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?

Answer 1

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:-

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

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

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

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