#### Question

A triode has mutual conductance of 2.0 millimho and plate resistance of 20 kΩ. It is desired to amplify a signal by a factor of 30. What load resistance should be added in the circuit?

#### Solution

Plate resistance, r_{p} = 20 kΩ

Mutual Conductance, g_{m }= 2.0 milli mho

g_{m }= 2 × 10^{−3} mho

Amplification factor, µ = 30

Load Resistance = R_{L} = ?

We know:-

\[A = \frac{\mu}{1 + \mathit{\frac{r_p}{R_L}}}\]

where A is the voltage amplification factor.

Also, amplification factor,

`mu=r_pxxg_m`

\[A = \frac{r_p \times g_m}{1 + \mathit{\frac{r_p}{R_L}}}\]

On substitution of all the given values, we get:-

\[ 30 = \frac{20 \times {10}^3 \times 2 \times {10}^{- 3}}{1 + \frac{20000}{R_L}}\]

\[ \Rightarrow 1 + \frac{20000}{R_L} = \frac{40}{30}\]

\[\frac{20000}{R_L} = \frac{1}{3}\]

\[ \Rightarrow R_L = 60000 \Omega = 60 k\Omega\]