SATHEE: Electronic Devices Question 252

Electronic Devices Question 252

Question: The plate current ip in a triode valve is given $i_{p} = K {(V_{p} + μ V_{g})}^{3 / 2}$ where ip is in milliampere and Vp and Vg are in volt. If rp = 104 ohm, and $g_{m} = 5 \times 10^{- 3} m h o,$ then for $i_{p} = 8 m A$ and $V_{p} = 300 v o l t,$ what is the value of K and grid cut off voltage

[Roorkee 1992]

Options:

A) - 6V, (30)3/2

B) $- 6 V, {(1 / 30)}^{3 / 2}$

C) + 6V, (30)3/2

D) + 6V, (1/30)3/2

Show Answer

Answer:

Correct Answer: B

Solution:

$μ = r_{p} g_{m} = 50$ From $i_{p} = K V_{p}^{^{3 / 2}}$

Therefore $\frac{Δ V_{p}}{Δ i_{p}} = r_{p} = \frac{2 i_{p}^{- 1 / 3}}{3 K^{2 / 3}}$

Therefore $g_{m} = \frac{μ}{r_{p}} = \frac{3 μ K^{2 / 3} i_{p}^{1 / 3}}{2}$

$= \frac{3}{2} μ K^{2 / 3} [K^{1 / 3} {(V_{p} + μ V_{g})}^{1 / 2}]$

$= \frac{3}{2} μ K {(V_{p} + μ V_{g})}^{1 / 2}$ = 75 K (ip/K)1/3

Because ip was in mA, gm is substituted as 5 m℧

Therefore $5 = 75 k^{2 / 3} i_{p}^{1 / 3}$

$= 75 k^{2 / 3} {(8)}^{1 / 3}$

Therefore $k = {(\frac{1}{30})}^{3 / 2}$

Cut off grid voltage $V_{G} = - \frac{V_{p}}{μ} = - \frac{300}{50} = - 6 V$

Electronic Devices Question 252

Options:

Answer:

Solution:

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Electronic Devices Question 252

Question: The plate current ip in a triode valve is given ip=K(Vp+μVg)3/2 where ip is in milliampere and Vp and Vg are in volt. If rp = 104 ohm, and gm=5×10−3mho, then for ip=8mA and Vp=300volt, what is the value of K and grid cut off voltage

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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