Hi everyone,
I have been doing some reading of the Art of Electronics and I have come across an explanation for the Emitter Follower configuration on page 66. The explanation serves to convince the reader that the emitter follower has a large input impedance compared to its output impedance. And it explains this with the help of some mathematics.
Now, I intuitively understand that the emitter follower has a large input impedance compared to its output impedance because of the fact that an emitter follower has no voltage gain but it has current gain. However, whilst I understand and accept what the book is trying to say, I can't quite grasp the mathematics of it. Here's what the book says:
\(\Delta\) IE = \(\Delta\)VB / R. (where R is the load).
I'm okay with that much. Since \(\Delta\)VB = \(\Delta\)VE it makes sense.
The author then makes a leap from that statement to this one;
\(\Delta\)IB = \(\frac{1}{hfe + 1}\) * \(\Delta\)IE.
= \(\Delta\)VB / R(hfe + 1)
Now, I must confess that my understanding has been lost on that jump. Is another more mathematically adept member able to explain it a little more fully for me?
Thanks very much in advance!
Brian
I have been doing some reading of the Art of Electronics and I have come across an explanation for the Emitter Follower configuration on page 66. The explanation serves to convince the reader that the emitter follower has a large input impedance compared to its output impedance. And it explains this with the help of some mathematics.
Now, I intuitively understand that the emitter follower has a large input impedance compared to its output impedance because of the fact that an emitter follower has no voltage gain but it has current gain. However, whilst I understand and accept what the book is trying to say, I can't quite grasp the mathematics of it. Here's what the book says:
\(\Delta\) IE = \(\Delta\)VB / R. (where R is the load).
I'm okay with that much. Since \(\Delta\)VB = \(\Delta\)VE it makes sense.
The author then makes a leap from that statement to this one;
\(\Delta\)IB = \(\frac{1}{hfe + 1}\) * \(\Delta\)IE.
= \(\Delta\)VB / R(hfe + 1)
Now, I must confess that my understanding has been lost on that jump. Is another more mathematically adept member able to explain it a little more fully for me?
Thanks very much in advance!
Brian