Fig. 101. Negative feedback from plate to grid makes the 6AQ5's characteristic
(shown in solid lines) look like those of a triode tube (broken lines).
BEFORE we consider the amplifier as a whole, it is rather instructive to see what happens to a single tube with feedback. For an example which is reasonably typical, we have in Fig. 101 the characteristics of a 6AQ5 tube: a miniature which will give you 3-4 watts. It is a pentode, and at the standard working point, Eb = 250 volts, E g =. -12.5 volts, I_b = 45 ma, the transconductance is about 4,000 u-mhos and the impedance about 50,000 ohms. This last figure is the real problem: the optimum load is 5,000 ohms, so that if you transform down to use a 15-ohm loudspeaker, the speaker, will be looking back at 150 ohms.
It might as well be an open circuit, as far as damping down speaker resonances goes.
Let us see what happens to the tube characteristics if we apply negative feedback. To make the calculations easy, let us apply feedback of 1/10 of any plate voltage change to the grid. Starting at the point E1 = 0, Eb = 300, let us reduce E, to -5 volts; then if Eb drops by 50 volts we shall feed back +5 volts to the grid, and we shall be on the same E, line as the starting point. Similarly, the point E1 = -10, Eb = 300-100 is on this Eg = 0 line. We can follow this same idea through and get a sufficient number of points to give us a smooth curve, as shown by the dashed lines in Fig. 101. The same technique can be employed to produce a family of curves.
The feedback tube curves appear as broken lines. If you will for get that we have a pentode you will see that the new characteristics are those of a triode, with an impedance of 2,100 ohms at the original working point. If you connect this plate-grid feedback around a pentode and put it in a black box, anyone measuring the characteristics will think they have a triode. There is only one difference: you can swing this triode down to a plate voltage of 50 without running into grid current; the corresponding limit when the screen and plate are joined together is about 170 volts.
Fig. 102. Negative current feedback will alter the curves of a triode
so they will take on the high-impedance characteristic of a pentode (broken
lines).
The results of Fig. 101 are well worth a closer study. In the area below E1 = 0 there is no grid current, because the feedback circuit will be an a.c. coupling. A pentode with negative feedback of this kind has the same characteristics as a triode, including the triode positive grid region, but you can work over the whole set of characteristics without grid current trouble. This is stressed because there are audio men who do not believe that tetrodes can sound as good as triodes.
Cast your eye over those curves and you can see they are better than triode curves! We can, to some extent, reverse this effect. A typical triode characteristic is shown in Fig. 102. This triode has a mu of about 50, and an impedance of about 20,000 ohms. By putting in a cathode resistor of 2,000 ohms we can make a change of 1 ma in the plate current feed back 2 volts to the grid. The effect of this is shown in part by the broken lines in Fig. 102. These are part of the characteristics of a "black box" tube having an impedance of about 105,000 ohms, which is getting on toward the pentode class. If you need a pentode, and only have a triode handy, this is one way of making the circuit think it sees a pentode.
Fig. 103. Voltage feedback network used for calculating the output impedance.
Fig. 104. Network using current feedback.
These two examples of modified tube characteristics have been considered because they provide a useful background to the general discussion of amplifier impedances. Ultimately any amplifier can be considered as a 4-terminal network in a box, and if you don't look inside you cannot be certain that it is not just one tube, with a transconductance of 1 amp/volt, perhaps! Now we can turn to the general amplifier circuit. Fig. 103 shows the general voltage feedback amplifier, with input short-circuited and output connected to a generator.
This generator produces a voltage of E0 at the output terminals, and the current flowing into the amplifier is I. The amplifier itself has an impedance of Ro and an open-circuit gain E2/E1 of Ko. (The gain of the amplifier without feedback is K). Notice that you are using the open-circuit gain which will be quite a lot higher than the usual loaded gain. For a triode the difference is 6 db, but for a pentode or tetrode it may be much more. The feedback network p is assumed to be of such high impedance that it does not affect the impedances.
In the output mesh, we have this equation:
E0 -E2 = 1 oRo
Now, E2 = K0E1 and since in this particular case the only input is that provided by the feedback network E1 = pE3 or, indeed, -flE0 since E3 and E0 are the same here.
Arranging our equations in order, we have:
Thus -E2 = KopEo and we can substitute in the previous equation:
Of course E0/I0 is the impedance seen by the generator connected to the output, and if the feedback were absent, p = 0, the impedance would be Ro. With negative feedback the impedance is reduced by the factor (1 + KO).
This feedback is voltage feedback. We could use a circuit like that of Fig. 104, in which the feedback voltage depends on the current in the output circuit. To keep the formulas very simple, the resistance across which the feedback is picked off will be assumed to be small, just as before we assumed that the /3 network was of infinite impedance.
In calculating this circuit we work in terms of current: the amplifier is assumed to have a transconductance of A, under short-circuit conditions, so that it produces an output current of A0E1 for input of E1.
