PG2DTA: Determining the operating point and cathode bias resistor

In an earlier post I discussed how to choose a tube and an appropriate output transformer. The primary value of the transformer is known as the load and is often referred to as RL. The load can be drawn as a straight line on the grid curve graph and is called the loadline. The slope of the loadline can be determined using Ohm's Law again:

I = V / R

Using 5K ohms, the RL we determined for the 6V6 tube, and 100 volts an arbitrary value for V that makes the math easy, I can be calculated with the following equations:

I = 100 / 5000
I = 0.02
I = 20mA

Which means that the slope of the line rises 20mA every 100 volts. Each of the lines on the following graph has a slope of 5K.


The loadline is useful for determining a number of things when designing an amplifier. So, how do you know where to draw the load line? There are many approaches, but the following seems to work. Start by plotting the maximum plate dissipation on the grid curves as was done in my earlier post, Choosing a Power Tube and Transformer. Then draw a loadline that is tangent to the maximum plate dissipation curve. Why tangent? Well, if you go over the maximum plate dissipation curve, then you risk damage to the tube, premature failure or at least distortion. If you go under the curve, then you are wasting power.


If you look closely, you will see that the point at which the loadline is tangent to the curve is around 250V and 48mA. Of course, the easier way to get these values would be to look at the tube data sheet. According to the 6V6 datas heet, the Plate Voltage and Plate Current should be 250V and 49.5 mA - very close to what we came up with above.

These values are known as the quiescence or operating point, sometimes referred to as Iq and Vq. Once you know your operating point, you can calculate the cathode resistor. This is done by looking at the operating point and determining the grid voltage. The operating point does not intersect one of the grid lines already on the graph so you have to guess. In this case the point falls in the center of the -10V and -15V grid lines so -12.5V is a reasonable value.

Once you have the grid voltage, you determine the cathode resistor value using Ohm's law again:

R = V / I
R = 12.5 / 49.5
R = 252.525... Ohms

Once you have calculated the value, then look for a standard resistor value that is close to the calculated value. In this case, 255 Ohms.

The next question is what power rating should the resistor have. This is determined by calculating the power dissipated by the resistor and then doubling the value (at least) for safety. The power dissipation can be calculated using several different but related formulas:

P = I * I * R
P = .0495 * .0495 * 255
P = 0.625 watts

or

P = I * V
P = .0495 * 12.5
P = 0.619 watts

The values are different because the first formula uses the standard resistor value instead of the value originally calculated. Regardless, the values are close enough so that if you double the results you will end up with around 1.2 watts. There aren't a lot of 1.2 watt resistors out there, so round up to the next highest rating - 2 watts should be fine while 3 watts will be extra reliable.