This is a quick instruction on how to calculate an essential triode cascade. It gives you values of Ra, Rc, and Rg. For example, a double triode tube – 6N2P
Using a datasheet for a specific tube, define the nominal (or max, if needed) parameters:
- The voltage on the plate (anode), Ua
- Anode current, Ia
These parameters are boundaries and two points on the chart of the load line above, see the red line.
6N2P:
Ua = 250V
Ua = 250V
Ia = 3,2mA
Anode current is a nominal (or max) current that will flow through the tube, and should not be exceeded. When the tube is in the cut-off region state, the internal resistance ri is high and almost all voltage from the power supply will be applied to the tube plate. Thus, the voltage from the power supply should be set close to the nominal anode voltage of the selected tube. If a DC-rectifier provides to high value, it should be reduced using a voltage drop resistor with a capacitor, which simultaneously formed an RC filter.
Ra with ri forms a voltage divider, so in the practice, some amount of voltage will drop on the load Ra resistor if a tube is not in the deep cut-off region.
Ra = Ua/Ia = 250V / 3,2mA = 78K
in the saturation state (tube is ON, the current is flowing)
voltage drop on the Ra:
78k * 3,2mA = 249,6 = 250V that is U, Ua
in the cutoff
voltage drop on the Ra:
78k * 0,25mA = 19,5V
*78k value of anode resistor gives us a non-linear amplifier, the best load for the linear 6n2p amp is 10-20K, see right char.
Operating point
Define the tube operating point on the chart, the operating point will move up and down on the load line. The best places are in the middle of the linear region. An operating point in practice defines an offset voltage on the grid, Ug.
For exp.:
Ug = -0,5V
look at the branch with the blue lines. Also, it could be 0,5, 0, -0,75 or -1V...
How can we get the offset voltage on the grid? It does an Rc resistor.
Rc = Ug/Ia (op. point) = -0,5V / 1,75mA = 0,29K
Grid leakage resistor
Rg is a grid leakage resistor, which is defined in the datasheet from 100 to 1M.
The grid has a negative charge. The flow of electrons could cause a huge negative voltage potential between grid and cathode, this process will move the operational point (bias) in the cutoff region, so the grid leak resistor is used as discharging path for accumulated electrons.
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Tube in the cutoff, Ia -> min
ri = Ua - (Ra * Ia) / Ia = 250v - (78k * 0,25mA) / 0,25mA = 922K
Tube in the saturation region, Ia -> max
ri = Ua - (Ra * Ia) / Ia = 250v - (78k * 3,2mA) / 3,2mA = 0,125K
