Question

Two triodes having amplification factor 30 and 21 and plate resistance $5k\Omega $ and $4k\Omega $ respectively are connected in parallel. The composite amplification factor of the system is
$\begin{align}
  & A.25 \\
 & B.50 \\
 & C.75 \\
 & D.100 \\
\end{align}$

Answer 1

Hint: Amplification factor of a triode is nothing but the factor by which an input signal is enhanced. For a triode, amplification factor is the rate of change of anode current with respect to grid voltage. It can be calculated as the product of transconductance and plate resistance. The effective transconductance of two triodes connected in parallel is simply the sum of individual transconductance. The effective plate resistance of two triodes connected in parallel can be calculated by the formula ${{r}_{P}}=\dfrac{{{r}_{{{P}_{1}}}}{{r}_{{{P}_{2}}}}}{{{r}_{{{P}_{1}}}}+{{r}_{{{P}_{2}}}}}$where, ${{r}_{{{P}_{1}}}}$ and ${{r}_{{{P}_{2}}}}$ are the individual plate resistances.

Complete answer:
A triode is an electronic device used for amplification purpose. It consists of a hot filament which acts as a cathode and it releases electrons when it is heated sufficiently. It also has a metal plate electrode which acts as anode and attracts the electrons emitted by cathode.
If $\mu $ is the amplification factor of a triode, ${{r}_{P}}$is the plate resistance then
$\mu ={{r}_{P}}{{g}_{m}}$
Where ${{g}_{m}}$ is the transconductance of the triode.
Therefore,
${{g}_{m}}=\dfrac{\mu }{{{r}_{P}}}$
Given that
$\begin{align}
  & {{\mu }_{1}}=30 \\
 & {{r}_{{{P}_{1}}}}=5k\Omega =5\times {{10}^{-3}}\Omega \\
\end{align}$
$\begin{align}
  & {{\mu }_{2}}=21 \\
 & {{r}_{{{P}_{2}}}}=4k\Omega =4\times {{10}^{-3}}\Omega \\
\end{align}$
Substituting this values in above equation, we get,
${{g}_{{{m}_{1}}}}=\dfrac{{{\mu }_{1}}}{{{r}_{{{P}_{1}}}}}=\dfrac{30}{5\times {{10}^{-3}}}=6\times {{10}^{-3}}mho$
and
${{g}_{{{m}_{2}}}}=\dfrac{{{\mu }_{2}}}{{{r}_{{{P}_{2}}}}}=\dfrac{21}{4\times {{10}^{-3}}}=5.25\times {{10}^{-3}}mho$
Therefore, effective transconductance is
$\begin{align}
  & {{g}_{m}}={{g}_{{{m}_{1}}}}+{{g}_{{{m}_{2}}}} \\
 & {{g}_{m}}=(6+5.25)\times {{10}^{-3}}mho \\
 & {{g}_{m}}=11.25\times {{10}^{-3}}mho \\
\end{align}$
Since triodes are connected in parallel, there effective resistance is given as
$\begin{align}
  & {{r}_{P}}=\dfrac{{{r}_{{{P}_{1}}}}{{r}_{{{P}_{2}}}}}{{{r}_{{{P}_{1}}}}+{{r}_{{{P}_{2}}}}} \\
 & {{r}_{P}}=\dfrac{5\times 4}{5+4}k\Omega \\
 & {{r}_{P}}=\dfrac{20}{9}k\Omega \\
 & {{r}_{P}}=\dfrac{20}{9}\times {{10}^{-3}}\Omega \\
\end{align}$
Therefore, the composite amplification factor is given by
$\begin{align}
  & \mu ={{r}_{P}}{{g}_{m}} \\
 & \mu =11.25\times {{10}^{-3}}\times \dfrac{20}{9}\times {{10}^{-3}} \\
 & \mu =25 \\
\end{align}$
Thus, the composite amplification factor is 25.
Answer - A. 25

Note: The amplification factor of a triode is independent of the voltage at its anode and cathode. It depends on its geometric conditions. It is the product of transconductance and its plate resistance. The value of amplification factor generally ranges from 10 to 100. The transconductance of grid plates is also termed as mutual conductance. Note that amplification factor is a unitless and dimensionless quantity.