# Homework Help: The Emitter Follower

1. Mar 16, 2014

### Duave

Thank you for any help that you can offer.

Did I answer all of the questions correctly and thoroughly? Can you please find any errors, and point them out to me so that I can fix them?

1. The problem statement, all variables and given/known data

(a) Confirm these quiescent values:

VB = 8.2V
IE = 1.0 mA

(b) show that Zin for the bias network alone is 50K.
(c) Show that for high frequency ac signals Zin measured at point P2 is 500k.
(d) Show that for high frequency ac signals Zin measured at point P1 is 45.5K.
(e) show that f3dB for the entire circuit is ~ 165Hz

https://scontent-a.xx.fbcdn.net/hphotos-prn2/t1.0-9/1901740_10151937081755919_967279941_n.jpg

2. Relevant equations

{R110k/(R110k + R91k)} x VCC = VB
...................................................................................................
VB - VBE = VE
............................................................................................
VE/RE = IE
....................................................................................................................
Zbias network = (R91k)(R110k)/(R91k) + (R110k)
............................................................................................
Zin,P2 = hFE{(re)^' + (re)}
............................................................................................
Zin,P1 = (Zbias network)(Zin,P2)/{(Zbias network) + (Zin,P2)}
.....................................................................................................................
f3dB = 1/(2)(pi)(ZP1 - 7.5k||15k){(C0.033uF)(C0.1uF)/{(C0.033uF + (C0.1uF)}}
.....................................................................................................................................

3. The attempt at a solution

(a)1 Confirm this quiescent value:

VB = 8.2V
.......................................................

{R110k/(R110k + R91k)} x VCC = VB
.........................................................................................
110k/(110k + 91k) x 15V = VB
.......................................
8.2V = VB
........................................

(a)2 Confirm this quiescent value:

IE = 1.0 mA
...........................

VB - VBE = VE
.......................................................
8.2V - 0.6V = VE
...........................
7.6V = VE
.....................
VE/RE = IE
...........................................
7.6V/7.5mA = IE
...........................................
1.03mA = IE
...........................................
1.00mA ~ IE
...........................................

(b) show that Zin for the bias network alone is 50K.

Zbias network = (R91k)(R110k)/(R91k) + (R110k)
............................................................................................
Zbias network = (91k)(110k)/(91k) + (100k)
.....................................................
Zbias network = 49.8k
...............................................
Zbias network ~ 50.0k
.............................................

(c) Show that for high frequency ac signals Zin measured at point P2 is 500k.

Zin,P2 = hFE{(re)^' + (re)
......................................................................
Zin,P2 = hFE{(25mV/(IC) + (R7.5k)(R15k)/(R7.5k + R15k)}
...........................................................................................................................
IC = IE
..............................
Zin,P2 = hFE{(25mV/[(VB - VBE)/RE] + (R7.5k)(R15k)/(R7.5k + R15k)}
.............................................................................................................................................................
Zin,P2 = hFE{25mV(RE)/[(VB - VBE)] + (R7.5k)(R15k)/(R7.5k + R15k)}
.............................................................................................................................................................
Zin,P2 = 100{(25mV[(7.5k)/{(8.2V - 0.6V)]) + (7.5k)(15k)/(7.5k + 15k)}
.................................................................................
Zin,P2 = 100{24.671(ohms) + 5k(ohms)}
.................................................
Zin,P2 = 502.4k(ohms)
................................
Zin,P2 ~ 500.0k(ohms)
................................

(d) Show that for high frequency ac signals Zin measured at point P2 is 45.5K.

Zin,P1 = (Zbias network)(Zin,P2)/{(Zbias network) + (Zin,P2)}
.....................................................................................................................
Zin,P1 = {(49.8k)(502.0k)/{(49.8k) + (502.0k)}
.........................................................
Zin,P1 = 45.3k
..........................

Z7.5k||15k = (7.5k)(15k)/(7.5k + 15k)
.............................................
Z7.5k||15k = 5k(ohms)
.............................................
Zin,P1 - Z7.5k||15k = ZP1 - 7.5k||15k
..................................................................
45.3k - 5k = ZP1 - 7.5k||15k
......................................
40.3k = ZP1 - 7.5k||15k
......................................

(e) show that f3dB for the entire circuit is ~ 165Hz

f3dB = 1/(2)(pi)(ZP1 - 7.5k||15k){(C0.033uF)(C0.1uF)/{(C0.033uF + (C0.1uF)}}
............................................................................................................................................
f3dB = 1/(2)(pi)(40.3k){(0.033uF)(0.1uF)/{(0.033uF + 0.1uF)}}
.........................................................................
f3dB = 1/[(2)(pi)(40.3k){2.48x10-8}}]
..................................................
f3dB = 159.17 Hz
.............................

I can see answer (e) is not 165 Hz. Where is the error? Are answers (a) - (d) correct? If not, where are the errors located?

2. Mar 17, 2014

### Staff: Mentor

For part (e), assume that the emitter follower itself is a buffer amplifier (high input impedance, low output impedance, gain very close to unity). You then have a cascade of two isolated low pass filters...

3. Mar 17, 2014

### Duave

gneil,

I will make corrections to (e) based on your comment. I am not asking for the answer, but does (a) - (d) look as though I have answered the questions?

4. Mar 17, 2014

### Staff: Mentor

I didn't go over the math in fine detail, but they look okay to me after a short perusal.

5. Mar 17, 2014

### Jony130

In this circuit we have two first order filter.
The first one
Fd1 ≈ 0.16/(Cin * Zin) = 0.16/(33nF * 45.5kΩ) = 106.5Hz
And the second one
Fd2 ≈ 0.16/(Cout * RL) = 0.16/(100nF * 15KΩ) = 106.6Hz

And as you can see in our case Fd1 = Fd2 and because of this the dominant lower cutoff frequency is increased by a factor of $\LARGE \frac{1}{\sqrt{ 2^{\frac{1}{n}}-1}}$
http://202.191.247.221/courses/imag..._Analysis/Resources/Multistage_AMplifiers.pdf

In our case n = 2 so we have $\frac{1}{\sqrt{ 2^{\frac{1}{2}}-1}} = 1.55377$

And finally

Fc = 106*1.55 = 164.3Hz