In an emitter-follower transistor amplifier, what is the relationship between Vin and Vout?

Answer 1

An emitter follower transistor amplifier essentially acts as a buffer. It has a high input impedance(or resistance) and a low output impedance (or resistance).

The output voltage Vout in this case is almost equal (a slight drop in output voltage occurs) to the input voltage Vin. The drop in voltage is due to the forward bias drop across the base-emitter junction.

hope that helps.

Answer 2

In the low frequency domain, you will always have a voltage gain (Vout/Vin) slightly less than unity. The basic equations are given in the link below.

If you need an emitter follower to operate at very high frequency, things can get very tricky. One little known fact (I know this from working as an EE for many years) is that emitter followers tend to have a peaked frequency response and may even become oscillators under the right conditions, which usually include driving a capacitive load from a high source resistance. Looking into the emitter, the capacitive load sees an inductive source, which creates an artificial LC resonant circuit, hence the tendency to have a ringing step response.

Answer 3

In an emitter follower transistor amplifier, Vin and Vout are same ( i.e. gain is unity).

Answer 4

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Answer 5

Vout is approximately = Vin

More accurately: Vout = (Beta-1)/Beta * Vin
which works out to be about 99% of Vin most of the time.

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Answer 6

About a diode junction drop between in and out. eg about 0.6V.

Answer 7

in phase and equal