When we apply an additional current of I o we have a current of (I0 A0E1 ) through Ro, so that the voltage across the output terminals is: E2 = Ro (Io A0E1)
The feedback network (including the small resistor across which E3 is produced) delivers a voltage plc , to the input terminals, and since the input has no other supply E1
Thus E2 = Ro (Io A019I0). The admittance seen at the output is 1 0/E2 and is:
1 Ro (1 Ale)
r -1 X 1 o Ro +
If there were no feedback (p = 0) this would be just 1/Ro. With current negative feedback, therefore, the output admittance is reduced by the factor (1 + A0 /3). This means, of course, that the impedance is increased by this factor.
Fig. 105. Six ways to pick off feed back F. a,b, and c, are voltage
feedback; others are current feedback.
Table for Phasing Feedback Connections --- Circuits b and f depend on transformer sense.
It must be noted that the factor Ao is not the same as Ko. This means that we must make a separate calculation when determining the output impedance. When using voltage feedback we must use the equation
M = R
for the gain of the last stage in determining the gain factor K, but Mo
= µ for the impedance factor Ko. If, however, the feedback network uses resistors of normal value, we may find it more accurate to write:
R/ +Ro
where R/ is the input impedance of the feedback network. We also ought to bring the output transformer losses into R/. Usually this means that K0 is about 3 times the usual value of K. By using positive feedback, the term (1 + K0 beta) can be made less than unity. Thus, positive voltage feedback increases the output impedance and positive current feedback reduces the output impedance. We can mix positive voltage feedback and negative current feedback to give a very high impedance, for example, without losing too much gain.
As examples of what can be done, one small amplifier uses a 12AT7, which has an impedance, at the high side of the output transformer, of something around 5 megohms. This is better than pentode performance. Another amplifier, using pentodes, gives an impedance of 0.1 ohm when designed to work into a 25-ohm load: this uses positive current and negative voltage feedbacks. There are good reasons for these designs: the pentode is needed to get power at low supply voltage, the 12AT7 to get gain from a single bottle, and the extreme impedances are "musts." The circuits for giving voltage and current feedback are summarized in Fig. 105. F is the feedback voltage in each case, and a, b, c are voltage feedback, d, e, f are current feedback. The table is a summary showing whether the feedback should go to cathode (k) or grid (g) of an earlier stage to be negative or positive. The plus signs show positive feedback, the minus signs negative. Thus in a 3-stage circuit you must feed back from cathode to cathode (circuit d) or from anode to grid (circuit e) to get negative current feedback.
For the sake of completeness we should examine the bridge feed back circuit. In this the feedback does not affect the impedance. Tube impedances are not so constant that anyone wants to keep them unaffected.
Now let us consider the input impedance. A typical input circuit is shown in Fig. 106, and although you will notice a resistor between grid and cathode, this could be a more complicated impedance. The biasing arrangements are also neglected. In this circuit, the input to the amplifier section K is V1, and the voltage developed by the feedback circuit is V2 = K ei. We must have E0 = E1 E2 = (1 + K/3) E1 and from this we could deduce that the gain would fall by a factor (1 ± Ki e) when feedback was added, because we now need an input of E0 to produce E1 at the grid.
The current which flows through R is given by I 0 = Ei/R = E0/R (1 ± KM. So far as any circuit connected to the input can tell, this current is produced by E0, so that the input impedance is:
E0 - 10 = R (1+ V1 p).
Fig. 106. Amplifier input circuit having series-connected feedback at
cathode.
The impedance therefore has been increased by a factor (1 ± Kf3). This may be 10 to 100, so that a resistance of 1 megohm, which is all we can normally use because of gas current in the tube, looks like 10-100 megohms to the external circuit. The input capacitance is reduced, too, which is important if you are using a crystal pickup.
Notice, however, that positive feedback in this connection will reduce the input impedance, since then (1 1-- Kfl) is less than unity.
An alternate way of connecting the feedback is shown in Fig. 107.
The current which flows into this circuit is
The R1 term is the impedance of the external generator, so that the input impedance, R2 without feedback, is reduced by the factor (1 + Kg) with negative feedback. As before, if the feedback is positive this is an increase in input impedance. It would seem to be quite possible to apply negative feedback to the cathode and positive to the grid to produce an extraordinarily high input impedance. The negative feed back would stabilize the gain so that the term (1 - Kfl) in the positive feedback equation could be made quite small.
Fig. 107. A parallel hookup for feedback.
In many applications it is necessary to provide a controlled value of input impedance, which involves the use of the circuit of Fig. 106 with an additional shunt resistor across the input. Often, too, we need a good 600-ohm output impedance, and then a high impedance shunted by 600 ohms is used.
This has only been a general survey of the problems of impedance control. If an amplifier is needed with some special impedance properties, a closer study may be required, but for almost all jobs the discussion here will be sufficient. Always the effect of feedback is to modify the apparent tube characteristics, and it is probably wise to point out one important thing. If you turn a tetrode into a triode by means of feedback, the optimum load is unchanged: you should not try to match this "triode." You can check this statement very care fully, both with modified characteristics and by actual experiment.
This is specially important if you are using a high degree of feedback to get a low impedance for damping a loudspeaker, or to get a high impedance for some other purpose. And don't forget, if you want a low impedance, to allow for the resistance of the output transformer windings. This is also important.