How Do Input/Output Impedance and Active Bandpass Filters Affect Node Voltages?

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Discussion Overview

The discussion revolves around calculating node voltages and impedances in circuits involving active bandpass filters and BJTs. Participants explore various calculations related to input and output impedance, as well as node voltages, while addressing potential errors in their approaches. The scope includes homework problems and technical reasoning related to circuit analysis.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents calculations for node voltages V1 and V2, questioning their correctness.
  • Another participant suggests that the circuit diagram may be incorrect, indicating that the ammeter is shorted out by a wire connecting two nodes.
  • Further contributions include suggestions for terms to include in the calculations for node voltages.
  • Participants discuss the input and output impedance calculations for a BJT, with one participant expressing uncertainty about their results and asking for assumptions to improve their understanding.
  • Another participant proposes a method to find output impedance using a specific voltage and current relationship.
  • Several participants express the need for verification of calculations and potential errors in the presented solutions.
  • One participant summarizes multiple calculations related to biasing in BJTs, asking for feedback on the correctness of their answers.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the calculations presented. There are multiple competing views regarding the accuracy of the circuit diagrams and the calculations of impedances and node voltages.

Contextual Notes

Some calculations depend on specific assumptions about circuit configurations and component values, which may not be fully clarified in the discussion. There are unresolved questions regarding the accuracy of the calculations and the interpretations of circuit elements.

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The Meaning Of An Active Bandpass Filter

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

Homework Equations



KVL and KCL calculation format.

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:
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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

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:
At node 1, two of the terms you will probably have are V1/10 and (V1 + 100)/30 plus some more.
 
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
https://scontent-a-pao.xx.fbcdn.net/hphotos-prn2/t1.0-9/1900003_10151939940590919_952631427_n.jpg

Homework Equations



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

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.
 
Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?
 
Last edited:
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Jony130 said:
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?
 
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;
 

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Answer

Jony130 said:
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)
.........
 
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
 

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  • #10
Jony130 said:
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?

So Zout = {R10k/(100 + 1)}||{R3.3k}
....................
 
  • #11
Excellent work
 
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  • #12
Jony130 said:
Very good, so Zout for your emitter follower circuit will be equal to Zout = ?

Jony130 said:
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?

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 BJThttps://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc1/t1.0-9/1932404_10151940102460919_638617745_n.jpg1(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 resistorIin = [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.jpg2(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.
 
  • #13
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?

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 BJThttps://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-frc1/t1.0-9/1932404_10151940102460919_638617745_n.jpg1(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 resistorIin = [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.jpg2(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.
 
  • #14
The ammeter is shorted out, as NO says. There is likely an error in your original post. Can you please check with the instructor?
 
  • #15
Thread closed while we sort out multiple posts of the same question...
 
  • #16
Thread re-opened after 2 threads merged.
 

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