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

  • Thread starter Duave
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    Impedance
In summary, the input impedance looking directly into the base of the BJT is 1010k ohms, and the output impedance including the 3.3k resistor is 96 ohms. VB including loading of the bias network by the BJT is 15V, VE is 14.4V, and IE is 1.92mA. Neglecting loading of the bias network by the BJT, VB is 8.03V, VE is 7.43V, and IE is 0.99mA.
  • #1
Duave
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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
 
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  • #2
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.
 
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  • #3
At node 1, two of the terms you will probably have are V1/10 and (V1 + 100)/30 plus some more.
 
  • #4
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.
 
  • #5
Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?
 
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  • #6
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?
 
  • #7
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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  • #8
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)
.........
 
  • #9
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.
 

1. What is input/output impedance?

Input/output impedance refers to the resistance of a circuit or system that is designed to transmit or receive electrical signals. It is a measure of how much the circuit or system resists the flow of current.

2. Why is input/output impedance important?

Input/output impedance is important because it affects the performance of a circuit or system. If the impedance is too high, it can cause signal loss and affect the accuracy of the transmitted or received signal. If the impedance is too low, it can cause excessive current flow and potentially damage the circuit or system.

3. How is input/output impedance measured?

Input/output impedance is typically measured using an impedance meter or by performing calculations using Ohm's law. The impedance is expressed in ohms (Ω) and can vary depending on the frequency of the signal being transmitted or received.

4. What is the difference between input and output impedance?

The main difference between input and output impedance is their function. Input impedance refers to the resistance that a signal encounters when entering a circuit or system, while output impedance refers to the resistance that a signal encounters when leaving a circuit or system.

5. How can input/output impedance be matched?

Input/output impedance can be matched by using impedance matching techniques, such as using a transformer or adding resistors to the circuit. This helps to minimize signal loss and ensure maximum power transfer between the source and load. Additionally, using components with similar impedance values can also help in matching the input/output impedance.

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