that we have quiescent current flowing of 4.5mA, then Ohm’s law will tell us that we need a value of:

R = V/I

R = 2 / 0.0045

R = 444 Ohms

Now we have technically increased our total load on the B+ voltage to the sum of the anode resistor and cathode resistor, so our original load line isn’t quite right anymore, but this value is very small relative to the 50k Ohm so we won’t worry about it, or you could re-calculate the load line with this larger value of 50,444 Ohms. (And of course 444 is an unusual number and we could round this cathode resistor value to one that is typically available.)

You will also notice a resistor Rg between grid and ground. This is called a grid leak resistor. We need something to reference the grid to the 0V DC potential, allowing the input AC signal to then be applied onto it to control the tube. We want this to be a high resistance value so that we don’t attenuate the input signal (keeping the input impedance high relative to the source impedance), but there are some other considerations. There is a very small amount of current that flows in the grid (“leaking”) and it could alter the bias of the grid. A value for this 12AT7 input stage could be 1M ohm.

In the next post, we will look at Coupling, AC load line, and Cathode Bypass Capacitor.

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