Say the plate had to be lowered by 40V to get the same current. 005/1 = 5000µmhos.Īmplification factor (symbol "µ" (Greek mu)) - Say you took the same change in grid voltage, +1V, and adjusted plate voltage so the current is the same as before. Say you increase the grid voltage by 1V, and plate current rose 5mA as a result (plate voltage is kept constant). Transconductance = Gm = mho (unit of conductance) or S (Seimens), usually in micro- or mili- units (3200µmhos = 3.2mS). Say you increase the plate voltage 20V while keeping the grid voltage the same if the plate current increased by 5mA, the plate resistance at this average operating point is 20/.005 = 4kohms. Rp = Ra = Plate Resistance, the effect a change in voltage has on current. Oh yeah, it sounds like you need to understand the basic parameters of tubes. The current or voltage may be too high to be useful for an RC-coupled stage. Using published operating points isn't usually a good starting point because they might've been drawn up for say, transformer coupling. In the mean time, a plate supply resistor should be more than 3 times the plate resistance at your chosen operating point (plate voltage, current, grid voltage), and should drop between half and 2/3 of the supply voltage for best linearity, leaving the other 1/2 to 1/3 of the supply voltage for the tube. Then all becomes clear and easy to calculate. It really helps to have an understanding of plate curves. I'm sure that they will clarify what I'm trying to say. I'm not in Frank's or Brett's League but I hope that I helped you instead of confusing you. So these are the operating points I am using for a 5687. These operating points will give me a pretty good circuit. Now to drop 150 volts at 10 mA I need a resistor value of 150/.01=15000. Which meand that my B+ is going to be 150 +150 = 300 volts. I like to drop half of the B+ across the plate resistor and the orher half across the tube/cathode resistor. So now you have the operating points of your circuit but you still need to determine the plate resistor value. This means that I will need a cathode resistor of 6.8/.01 = 680 ohms. Looking at the curves on the datasheet I see that in order to have 10 mA current flow with a plate voltage of 150 volts I need a negative voltage (or a voltage drop across the cathode resistor) of about 6.8 volts or so. The circuit that I am using requires 150 volts on the plate at 10 mA. Let's take the 5687 circuit for an example. The DC function is to drop the B+ to the appropriate value for the tube at a given level of current. Remember that that is the AC function of a plate resistor. The value of the plate resistor determines how much signal will flow through the tube to ground and how much will flow through the next stage. If you don't use a big enough plate resistor then all of your signal is going to go through the tube straight to ground and no signal will flow to the next stage. The plate resistor determines how much signal is going to the next stage or component in the same way. If one resistor is 10 ohms and the other is 1 ohm then 90% of the current(signal) is going to flow through the 1 ohm resistor and only 10% through the 10 ohm resistor. The plate resistor is what makes the tube work. While I'm certainly not an expert the rule of thumb that I have followed is to use a plate resistor of at least three time the plate resistance of a triode.
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