Semiconducting Devices Question 56
Question: The plate current ip in a triode valve is given $ i _{p}=K{{(V _{p}+\mu 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}}mho, $ then for $ i _{p}=8mA $ and $ V _{p}=300volt, $ what is the value of K and grid cut off voltage
[Roorkee 1992]
Options:
A) - 6V, (30)3/2
B) $ -6V,{{(1/30)}^{3/2}} $
C) + 6V, (30)3/2
D) + 6V, (1/30)3/2
Show Answer
Answer:
Correct Answer: B
Solution:
$ \mu =r _{p}g _{m}=50 $ From $ i _{p}=KV _{p}^{^{3/2}} $
Therefore $ \frac{\Delta V _{p}}{\Delta i _{p}}=r _{p}=\frac{2i _{p}^{-1/3}}{3{{K}^{2/3}}} $
Therefore $ g _{m}=\frac{\mu }{r _{p}}=\frac{3\mu {{K}^{2/3}}i _{p}^{1/3}}{2} $
$ =\frac{3}{2}\mu {{K}^{2/3}}[ {{K}^{1/3}}{{(V _{p}+\mu V _{g})}^{1/2}} ] $
$ =\frac{3}{2}\mu K{{(V _{p}+\mu 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}}{\mu }=-\frac{300}{50}=-6V $
