In this article we will explore establishing a loadline on a triode vacuum tube (valve), determine the power available from this loadline, and the distortion predicted from this loadline. This will allow the reader a step by step way to calculate the parameters for his design. I have chosen to "invent" a tube for illustrative purposes. My invention is "spec'd" as a triode with the following parameters:
For this article, I will stick with a single ended transformer loaded Class A amplifier, as one of the simplest vehicles that can convey this knowledge. Most of the information is appropriate for push-pull operation, once a composite load line is established (may do that in another article).
- Get the tube characteristics for your design (shown above for my example). If not already on the characteristics, plot the maximum power dissipation curve (voltage * current = spec'd power).
- Select a reasonable quiescent operating point for your design. Often this is described in the manufacturers typical operation. For instance, a reasonable starting point for my example would be 300 volts and 60 mA. (This is the orange "dot" on the characteristics above.) Whatever you pick MUST BE within the tube's ratings. For this example, I'll choose to ignore the manufacturers recommendation, and use 300 volts and 65 mA. Put a dot on the tube characteristics at this point. (I am expecting that the "recommended point" is somewhere near optimum, but I want to run the tube a little "hotter", perhaps gaining more power output, lower distortion, or both.)
- Now start drawing a potential load line. This is a straight line whose "slope" is the primary "impedance" of the transformer. How do you do this? Pick a point somewhere about DOUBLE the quiescent voltage and zero plate (anode) current. Draw a line (called "loadline" in the curve below) thru these 2 points extending to the 0 plate volt axis. Did your loadline go above the max power curve you drew? Not good. Go back, pick a different zero current point. Is the voltage at no current more than twice the "rated" voltage? Not good. Go back and pick another point. Within spec? Good. Now determine the impedance. Z=(maxvoltage-quiescent voltage)/quiescent current. Alternately, you can pick an impedance to suit available plate transformers. In this case, if you go outside the safe boundaries, you must choose a new quiescent point. The curve below chooses a 4k load line.
- Having now chosen a tentative loadline, we need the following information from it......
- The required grid bias (you'll need to supply the grid with this value of DC voltage).
- The quiescent voltage (called Vq)
- Voltage where the grid bias voltage is zero (called Va).
- Voltage where the grid bias is double the quiescent value (called Ve). Note... if the plate current reaches zero before you get to this bias, you've got a bad operating point, go back to step 3 and try again.
- Current at 0 volt grid bias (called Ia). If this current is above the maximum rated tube current, go back to step 3 and try again. Hint: If its off the scale, chances are the current is too high.
- Current at half the quiescent bias voltage. (Called Ib.)
- Quiescent Current.(Called Ic.)
- Current at 1.5 times the quiescent bias voltage (Called Id).
- Current at twice quiescent bias voltage. (Called Ie).
Note that these are all plotted out on the following set of tube curves for our example.
For our example, the data is as follows: Va= 99V, Vq=300V, Ve=465V, Ia=115mA, Ib=89mA, Ic=65mA, Id=42mA, Ie=23mA. In addition, 40V p-p are required from the driver stage, and a 4k plate transformer is required.
In any device with even order distortion (not just second harmonic as sometimes stated), the average current of a Class A stage will change depending on the signal level. If the distortion is relatively low, the effect is relatively unimportant, if the distortion is high, this effect becomes increasingly important. (It is also more important for self bias than fixed bias, as the bias voltage is a function of the average current). This effect modifies all tube operating characteristics. The degree can be seen by comparing the quiescent current (65 mA in our example) with the average current as taken from the "extremes": that is, (Ia+Ie)/2. In our example that value calculates to 69mA (note that is not substantially different). The effect requires an iterative plot of the load line on the tube characteristics. The "worst case" is shown in "pink" in the constructed load lines above. The real case will be between the two extremes. Notice that the "curve" includes some portion above the maximum dissipation. In this case, the power dissipated in the tube is (300*0.069 - power output), or in this case 16.5 watts. This is still LOWER than the 20 watt max dissipation case, so there's no problem. Note the quiescent dissipation is still 300*0.65, or in this case 19.5 watts. The remainder of this document ignores this effect: it is important in high distortion cases only, but high distortion cases are not of much interest.
Po = (Ve-Va)*(Ve-Va)/(8*loadimpedance). In our example this is (465-99)*(465-99)/(8*4000)= 4.18 watts. This number is an approximation in that it assumes low distortion. HD2(%) = 75*(Ia + Ie - 2*Ic)/(Ia + Ib - Id - Ie) In our example this is 75*(115 + 23 - 2*65)/(115 + 89 - 42 - 23) = 4.3% HD3(%) = 50*(Ia - (2*Ib) + (2*Id) - Ie)/(Ia + Ib - Id - Ie) In our example this is 50*(115 - 178 + 84 - 23)/(115 + 89 - 42 - 23) = -0.72% Notice the minus sign. This indicates that the harmonic content subtracts from the fundamental (flattening it) when the fundamental is at its crest. This *usually* happens on third harmonic distortion in tubes. HD4(%) = 25*(Ia - (4*Ib) + (6*Ic) - (4*Id) + Ie)/(Ia + Ib - Id - Ie) In our example this is 25*(115 - 356 +390 -168 + 23)/(115 + 89 -42 - 23) = 0.72%
If you don't get numbers you like from the above, you are free to choose different operating points and loads to alter the characteristics of the stage. This may alter the distortion, power output or both. You may freely experiment on paper until you get a "design" you like. You can then correlate the results with your own listening tests. Over time, you'll find there is a correlation between what you predict, and what you hear. Then, you will be able to come up with new amplifiers more rapidly, as you will be able to zero in on what you like by doing a few "paper designs" up front, before committing to a particular tube, transformer, or power supply. This procedure also works for small signal tubes for driver applications, preamps, etc.
Steve |