Modern Physics 6 Question 39

44. A triode has plate characteristics in the form of parallel lines in the region of our interest. At a grid voltage of 1V the anode current I (in mA ) is given in terms of plate voltage V by the algebraic relation :

I=0.125V7.5

(1987, 7M)

For grid voltage of 3V, the current at anode voltage of 300V is 5mA. Determine the plate resistance (rp) transconductance (gm) and the amplification factor (μ) for the triode.

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Solution:

  1. Given, at Vg=1V

Ip=(0.125Vp7.5)×103AdIpdVp|Vg=1 volt =0.125×103A/Vrp=dVpdIp|Vg=1 volt =10.125×103Ω=8×103Ω=8kΩ

From the given equation,

Ip=(0.125×3007.5)mA( at Vg=1V and Vp=300V)=30mA

Now,

gm=ΔIpΔVg|Vp= constant =(305)(1)(3)|Vp=300 volt =12.5×103A/V

Amplification factor μ=rp×gm=100



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