At first glance, the resistor would seem the poorest choice as, unlike most of the other devices, it does not present a fixed voltage. But in fact, the resistor's current-dependent voltage drop is just what is needed to ensure the most consistently stable idle current.
    An individual triode differs from other triodes of the same type and, over time, it even differs from itself. In other words, the fixed voltage relationship between the cathode and grid that worked perfectly with one tube may not work so well with another triode of the same type or even with the same triode two years from now.
    By using a cathode resistor, however, we introduce a feedback mechanism into the circuit. As the triode increases in conduction, a larger voltage drop develops across the cathode resistor, which in turn makes the grid less positive relative to the cathode, reducing the triode's conduction. Conversely, as the triode decreases in conduction, a smaller voltage drop develops across the cathode resistor, which in turn makes the grid more positive relative to the cathode, increasing the triode's conduction. This is feedback of a truly negative sort. (In contrast to everyday speech, for most of electronic practice,
negative feedback is desired and positive feedback feared, as positive feedback can lead to dangerous oscillations.) The negative feedback results in a welcome increase in consistency.
     Okay, cathode bias seems like a good idea, but how do we chose the right value for the cathode resistor? Five approaches present themselves before the tube circuit practitioner.
     First,
look up the value (in a book or magazine -- or ask a friend).  While this is the most popular method, it is the least universally applicable, as many triodes are never mentioned and of those that are, only a few bias points are given.
     Second,
work out the value from a single simple formula. This does have a universal application, but it relies on an idealized model of the triode.
    Third,
extrapolate the value from inspecting the triode's plate curves. This method has the

     How can we alter this circuit to decrease the idle current, while retaining the same Vp? Two avenues present themselves. The first technique is to make the grid more negative than the cathode by connecting the grid to a negative power supply, such as the battery from the previous example. The advantage of a negative power supply is that a simple potentiometer can be used to adjust the idle current. How negative should this power supply be?
    Consider this: because the grid is mu times more effective than the plate in controlling current, making the grid voltage equal the B+ voltage divided by the mu and made negative, should turn the triode completely off. Expressed mathematically:
      Iq = 0 when Vgk = -Vb/mu.
(If triodes were perfect, then "should" would have been replaced with "will." Unfortunately, as good as triodes are, they are not perfect, particularly at the bottom of their conduction near cutoff. Still, this is a handy equation to memorize.)  Thus, the negative power supply need only be 1/mu times as large as the B+ power supply to ensure a wide range of current control.

      Cathode-Biased Voltage Dropping Devices

   The second technique to decrease the idle current is to make the cathode more positive than the grid by inserting a voltage drop between the cathode and ground. We could use a battery, an LED, a zener, a solid-state diode, vacuum tube diode, or even a second power supply, but a resistor is the most common choice. The larger the cathode resistor's value, the greater the effective negative bias voltage.
     So could a cathode resistor's value ever be so great as to turn off a triode? No, it never could be so large, as some current must flow to define a voltage drop across the resistor, so the triode could not be turned off.

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