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Why? What has happened is that product of this triode's mu (about 10) against its grid's -6 volts (-60 volts) has been added to the 100 volts to yield 40 volts as the effective plate voltage, which when divided by 714 ohms results in a current draw of 56 mA. (The grid line curves enough to make up the 6 mA difference.) Had we focused on the 0 volt grid line instead, then the plate resistance would have been much more accurately directly deduced from dividing the plate voltage by the plate current along any point on the gridline. (Once again, the lines curve enough to introduce a small error.)
Determining the value of the required cathode resistor to set the idle to 50 mA at 100 volts seems obvious enough: just divide 6 volts by 50 mA; unfortunately, this method gives an inaccurately high resistor value, as the full 100 volts will no longer be available to the triode once the cathode resistor displaces 6 volts of the B+ voltage. So the set of curves should be mentally shifted -6 volts to the right. For most beginners this is asking too much. The quick workaround is to take the apparent bias voltage (-6 volts, in this example) and divide it by the reciprocal of the triode's mu (10, in this example) and then subtract this value from the apparent negative bias voltage, which when simplified and expressed as a formula becomes:
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Vbias´ = Vbias(1- 1/mu)
Rk = Vbias´ / Iq
In this case, the -6 volts becomes 5.4 volts, which divided by 50 mA equals 108 ohms, the correct value. Obviously, the higher the mu, the smaller the adjustment becomes; consequently, with a high-mu tube like the 12AX7, often no workaround is sought, but it is essential with low mu triodes, such as the 2A3, 6AS7, 6BX7, 6C33, 300B and 845.
The previous examples relied on circuits without plate resistors. Factoring in the role played by the plate resistor only slightly increases the complexity of the task. The formula for determining the cathode resistor's value must be modified:
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where, once again, a resulting negative value betrays that the specified idle current cannot be met under the cathode bias configuration and a lower-rp triode or a lower-valued plate resistor is required.
Graphing the added plate resistor is easy enough, as it relies on the method used to plot the two resistors in series. We start at the B+ voltage, which in this example we will set at 200 volts, and place our first point.
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