Electronics: Understanding the Emitter Follower

This article focuses on the emitter follower circuit shown in figure one. It defines the operation of the emitter follower circuit and defines the circuit's input resistance and output resistance.

The voltage gain of the emitter follower circuit shown in figure one is equal to Ve/Vb

where Ve is the emitter voltage or output voltage and

Vb is the base voltage or input voltage.

Since the transistor is always on, the voltage gain must be less than 1. This is due to the voltage drop Vbe from base to emitter.

The current gain of the emitter follower is equal to the ratio of Ie to Ib where Ie is the output current and Ib is the input current..

Assuming that Ic/Ie = 1

Ie = Ic + Ib

where

Ic = collector current

Ib = base current and

Ie = emitter current and

We know that

Ic = Hfe* Ib

where

Hfe is the base to collector current gain

Hence

Ie = = Ic + Ib = Hfe*Ib + Ib

Factoring out the base current Ib from the right side of the equation

Ie = Hfe*Ib + Ib

we get

Ie = (Hfe+1)*Ib

Hence, the current gain of the emitter follower equals Ie/Ib or (Hfe + 1)

Ie/Ib = Hfe + 1

Now we will calculate in the input resistance and the output resistance.

In order for the transistor to remain on, the emitter base junction must be forward biased. In order for the base emitter junction to remain forward biased, any change in the emitter voltage must be accompanied by an equal change in the base voltage.

d(Ve) = d(Vb)

where

d stands for delta or change

d(Ve) means change in emitter voltage

d(Vb) means change in base voltage

d(Ie) means change in emitter current

d(Ie) = d(Ve)/R1

Since d(Ve) = d(Vb)

d(Ie) = d(Ve)/R1 = d(Vb)/R1

where

Ve = emitter voltage

Vb = base voltage

Ie = emitter current

Assuming that the ratio of collector current to emitter current is one.

Ic/Ie = 1

The change in the emitter current is equal to the sum of the changes in the base current and the collector current.

d(Ie) = d(Ic) + d(Ib)

and the collector current is equal to the base current multiplied by the current gain.

Ic = Hfe * Ib

therefore

d(Ie) = d(Ic) + d(Ib) = Hfe*d(Ib) + d(Ib)

factoring we get

d(Ie) = (Hfe + 1)*d(Ib)

therefore

d(Ie)/(Hfe+1) = d(Ib)

We already established that

d(Ie) = d(Vb)/R1

multiplying both sides of the equation by 1/(Hfe + 1) we get

d(Ie)/(Hfe+1) = d(Vb)/(Hfe +1)* R1

we know that

d(Ie)/(Hfe+1) = d(Ib)

substituting we get

d(Ie)/(Hfe+1) = d(Ib) = d(Vb)/((Hfe +1)* R1)

d(Ib) = d(Vb)/((Hfe +1)* R1)

or

d(Vb)/(d(Ib)) = (Hfe+1)*R1

Hence, measuring the input resistance (Ri) of the emitter follower we find that

Ri = d(Vb)/(d(Ib)) = (Hfe+1)*R1

The output resistance Ro is equal to R1.

Ro = R1

The emitter follower circuit has a high input resistance and a low output resistance.

This concludes this article on the emitter follower.

The next article on the emitter follower will cover practical applications.

References:

I have a Bachelor of Science In Electrical Engineering and worked as an Electronics Technician.

Electronic Principles: Third Edition
ISBN 0-07-039912-3