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 $-1 V$ the anode current $I$ (in $mA$ ) is given in terms of plate voltage $V$ by the algebraic relation :
$$ I=0.125 V-7.5 $$
(1987, 7M)
For grid voltage of $-3 V$, the current at anode voltage of $300 V$ is $5 mA$. Determine the plate resistance $\left(r _p\right)$ transconductance $\left(g _m\right)$ and the amplification factor $(\mu)$ for the triode.
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Solution:
- Given, at $V _g=-1 V$
$$ \begin{aligned} & I _p=\left(0.125 V _p-7.5\right) \times 10^{-3} A \\ & \left.\frac{d I _p}{d V _p}\right| _{V _g=-1 \text { volt }}=0.125 \times 10^{-3} A / V \\ & \therefore \quad r _p=\left.\frac{d V _p}{d I _p}\right| _{V _g=-1 \text { volt }}=\frac{1}{0.125 \times 10^{-3}} \Omega \\ & =8 \times 10^{3} \Omega=8 k \Omega \end{aligned} $$
From the given equation,
$$ \begin{aligned} I _p & =(0.125 \times 300-7.5) mA \\ \left(\text { at } V _g\right. & \left.=-1 V \text { and } V _p=300 V\right)=30 mA \end{aligned} $$
Now,
$$ \begin{aligned} g _m & =\left.\frac{\Delta I _p}{\Delta V _g}\right| _{V _p=\text { constant }} \\ & =\left.\frac{(30-5)}{(-1)-(-3)}\right| _{V _p=300 \text { volt }} \\ & =12.5 \times 10^{-3} A / V \end{aligned} $$
Amplification factor $\mu=r _p \times g _m=100$
