# Transfer-Function: Vout/Vin

1. Feb 24, 2014

### Duave

Transfer-Function: "Vout/Vin"

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

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

2. Relevant equations

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

3. The attempt at a solution

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

Thank you

Last edited: Feb 24, 2014
2. Feb 24, 2014

### Staff: Mentor

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

3. Feb 25, 2014

### 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?

4. Mar 17, 2014

### Duave

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?

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

Calculate the input impedance looking directly into the base of the BJT
Calculate the output impedance including the 3.3k resistor

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

...................................................
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)}
.....................................

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

Iin = [IB + (hFE)(IB)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
...................................................................
Iin = [(hFE + 1)(Ib)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
...................................................................
Ib = [(Vin)/(Rb)]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
...................................................................
Iin = [(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
.........................................................................
Zout = {(Vin)/(Iin)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
.........................................................................
Zout = (Vin)/[(hFE + 1){(Vin)/(Rb)}]{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
.............................................................................................
Zout = {(Rb)/(hFE + 1)}{R3.3k}/{(R10k)/(100 + 1)} + {R3.3k}
..................................................................................................................................
Zout = {(10k)/(100 + 1)}{3.3k}/{(10k)/(100 + 1)} + {3.3k}
.....................................................................
Zout = 96(ohms)
..................................................

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

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
............................

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

IE = IE/R7.5k
................................................
IE = 14.4V/7.5k
................................................
IE = 1.92mA
................................................

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

{R150k/(R150k + R130k)} x VCC = VB
.........................................................................................
150k/(150k + 130k) x 15V = VB
.......................................
8.03V = VB
........................................

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

VE = VB - VBE
................................................
VE = 8.03V - 0.6V
............................
VE = 7.43V
............................

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

IE = IE/R7.5k
................................................
IE = 7.43V/7.5k
................................................
IE = 0.99mA
................................................

Are there any errors?

5. Mar 18, 2014

### Jony130

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.

6. Mar 18, 2014

### Duave

I do not know how to calculate for BJT's

I think now, that the RL = 3.3k

I thought 15V was strange myself,

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

7. Mar 18, 2014

### Duave

Which answers have errors? I would like to fix them

8. Mar 18, 2014

### Jony130

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

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Last edited: Mar 18, 2014
9. Mar 18, 2014

### Duave

Thanks Jony130,

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

Thank you