50 ohm output impedance

Question

I saw some circuits that uses a emitter follower as a buffer for output. The emitter resistance is 470\$\Omega\$. The author says that, since the ac emitter resistance is very small, he put a 50\$\Omega\$ resistor in series with the output terminal, so that the circuit would look like 50\$\Omega\$ from the output.

But what I don't understand is, if you now connect the output to a 50\$\Omega\$ load, the emitter resistance then becomes (470\$\Omega\$ || 50+50\$\Omega\$) = 82\$\Omega\$! This could affect a lot of things, for example the collector current would suddenly change 5-fold, the signal source would see a 80% reduction of load, everything could go wrong. What am I missing here?

Answer 1

The output impedance of an emitter (or source) follower is not the load resistor, as it would (approximately) be in a common emitter circuit. In fact, it is much lower- roughly the impedance the base sees divided by (hfe + 1), in parallel with the load resistor.

To see this intuitively, think of the large signal case with 10V on the collector, a stiff 5.6V on the base, and 470 ohm load. The emitter voltage is 5.0V. Now put a 470 ohm load on the emitter to ground- does the voltage drop to 2.5V as would be expected with a 470 ohm output impedance? No, it hardly drops at all- which tells you the impedance is very low.

Naturally, there must be a series capacitor or the DC emitter bias will cause problem with the input transformer or whatever it is connected to, but that's true with or without the 50 ohm resistor.

Answer 2

Why do you think you are missing something? I think you pin pointed a valid issue with the emitter follower circuit.

Many 50Ω circuits are designed for an AC signal and thus you could solve part of your issue by using a output series capacitor. At least the output load won't affect the DC operating point. Also an extra input stage can help in increasing input impedance, but you may consider not using the emitter follower.