Solving The Emitter Follower Quiz

[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?

Homework Statement

(a) Confirm these quiescent values:

V_B = 8.2V
I_E = 1.0 mA

(b) show that Z_in for the bias network alone is 50K.
(c) Show that for high frequency ac signals Z_in measured at point P2 is 500k.
(d) Show that for high frequency ac signals Z_in measured at point P1 is 45.5K.
(e) show that f_3dB for the entire circuit is ~ 165Hz


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

Homework Equations

{R_110k/(R_110k + R_91k)} x V_CC = V_B
...................
V_B - V_BE = V_E
.................
V_E/R_E = I_E
.....................
Z_{bias network} = (R_91k)(R_110k)/(R_91k) + (R_110k)
.................
Z_in,P2 = h_FE{(r_e)^' + (r_e)}
.................
Z_in,P1 = (Z_{bias network})(Z_in,P2)/{(Z_{bias network}) + (Z_in,P2)}
.....................
f_3dB = 1/(2)(pi)(Z_{P1 - 7.5k||15k}){(C_0.033uF)(C_0.1uF)/{(C_0.033uF + (C_0.1uF)}}
........................

The Attempt at a Solution

(a)₁ Confirm this quiescent value:

V_B = 8.2V
..........

{R_110k/(R_110k + R_91k)} x V_CC = V_B
.................
110k/(110k + 91k) x 15V = V_B
.......
8.2V = V_B
.......


(a)₂ Confirm this quiescent value:

I_E = 1.0 mA
......

V_B - V_BE = V_E
..........
8.2V - 0.6V = V_E
......
7.6V = V_E
.....
V_E/R_E = I_E
.........
7.6V/7.5mA = I_E
.........
1.03mA = I_E
.........
1.00mA ~ I_E
.........


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

Z_{bias network} = (R_91k)(R_110k)/(R_91k) + (R_110k)
.................
Z_{bias network} = (91k)(110k)/(91k) + (100k)
...........
Z_{bias network} = 49.8k
.........
Z_{bias network} ~ 50.0k
..........



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

Z_in,P2 = h_FE{(r_e)^' + (r_e)
...............
Z_in,P2 = h_FE{(25mV/(I_C) + (R_7.5k)(R_15k)/(R_7.5k + R_15k)}
.......................
I_C = I_E
........
Z_in,P2 = h_FE{(25mV/[(V_B - V_BE)/R_E] + (R_7.5k)(R_15k)/(R_7.5k + R_15k)}
............................
Z_in,P2 = h_FE{25mV(R_E)/[(V_B - V_BE)] + (R_7.5k)(R_15k)/(R_7.5k + R_15k)}
............................
Z_in,P2 = 100{(25mV[(7.5k)/{(8.2V - 0.6V)]) + (7.5k)(15k)/(7.5k + 15k)}
...............
Z_in,P2 = 100{24.671(ohms) + 5k(ohms)}
..........
Z_in,P2 = 502.4k(ohms)
......
Z_in,P2 ~ 500.0k(ohms)
......



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

Z_in,P1 = (Z_{bias network})(Z_in,P2)/{(Z_{bias network}) + (Z_in,P2)}
.....................
Z_in,P1 = {(49.8k)(502.0k)/{(49.8k) + (502.0k)}
...........
Z_in,P1 = 45.3k
......



Z_7.5k||15k = (7.5k)(15k)/(7.5k + 15k)
..........
Z_7.5k||15k = 5k(ohms)
..........
Z_in,P1 - Z_7.5k||15k = Z_{P1 - 7.5k||15k}
............
45.3k - 5k = Z_{P1 - 7.5k||15k}
.........
40.3k = Z_{P1 - 7.5k||15k}
.........

(e) show that f_3dB for the entire circuit is ~ 165Hz


f_3dB = 1/(2)(pi)(Z_{P1 - 7.5k||15k}){(C_0.033uF)(C_0.1uF)/{(C_0.033uF + (C_0.1uF)}}
..........................
f_3dB = 1/(2)(pi)(40.3k){(0.033uF)(0.1uF)/{(0.033uF + 0.1uF)}}
..............
f_3dB = 1/[(2)(pi)(40.3k){2.48x10-8}}]
.........
f_3dB = 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?

Thanks again for your help.

 

[gneill]

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...

 

[Duave]

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...

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?

 

[gneill]

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?

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

 

[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