NEGATIVE VOLTAGE FEEDBACK AND THE STABILIZED GAIN AMPLIFIER 



grid, ions hitting the grid, leakage of current through surface moisture on 

 the valve envelope, and the whole lot combine to produce a curve having 

 the general shape of Figure 11.10. The slope of the curve at any particular 



HT+ 



Out of 

 valve 



Figure 11.10 



Figure 11.11 



value of bias gives the value of the input resistance. The input resistance is 

 positive between Q and R, infinite at Q and negative between P and Q. 



What does negative resistance mean? The power absorbed by a positive 

 resistance is V'^fR, so the power absorbed by a negative resistance is — V^jR, 

 that is, it does not absorb power, it generates it. We cannot have spurious 

 generators associated with the input circuits of measuring devices, and the 

 net input resistance must somehow be made positive. The spurious generator 

 is most powerful when the negative resistance is lowest. If this value is 

 — -Rj-min, then if we connect between grid and cathode an ordinary resis- 

 tance Rg, the resistance of the parallel combination is 



-R.R. 



Rg — Ri 



and this is positive if Rg < R^. 



This is the explanation for the maximum value of grid resistance permis- 

 sible in amplifying stages being stated by the valve manufacturers. It means 

 that the valve can be operated over the bias range corresponding to the PQ 

 section of the grid current characteristic without instability. Over the QR 

 range there is no difficulty anyway, as the input resistance is positive. 



Now let us see how this affects the cathode follower. Suppose we are 

 working in the region where the input resistance is positive. If we make the 

 grid positive, say, an amount dVg, then in an ordinary amplifying stage the 

 change in grid current is dig and the input resistance is R^ = dVgjdlg. In a 

 cathode follower having a gain of, say, 0-95, when we apply dVg we also pro- 

 duce a change of cathode potential dVj, = 0-95 dVg, and so the change of grid 

 current is only 0-05 dig. The apparent input resistance is thus <5KJ(0-05 dig), 

 20 times higher than when the valve is used as an amplifier. By a similar 



170 



