# Solving The Emitter Follower Quiz

• Duave
In summary: Hz that we got before.In summary, this conversation involved confirming quiescent values for VB and IE, finding Zin for the bias network, determining Zin for high frequency ac signals at point P2 and P1, and finding the f3dB for the entire circuit. After making corrections based on a comment about the emitter follower being a buffer amplifier, it was determined that the dominant lower cutoff frequency for the two first-order filters in the circuit is approximately 164.3Hz.
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:

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

## Homework 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)}}
........................

## 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?

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...

1 person
gneill said:
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?

Duave said:
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.

1 person
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

1 person

## 1. What is an emitter follower?

An emitter follower, also known as a common collector amplifier, is a type of circuit in which the emitter terminal of a transistor is connected directly to the output, while the collector terminal is connected to the power supply.

## 2. How does an emitter follower work?

In an emitter follower circuit, the input signal is applied to the base terminal of the transistor, causing a change in the collector current. This change in current is then transmitted to the emitter, resulting in an output signal that is a replica of the input signal but with a lower voltage.

## 3. What are the advantages of using an emitter follower?

An emitter follower has a high input impedance, which means it does not load the previous stage of the circuit and allows for better signal transfer. It also has a low output impedance, making it a good buffer between high and low impedance circuits. Additionally, it has a high current gain, providing a high output current with minimal distortion.

## 4. What are the limitations of an emitter follower?

One limitation of an emitter follower is that it has a voltage gain of less than 1, meaning the output voltage is always lower than the input voltage. It also has a tendency to oscillate at high frequencies, which can cause distortion in the output signal. Additionally, it requires a stable power supply and can be affected by temperature changes.

## 5. How can I improve the performance of an emitter follower circuit?

To improve the performance of an emitter follower, you can use a Darlington pair, which is essentially two transistors connected together. This increases the current gain and reduces distortion. You can also use negative feedback to stabilize the circuit and reduce oscillations. Using higher quality components and careful circuit design can also improve the overall performance of an emitter follower.