Semiconducting Devices Question 63
Question: The plate current in a triode is given by $ I _{p}=0.004\ {{(V _{p}+10V _{g})}^{3/2}}mA $ where Ip, Vp and Vg are the values of plate current, plate voltage and grid voltage, respectively. What are the triode parametersm, rp and gm for the operating point at $ V _{p}=120volt $ and $ V _{g}=-2volt $ ?
Options:
A) 10, 16.7 kW, 0.6 m mho
B) 15, 16.7 kW, 0.06 m mho
C) 20, 6 kW, 16.7 m mho
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I _{p}=0.004{{(V _{p}+10V _{g})}^{3/2}} $
Therefore $ \frac{\Delta I _{p}}{\Delta V _{g}}=0.004[ \frac{3}{2}{{(V _{p}+10V _{g})}^{1/2}}\times 10 ] $
Therefore $ g _{m}=0.004\times \frac{3}{2}{{(120+10\times -2)}^{1/2}}\times 10 $
Therefore $ g _{m}=6\times {{10}^{-4}}mho=0.6mmho $
Comparing the given equation of Ip with standard equation $ I _{p}=K{{(V _{p}+\mu V _{g})}^{3/2}} $
we get m = 10 Also from m = rp x gm
Therefore $ r _{p}=\frac{\mu }{g _{m}}=\frac{10}{0.6\times {{10}^{-3}}} $
Therefore $ r _{p}=16.67\times 10^{3}\Omega =16.67k\Omega .$
