SATHEE: Electronic Devices Question 265

Electronic Devices Question 265

Question: On applying a potential of - 1 volt at the grid of a triode, the following relation between plate voltage Vp (volt) and plate current $I_{p} (in m A)$ is found $I_{p} = 0.125 V_{p} - 7.5$ If on applying - 3 volt potential at grid and 300 V potential at plate, the plate current is found to be 5mA, then amplification factor of the triode is

Options:

A) 100

B) 50

C) 30

D) 20

Show Answer

Answer:

Correct Answer: A

Solution:

At $V_{g} = - 3 V, V_{p} = 300 V$ and $I_{p} = 5 m A$ At $V_{g} = - 1 V,$ for constant plate current i.e.

$I_{p} = 5 m A$ From $I_{p} = 0.125 V_{p} - 7.5$

Therefore $5 = 0.125 V_{p} - 7.5$

Therefore $V_{p} = 100 V$

hange in plate voltage $Δ V_{p} = 300 - 100 = 200 V$

Change in grid voltage $Δ V_{g} = - 1 - (- 3) = 2 V$ So, $μ = \frac{Δ V_{p}}{Δ V_{g}} = \frac{200}{2} = 100$

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Electronic Devices Question 265

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