[Footnote 9: There is always some amplification taking place in an audion detector but the regenerative circuit amplifies over and over again until the signal is as large as the tube can detect.] LETTER 19 THE AUDION AMPLIFIER AND ITS CONNECTIONS DEAR SON: In our use of the audion we form three circuits. The first or A-circuit includes the filament. The B-circuit includes the part of the tube between filament and plate. The C-circuit includes the part between filament and grid. We sometimes speak of the C-circuit as the "input" circuit and the B-circuit as the "output" circuit of the tube. This is because we can put into the grid-filament terminals an e. m. f. and obtain from the plate-filament circuit an effect in the form of a change of current. [Illustration: Fig 96] Suppose we had concealed in a box the audion and circuit of Fig. 96 and that only the terminals which are shown came through the box. We are given a battery and an ammeter and asked to find out all we can as to what is between the terminals _F_ and _G_. We connect the battery and ammeter in series with these terminals. No current flows through the circuit. We reverse the battery but no current flows in the opposite direction. Then we reason that there is an open-circuit between _F_ and _G_. As long as we do not use a higher voltage than that of the C-battery which is in the box no current can flow. Even if we do use a higher voltage than the "negative C-battery" of the hidden grid-circuit there will be a current only when the external battery is connected so as to make the grid positive with respect to the filament. Now suppose we take several cells of battery and try in the same way to find what is hidden between the terminals _P_ and _F_. We start with one battery and the ammeter as before and find that if this battery is connected so as to make _P_ positive with respect to _F_, there is a feeble current. We increase the battery and find that the current is increased. Two cells, however, do not give exactly twice the current that one cell does, nor do three give three times as much. The current does not increase proportionately to the applied voltage. Therefore we reason that whatever is between _P_ and _F_ acts like a resistance but not like a wire resistance. Then, we try another experiment with this hidden audion. We connect a battery to _G_ and _F_, and note what effect it has on the current which our other battery is sending through the box between _P_ and _F_. There is a change of current in this circuit, just as if our act of connecting a battery to _G-F_ had resulted in connecting a battery in series with the _P-F_ circuit. The effect is exactly as if there is inside the box a battery which is connected into the hidden part of the circuit _P-F_. This concealed battery, which now starts to act, appears to be several times stronger than the battery which is connected to _G-F_.