1.

Figure 33.5 depicts the grid characteristics of a triode plotted at anode voltages of 450 V and 600 V. Find the triode's internal resistance R_(i) in the linear section of the characteristic and its amplification factor mu, i.e. the ration of the change in the anonde voltage to the change in the grid voltage which cauases a given change in the anode current.

Answer»


Solution :Calculations should be done for the linear section of the CHARAC. teristic CURVE. In this case a change in the anode voltage `DeltaU_(a)=150` V, with a constant grid voltage (for instance, with `U_(g)=0` ), causes a change in the anode current of `Deltai_(a)=75mA`. The tube.s internal resistance is `R_(i)=(DeltaU_(a))/(Deltai_(a))` The tube.s AMPLIFICATION factor is `mu=(DeltaU_(a))/(DeltaU_(g))`, i.e. `mu` is the RATIO of the change in the anode voltage to the change in the grid voltage which causes a given change in the anode current. It follows from Fig. that a change of anode current equal to 75 mA may be obtained either by changing the anode voltage by `DeltaU_(a)=150V`, or by changing the grid voltage by `DeltaU_(g)=7.5` V. Therefore the grid is 20 times more effective in controlling the current flowing through the tube on the linear section of its characteristic curve), i.e. amplification factor `mu=20`


Discussion

No Comment Found

Related InterviewSolutions