06-11-2016, 05:40 PM
Hmmm... I'm not quite sure what you're asking Lawrence!
The mutual conductance is change in anode current, divided by change in grid voltage, for constant Va. So, go vertically from Q, each side, you get just over 4mA change for 4V grid change, and that's just over 1mA/V.
The anode resistance is change in anode voltage divided by change in anode current, for constant grid voltage. That's the same as the slope of the tangent at Q. The fact that the tangents at A, at Q, and at B are not quite parallel, says that ra isn't quite constant.
The mu is just ra x gm, which you can work out by multiplication. Alternatively, it's just change in anode voltage divided by change in grid voltage, for constant anode current. It is given by the intersections of the tangents with the horizontal, constant-current line at the operating point Q, divided by the grid voltage change needed to shift from one tangent to another. But it's a rather complicated way when just simple multiplication would be easier!
The mutual conductance is change in anode current, divided by change in grid voltage, for constant Va. So, go vertically from Q, each side, you get just over 4mA change for 4V grid change, and that's just over 1mA/V.
The anode resistance is change in anode voltage divided by change in anode current, for constant grid voltage. That's the same as the slope of the tangent at Q. The fact that the tangents at A, at Q, and at B are not quite parallel, says that ra isn't quite constant.
The mu is just ra x gm, which you can work out by multiplication. Alternatively, it's just change in anode voltage divided by change in grid voltage, for constant anode current. It is given by the intersections of the tangents with the horizontal, constant-current line at the operating point Q, divided by the grid voltage change needed to shift from one tangent to another. But it's a rather complicated way when just simple multiplication would be easier!
