# How to calculate output resistance in collector circuit?

Calculate output resistance $r_a$.

Here is the solution:

$$r_a=R_E || \left(\frac{1}{g_m}+r_a'\right)$$

$$r_a'= \frac{U_e}{-i_c}=\frac{U_e}{-\beta i_B}=\frac{R_G || R_B || R_V}{\beta}$$

I don't understand the solution. What is $r_a'$ here? Can I draw $r_a'$ as resistor instead of $1/g_m$ and $R_E$? Can someone explain this solution?

The output resistance $r_a$ is the parallel combination of $R_E$ and the resistance looking into the emitter of the transistor. As shown in the solution drawing, the resistance looking into the emitter of the transistor is $1/g_m$ plus some other resistance due to the source resistance $R_G$ and the biasing resistors $R_B$ and $R_V$. That "other resistance" caused by $R_G$, $R_B$, and $R_V$ is denoted by a convenience term that the solution calls $r_a'$. In equation form, this means that $$r_a = R_E || \left(\frac{1}{g_m}+r_a'\right)$$
Now we need to calculate $r_a'$. In small signal analysis the DC sources are turned off so $U_0 = U_G = 0$, and this means that $R_G$, $R_B$, and $R_V$ end up in parallel with each other. The parallel combination of these three resistors is shown in the solution in the dotted box with the value $R$. $r_a'$ is the parallel combination of these resistors but divided by a factor of approximately $\beta$, the small signal current gain (since the current through the three resistors is $i_b$ but we were calculating the resistance looking into the emitter). Therefore $$r_a'= \frac{R_G || R_B || R_V}{\beta}$$
You can substitute the equation for $r_a'$ into the equation for $r_a$ to get $$r_a = R_E || \left(\frac{1}{g_m}+\frac{R_G || R_B || R_V}{\beta}\right)$$
The transistor model this solution uses is a little different than the commonly used hybrid-$\pi$ model which uses a resistance from base to emitter $r_{\pi} = \beta/g_m$. You might want to calculate the solution using the hybrid-$\pi$ model since this circuit's output resistance is a little easier to express in terms of $r_{\pi}$.