I am just learning about the emitter-follower amplifier, but i'm struggling to understand the output resistance. My book says that it is roughly

\(R_{out}\approx (\frac{R_s}{\beta_{ac}})||R_E \approx \frac{R_s}{\beta_{ac}}\)

but I don't understand why it's not just \(R_{out} = \frac{V_e}{I_e} = R_E\)?

The voltage gain is

\(A_v = \frac{V_b}{V_e} = \frac{I_eR_E}{I_e(r'_e+R_E)}\)

which is roughly equal to 1 when the emitter resistor is much larger than the internal emitter resistance. If I now connect a load resistor \(R_L\) between the emitter and ground, we will have to replace \(R_E\) by \(R_E||R_L\) in the gain equation above, so it will be the load resistor's size

*relative to the emitter resistor*which will determine how much the gain is decreased by.

But I thought the whole point in knowing the output resistance was so you know how small a load you can connect without affecting the circuit too much? If I follow my book's equation

\(R_{out}\approx \frac{R_s}{\beta_{ac}}\)

then I would think I could connect a much smaller load without affecting the gain?

Could someone help me to understand a little better?

Thanks