How to Calculate Vout/Vin for High-Pass Filter Using Transfer-Function?

[Duave]

Transfer-Function: "Vout/Vin"

Homework Statement

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash3/t1/1656161_10151902951360919_2042068460_n.jpg

Homework Equations

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-prn1/t1/16672_10151902951355919_965456274_n.jpg

The Attempt at a Solution

"Answer #1: Calculation Of Vout/Vin"

The very last equation is my answer. Am I correct?
https://scontent-a.xx.fbcdn.net/hphotos-ash3/t1/1969246_10151902957535919_1007404817_n.jpg


"Answer #2: Display of log amplitude vs. log frequency"

Does this diagram give what was asked for?
https://scontent-a.xx.fbcdn.net/hphotos-frc1/t1/1958500_10151902957555919_1931978744_n.jpg


"Answer #3: phase shift vs. log frequency"

Does this diagram give what was asked for?
https://scontent-a.xx.fbcdn.net/hphotos-ash3/t1/1901948_10151902957540919_851749761_n.jpg


Did I answer everything about this high-pass filter?

Thank you

 

[berkeman]

The answers look fine. How did you make the Bode plot diagrams?

 

[FOIWATER]

It looks to be a picture, since it is slightly transparent you can see the next page through it.

I guess it was more of a 'match the bode plot to the circuit' type of question?

 

[Duave]

It looks to be a picture, since it is slightly transparent you can see the next page through it.

I guess it was more of a 'match the bode plot to the circuit' type of question?

FOIWATER,

This is a whole problem. Can you please look at all eight questions and answers and tell me if you see errors?

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

Calculate the input impedance looking directly into the base of the BJT
Calculate the output impedance including the 3.3k resistor
Calculate V_B neglecting loading of the bias network by the BJT
Calculate V_E neglecting loading of the bias network by the BJT
Calculate I_E neglecting loading of the bias network by the BJT
Calculate V_B including loading of the bias network by the BJT
Calculate V_E including loading of the bias network by the BJT
Calculate I_E including loading of the bias network by the BJT


https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc1/t1.0-9/1932404_10151940102460919_638617745_n.jpg


1(a): Calculate the input impedance looking directly into the base of the BJT

Z_in = (h_FE + 1){Z_Load}
..........
Z_in = (h_FE + 1){R_10k
..........
Z_in = (100 + 1){10 x 10^3(ohms)}
..........
Z_in = (101){10 x 10^3(ohms)}
..........
Z_in = {10.1 x 10^5(ohms)}
........
Z_in = {1010k(ohms)}
........


1(b): Calculate the output impedance including the 3.3k resistor


I_in = [I_B + (h_FE)(I_B)]{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
.............
I_in = [(h_FE + 1)(I_b)]{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
.............
I_b = [(V_in)/(R_b)]{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
.............
I_in = [(h_FE + 1){(V_in)/(R_b)}]{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
..............
Z_out = {(V_in)/(I_in)}{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
..............
Z_out = (V_in)/[(h_FE + 1){(V_in)/(R_b)}]{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
..................
Z_out = {(R_b)/(h_FE + 1)}{R_3.3k}/{(R_10k)/(100 + 1)} + {R_3.3k}
.......................
Z_out = {(10k)/(100 + 1)}{3.3k}/{(10k)/(100 + 1)} + {3.3k}
..............
Z_out = 96(ohms)
.........


https://scontent-a-pao.xx.fbcdn.net/hphotos-prn1/t1.0-9/1964785_10151940672820919_881396491_n.jpg


2(a): Calculate V_B including loading of the bias network by the BJT

r_in = (h_FE + 1){R}
..........
r_in = (h_FE + 1){R_7.5k
..........
r_in = (100 + 1){7.5 x 10^3(ohms)}
..........
r_in = (101){7.5 x 10^3(ohms)}
..........
r_in = {7.575 x 10^5(ohms)}
........
r_in = {757.5k(ohms)}
........


I_B = I_E/(h_FE + 1)
..........
I_B = {(V_CC/R_7.5k)/(h_FE + 1)}
...............
I_B = {(15V/7.5k/(100 + 1)}
.........
I_B = {(15V/7.5k/(100 + 1)}
.........
I_B = 1.98 x 10^-5
.........
I_B = 19.8uA
.........
V_B = (r_in)(I_B)
.........
V_B = (757.5k)(19.8uA)
........
V_B = (757.5k)(19.8uA)
........
V_B = 15V
.......


2(b): Calculate V_E including loading of the bias network by the BJT

V_E = V_B - V_BE
.........
V_E = 15V - 0.6V
......
V_E = 14.4V
......


2(C): Calculate I_E including loading of the bias network by the BJT

I_E = I_E/R_7.5k
.........
I_E = 14.4V/7.5k
.........
I_E = 1.92mA
.........


2(D): Calculate V_B neglecting loading of the bias network by the BJT

{R_150k/(R_150k + R_130k)} x V_CC = V_B
.................
150k/(150k + 130k) x 15V = V_B
.......
8.03V = V_B
.......


