# Impedance matching with emitter follower

Discussion in 'General Electronics Chat' started by Eduard Munteanu, Sep 9, 2007.

1. ### Eduard Munteanu Thread Starter Active Member

Sep 1, 2007
86
0
Hello. Common-collector BJT amplifiers (aka emitter followers) are said to be good at matching impedance. But, as far as I can see, they do horrible when it comes to parameter spread.

Problem
I have a common-emitter amplification stage, with Zout = Rc = 6k, which needs to drive a 4 or 8 ohm speaker. Adding an emitter follower immediately comes to mind, as it also provides some current gain and they're easy to make (as opposed to transformers).

But theory says...
Theory says that the impedances for an emitter follower are:
$Z_{in} \approx \beta_F R_E\\
Z_{out} \approx {1 \over g_m} + {R_{source} \over \beta_F}$

(from Wikipedia)
In the output impedance, the second term can be neglected if we approximate the power source as an ideal voltage source. So:
$Z_{out} \approx {1 \over g_m} = {V_{in} \over I_{out}} = {1 \over \beta_F} {V_{in} \over I_{in}} = {Z_{in} \over \beta_F} = R_E$
Looks like $Z_{out}$ is independent of $\beta_F$. But $Z_{in}$ is not!

Question
In the CE amplifier you had a way to select input and output impedances, as well as other parameters, to be almost independent of the transistor's characteristics. The emitter-follower seems to be awfully affected by $\beta_F$ spread, which is very annoying if you want to match impedances for maximum power transfer. Is there anything there can be done, any hope?

I tried adding a collector resistor. At the first glance, you might think it can limit $\beta_F$ if you set it at $Rc = {V_c \over I_{c(desired)}} = {V_{source} \over {\beta_{F(desired)} I_b}}$. But that's not right, the output waveform just clips. That resistor, along with the voltage source, behaves like a (bad) current source and it won't operate the transistor linearly.

What I think it would work might be some kind of negative feedback, as in the CE amplifier, so that you can limit the current gain, thus using enforcing an upper bound for beta, thus having a fixed input impedance.

Any thought on this?

2. ### Eduard Munteanu Thread Starter Active Member

Sep 1, 2007
86
0
Looks like Wikipedia has other means for $g_m$ (I thought it was transconductance, as in $g_m = {I_{out} \over V_{in}}$). Somehow, the same problem might also arise for $Z_{out}$, provided they're correct, since $g_m = {I_c \over V_T} = \beta {I_b \over V_T}$.