FREE BOOKS

Author's List




PREV.   NEXT  
|<   69   70   71   72   73   74   75   76   77   78   79   80   81   82   83   84   85   86   87   88   89   90   91   92   93  
94   95   96   97   98   99   100   101   102   103   104   105   106   107   108   109   110   111   112   113   114   115   116   117   118   >>   >|  
ignals. There are just three possible answers to this question. The first answer is: "No, there is a smaller number of electrons passing through the plate circuit each second if the grid is being affected by an incoming signal." The second is: "The signal doesn't make any difference in the total number of electrons which move each second from filament to plate." And the third answer is: "Yes, there is a greater total number each second." [Illustration: Fig 56] Any one of the three answers may be right. It all depends on the characteristic of the tube as we are operating it, and that depends not only upon the type and design of tube but also upon what voltages we are using in our batteries. Suppose the variations in the voltage of the grid are as represented in Fig. 55, and that the characteristic of the tube is as shown in the same figure. Then obviously the first answer is correct. You can see for yourself that when the grid becomes positive the current in the plate circuit can't increase much anyway. For the other half of the cycle, that is, while the grid is negative, the current in the plate is very much decreased. The decrease in one half-cycle is larger than the increase during the other half-cycle, so that on the average the current is less when the signal is coming in. The dotted line shows the average current. Suppose that we take the same tube and use a B-battery of lower voltage. The characteristic will have the same shape but there will not be as much current unless the grid helps, so that the characteristic will be like that of Fig. 56. This characteristic crosses the axis of zero volts at a smaller number of mil-amperes than does the other because the B-batteries can't pull as hard as they did in the other case. [Illustration: Fig 57] You can see the result. When the grid becomes positive it helps and increases the plate current. When it becomes negative it opposes and decreases the plate current. But the increase just balances the decrease, so that on the average the current is unchanged, as shown by the dotted line. On the other hand, if we use a still smaller voltage of B-battery we get a characteristic which shows a still smaller current when the grid is at zero potential. For this case, as shown in Fig. 57, the plate current is larger on the average when there is an incoming signal. If we want to know whether or not there is any incoming signal we will not use the tube in the second
PREV.   NEXT  
|<   69   70   71   72   73   74   75   76   77   78   79   80   81   82   83   84   85   86   87   88   89   90   91   92   93  
94   95   96   97   98   99   100   101   102   103   104   105   106   107   108   109   110   111   112   113   114   115   116   117   118   >>   >|  



Top keywords:
current
 

characteristic

 

signal

 
smaller
 

average

 

number

 

answer

 

increase

 
voltage

incoming
 

batteries

 

larger

 

battery

 

negative

 

positive

 

decrease

 

dotted

 

Suppose


Illustration

 
circuit
 
answers
 

electrons

 
depends
 

amperes

 

question

 

crosses

 

potential


ignals
 
unchanged
 

result

 
increases
 

opposes

 

balances

 

decreases

 

filament

 

greater


represented

 

figure

 

correct

 

difference

 

coming

 

operating

 

design

 
passing
 

affected


voltages

 

decreased

 

variations