2(E): Calculate V_E neglecting loading of the bias network by the BJT

V_E = V_B - V_BE
.........
V_E = 8.03V - 0.6V
......
V_E = 7.43V
......

2(F): Calculate I_E neglecting loading of the bias network by the BJT

I_E = I_E/R_7.5k
.........
I_E = 7.43V/7.5k
.........
I_E = 0.99mA
.........

Are there any errors?

Thanks again for your help.

 

[Jony130]

FOIWATER,

Are there any errors?

Thanks again for your help.

You made many mistakes in your answers

1(a): Calculate the input impedance looking directly into the base of the BJT

Zin = (hFE + 1){ZLoad}
..........
Zin = (hFE + 1){R10k
..........
Zin = (100 + 1){10 x 10^3(ohms)}
..........
Zin = (101){10 x 10^3(ohms)}
..........
Zin = {10.1 x 10^5(ohms)}
........
Zin = {1010k(ohms)}
........

Are you sure that RL = 10K?

2(a): Calculate VB including loading of the bias network by the BJT

rin = (hFE + 1){R}
..........
rin = (hFE + 1){R7.5k
..........
rin = (100 + 1){7.5 x 10^3(ohms)}
..........
rin = (101){7.5 x 10^3(ohms)}
..........
rin = {7.575 x 10^5(ohms)}
........
rin = {757.5k(ohms)}
........


IB = IE/(hFE + 1)
..........
IB = {(VCC/R7.5k)/(hFE + 1)}
............. ...
IB = {(15V/7.5k/(100 + 1)}
.........
IB = {(15V/7.5k/(100 + 1)}
.........
IB = 1.98 x 10^-5
.........
IB = 19.8uA
.........
VB = (rin)(IB)
.........
VB = (757.5k)(19.8uA)
........
VB = (757.5k)(19.8uA)
........
VB = 15V
.......
2(b): Calculate VE including loading of the bias network by the BJT

VE = VB - VBE
.........
VE = 15V - 0.6V
......
VE = 14.4V
......

Very interesting VB = 15V?? Are you training to say that voltage divider magically disappeared?
It seems that you completely don't understand what this loading effect of the bias network by the BJT means.
If you remove BJT from the circuit this two 130k and 150K resistors form a unloaded voltage divider. But in your circuit the base current is loading our voltage divider. Simply you must include the base current in your calculation. Because now upper resistor provide current for the lower resistor and for the base. All you need here is to use KVL,KCL and maybe a Thevenine.

 

[Duave]

You made many mistakes in your answers


Are you sure that RL = 10K?


Very interesting VB = 15V?? Are you training to say that voltage divider magically disappeared?
It seems that you completely don't understand what this loading effect of the bias network by the BJT means.
If you remove BJT from the circuit this two 130k and 150K resistors form a unloaded voltage divider. But in your circuit the base current is loading our voltage divider. Simply you must include the base current in your calculation. Because now upper resistor provide current for the lower resistor and for the base. All you need here is to use KVL,KCL and maybe a Thevenine.

I do not know how to calculate for BJT's

I think now, that the R_L = 3.3k

I thought 15V was strange myself,

I don't know how to use KCL, or KVL on a transistor.

 

[Duave]

You made many mistakes in your answers


Are you sure that RL = 10K?


Very interesting VB = 15V?? Are you training to say that voltage divider magically disappeared?
It seems that you completely don't understand what this loading effect of the bias network by the BJT means.
If you remove BJT from the circuit this two 130k and 150K resistors form a unloaded voltage divider. But in your circuit the base current is loading our voltage divider. Simply you must include the base current in your calculation. Because now upper resistor provide current for the lower resistor and for the base. All you need here is to use KVL,KCL and maybe a Thevenine.

Which answers have errors? I would like to fix them

 

[Jony130]

Which answers have errors? I would like to fix them

1a, 2a, 2b, 2c, are all wrong.

I don't know how to use KCL, or KVL on a transistor.

You do KVL and KCL exactly the same as for normal circuit with resistors and sources.
You simply treat Vbe as voltage source equal to 0.6V. And remember that KCL for BJT
Ie = Ib + Ic; and Ic = Ib*β (you treat the transistor as CCCS) that's all.
For example for this circuit

Vcc = Ib*Rb + Vbe + Ie*Re and since Ie = Ib + Ic = Ib + Ib*β = Ib*(β + 1)

So we have

Vcc = Ib*Rb + Vbe + Ib*(β + 1)*Re and simply solve for Ib

 

[Duave]

1a, 2a, 2b, 2c, are all wrong.

You do KVL and KCL exactly the same as for normal circuit with resistors and sources.
You simply treat Vbe as voltage source equal to 0.6V. And remember that KCL for BJT
Ie = Ib + Ic; and Ic = Ib*β (you treat the transistor as CCCS) that's all.
For example for this circuit

Vcc = Ib*Rb + Vbe + Ie*Re and since Ie = Ib + Ic = Ib + Ib*β = Ib*(β + 1)

So we have

Vcc = Ib*Rb + Vbe + Ib*(β + 1)*Re and simply solve for Ib

Thanks Jony130,

you're a big help, I am going to calculate values based on your advice, and I will respond.

Thank you