Emitter follower/Common collector confusion

Question

I'm in the need of a buffering circuit for a project and I would like to use a common collector amplifier or emitter follower for this purpose. Above is a schematics of the circuit I'm looking into.

However, I'm having slight trouble understanding this circuit, specifically why the voltage at the output is the input voltage minus the base-emitter voltage drop (around 0.7 volts). Doesn't the capacitor C1 cause a voltage drop? It has capacitance, therefore it has impedance. If this circuit is used as a unity-gain current amplifier (buffer), I want only a small amount of current to enter at the input. Therefore, C1 should be quite small capacitance (because current is the voltage divided by the impedance, which in turn is inversely proportional to the capacitance). But if it has large impedance, I think it would have a larger voltage drop across it, and the output voltage would be input voltage minus the capacitor drop minus the diode voltage drop.

Somehow I think I'm thinking about this circularly, and I can't really make sense of what's happening. Can somebody explain to me why the output voltage is (almost) the same value as the input, and how the capacitor values are chosen?

EDIT: What if there is a load across the output? Say, a small-ish resistor. If the input capacitor is also small impedance, would there now be a quite large current through the circuit from input to output, defeating the whole purpose of the buffer?

Answer 1

You have shown an AC amplifier. The average input current is essentially zero, and the average voltage is whatever your load drags it to. The emitter of the transistor sits at roughly \$V_{Supply} \cdot \frac{R2}{R1+R2} -0.7V\$ if R3 isn't too low value, but that emitter voltage does not make its way to the output except momentarily when power is applied.

The input impedance should be about R1||R2 for frequencies in which the amplifier is useful, that's ignoring the series impedance of the capacitor and the loading of the base, both of which should have much less effect than R1||R2.

Answer 2

C1 is used to superimpose your signal on top of the base bias voltage. C2 acts as a dc blocking component which blocks the DC component of the emitter voltage and lets through the buffered signal. C1 in connection with R1||R2||(Input impedance of the transistor) forms a high pass filter.

Neglecting C2 for now here's how you should pick up a value for capacitor C1. The input impedance of an emitter follower is BetaRE, where Beta is the current gain of the transistor, and RE the resistance measured across the emitter and ground. As I said, C1 together with R1||R2||BetaRE forms a high pass filter. If Beta*RE is large compared to R1||R2 then you can write simply

$$H(w)=\frac{R_1||R_2}{R_1||R_2+1/jwC_1}$$

where its absolute can be written as

$$|H(w)|=\frac{R1||R2 \times wC_1}{\sqrt(1+R_1||R_2 \times wC_1)}$$

Now if you want to pass all signals of interest then you should choose C1 so that

$$C1 >\frac{1}{2\pi f R_1||R_2},$$

where f is the frequency of the signal of interest. For example, for R1=10k, R2=110k, RE=100k, and f=400Hz, you'll need to set C1 equal to 0.1uF or so.