Loudspeaker (Jan-Aug 1931)

tion curve. Not only are the low notes amplified more because there is no DC in the primary, but due to the resonant circuit we can increase them still more, and just where we want to, because we can shift this hump by varying the size of the condenser. Of course we cannot do so ourself in this particular transformer, but the manufacturer has taken advantage of this fact and adjusted the various values so that the humn will fall where it will do the most good, somewhere below 100 cycles. This is in the region where the amplifier becomes less efficient and the loudspeaker response also falls off, so that the rise due to the transformer serves to straighten out the curve of the entire system at the low end. Now that we know how the Clough transformer is built and wired let us see what the action is. DC cannot flow through the transformer because the condenser stops the flow of direct current, and when the tube is operating as it should, there is no current flowing between the grid and filament. Thus there is no DC whatever in the transformer, so the core does not become saturated. To get the signal into the transformer a current must be made to flow in the primary, from the tap to the terminal marked minus C in Fig. 2. Since we have eliminated the DC we have only the signal current to contend with, and it is AC. The isolating condenser readily passes the AC. If we impress an audio signal across the condenser and the negative C terminal it will be stepped up and transferred to the grid of the following tube. Since the source of C potential is bypassed by a large condenser, connected by the dotted lines from C minus to the filament, we can connect one terminal of the audio input to the filament of the tube instead of C minus. As you will remember the variations of the plate current constitute the signal, and these must be made to vary the charge on the isolating condenser. This is where the coupling resistance comes in. The first tube in Fig. 2 draws a plate current of 2.5 mils, which also flows through the coupling resistance. Whenever a current flows through a resistance a drop in voltage takes place. According to Ohm’s law the drop equals the current times the resistance; 2.5 mils (.0025 amperes) times 20,000 ohms. This gives us 50 volts. The B supply is furnishing 90 volts to this circuit, 50 are consumed in the coupling resistance, so there are 40 volts left to supply the plate of the tube. That means a difference of 40 volts exists between the plate and the filament of the tube. One terminal of the isolating condenser being connected to the plate is also at a potential of 40 volts in respect to the filament. The other side of the condenser is 40 volts negative in respect to the filament. The negative 40 volts is due to the bias put on the grid of the following tube. (40 volts is the proper bias for a ’71 type tube). These two voltages being in series add up, making 80 volts between the two sides of the. condenser. Since bias voltage in itself does not affect the operation of the transformer we will ignore it, as it will simplify the explanation. So we say there are 40 volts across the condenser. Let us see just what takes place. When the amplifier is dead, there is FIG. 3 no charge on the condenser. Zero voltage across it. When we light the amplifier a current flows from one (Continued on Page 46) T zv e n ty -one