Two identical triode tubes are connected in parallel, as shown in...

For a single triode, the small change in plate current $\Delta i_p$ is given by:

$\Delta i_p = g_m \Delta v_g + \dfrac{1}{r_p} \Delta v_p$

When two identical triodes are connected in parallel, their grids, plates, and cathodes are joined together. Therefore, they share the same grid voltage change $\Delta v_g$ and plate voltage change $\Delta v_p$.

The total plate current change $\Delta i_{p,eq}$ is the sum of the individual plate current changes:

$\Delta i_{p,eq} = \Delta i_{p1} + \Delta i_{p2}$

Since the triodes are identical, substituting the expressions gives:

\Delta i_{p,eq} = 2 \left( g_m \Delta v_g +\dfrac{1}{r_p} \Deltav_p \right)

\Delta i_{p,eq} = (2g_m) \Delta v_g + \left(\dfrac{2}{r_p} \right) \Delta$v_p$

For the equivalent triode, the equation is:

$\Delta i_{p,eq} = g_{m,eq} \Delta v_g +$\dfrac{1}{r_{p,eq}} \Delta$v_p$

Comparing the coefficients, we obtain:

$g_{m,eq} = 2g_m$

\dfrac{1}{r_{p,eq}} = \dfrac{2}{r_p} \Rightarrow$r_{p,eq} =$\dfrac{r_p}{2}

The equivalent amplification factor $\mu_{eq}$ is given by the product of equivalent mutual conductance and equivalent plate resistance:

$\mu_{eq} = g_{m,eq} \times r_{p,eq}$

\mu_{eq} = (2g_m) \times \left(\dfrac{r_p}{2} \right) =$g_m r_p$

Since $\mu = g_m r_p$ for a single triode, we have:

$\mu_{eq} = \mu$

Thus, the equivalent parameters are \dfrac{r_p}{2}, $2g_m$, and $\mu.$