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Input/Output Impedance

  • Thread starter Duave
  • Start date
81
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The Meaning Of An Active Bandpass Filter

1. Homework Statement

Calculate the node voltages V1 and V2. Assume that the 0 ohm resistor is an ammeter, and that the zero current element is a voltmeter.

https://scontent-b.xx.fbcdn.net/hphotos-ash3/t1/1898156_10151912573775919_1335607656_n.jpg

2. Homework Equations

KVL and KCL calculation format.

3. The Attempt at a Solution


Can you please tell me if my calculations are correct.

Node 1: V1/0ohm -V1/30ohm + 100V/10ohm = 4

-V1 + 300V = 120V

V1 = 180V


Node 2: V2 = (4A)(7.5ohm)

V2 = 180V


Thank You
 
Last edited:

NascentOxygen

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That's an interesting choice of title for your thread, Duave. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken]

I'm thinking you have not drawn the circuit correctly. You show the ammeter shorted out by a connecting wire. That wire also connects node 1 to node 2, making them one node. Please check and correct.
 
Last edited by a moderator:

NascentOxygen

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At node 1, two of the terms you will probably have are V1/10 and (V1 + 100)/30 plus some more.
 
81
0
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. Homework Statement


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



https://scontent-a-pao.xx.fbcdn.net/hphotos-prn2/t1.0-9/1900003_10151939940590919_952631427_n.jpg

2. Homework Equations

Zin = (hFE + 1){ZLoad}
...................................................
Zout = {(R10k)(R3.3k)/(R10k) + (R3.3k)}/(hFE+1)
.................................................................................................................


3. The Attempt at a Solution

1(a)

Calculate the input impedance looking directly into the base of the BJT

Zin = (hFE + 1){ZLoad}
...................................................
Zin = (hFE + 1){RL
...................................................
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

Zout = (R10k)(R3.3k)/(R10k) + (R3.3k)}/(hFE+1)
.................................................................................................................
Zout = {(10 x 10^3(ohms))(3.3 x 10^3(ohms))/(10 x 10^3(ohms)) + (3.3 x 10^3(ohms)}/(100 + 1)
.................................................................................................................
Zout = {2.481 x 10^3(ohms)}/(100 + 1)
.................................................
Zout = 24.56(ohms)
..............................

Are there any errors?

Thank you.
 
623
137
Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?
 
Last edited:
81
0
Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?
Jony130,

Okay, no, I am not sure about Zout. Seeing that my answers are not right. What assumptions can I make to improve my next attempt? What kind of questions does one ask to tackle a problem like this?
 
623
137
For this simple circuit with CCCS

attachment.php?attachmentid=67738&stc=1&d=1395073484.png


try to find Zout = Vin/Iin for Vin = 1V and β = 100;
 

Attachments

81
0
Answer

For this simple circuit with CCCS

attachment.php?attachmentid=67738&stc=1&d=1395073484.png


try to find Zout = Vin/Iin for Vin = 1V and β = 100;
This is my answer to the question

Iin = IB + (hFE)(IB)
...................................................................
Iin = (hFE + 1)(Ib)
...................................................................
Ib = (Vin)/(Rb)
...................................................................
Iin = (hFE + 1){(Vin)/(Rb)}
.........................................................................
Zout = (Vin)/(Iin)
.........................................................................
Zout = (Vin)/[(hFE + 1){(Vin)/(Rb)}]
.............................................................................................
Zout = (Rb)/(hFE + 1)
........................................................
Zout = (10k)/(100 + 1)
........................................................
Zout = 99(ohms)
..................................................
 
623
137
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
 

Attachments

81
0
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
So Zout = {R10k/(100 + 1)}||{R3.3k}
............................................................................................................
 
623
137
Excellent work
 
81
0
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
Excellent work
Okay this is a revision of the whole problem. Can you please look at all eight questions and answers? Thanks Jony130.

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. Homework Statement

Calculate the input impedance looking directly into the base of the BJT
Calculate the output impedance including the 3.3k resistor
Calculate VB neglecting loading of the bias network by the BJT
Calculate VE neglecting loading of the bias network by the BJT
Calculate IE neglecting loading of the bias network by the BJT
Calculate VB including loading of the bias network by the BJT
Calculate VE including loading of the bias network by the BJT
Calculate IE 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

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


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?

Thanks again for your help.
 
81
0
NascentOxygen,

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. Homework Statement

Calculate the input impedance looking directly into the base of the BJT
Calculate the output impedance including the 3.3k resistor
Calculate VB neglecting loading of the bias network by the BJT
Calculate VE neglecting loading of the bias network by the BJT
Calculate IE neglecting loading of the bias network by the BJT
Calculate VB including loading of the bias network by the BJT
Calculate VE including loading of the bias network by the BJT
Calculate IE 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

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


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?

Thanks again for your help.
 

berkeman

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The ammeter is shorted out, as NO says. There is likely an error in your original post. Can you please check with the instructor?
 

berkeman

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Thread closed while we sort out multiple posts of the same question...
 

berkeman

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Thread re-opened after 2 threads merged.
 

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