# Input/Output Impedance

#### Duave

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:
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#### NascentOxygen

Mentor
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

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

#### Duave

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

...................................................
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){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.

#### Jony130

Are you sure about Zout ?? Also check your numbers in Zin; ZL = ?

Last edited:

#### Duave

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?

#### Jony130

For this simple circuit with CCCS

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

#### Attachments

• 2 KB Views: 342

#### Duave

For this simple circuit with CCCS

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)
..................................................

#### Jony130

Very good, so Zout for your emitter follower circuit will be equal to Zout = ?

#### Attachments

• 5.2 KB Views: 240

#### Duave

Very good, so Zout for your emitter follower circuit will be equal to Zout = ?
So Zout = {R10k/(100 + 1)}||{R3.3k}
............................................................................................................

Excellent work

#### Duave

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

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?

#### Duave

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

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?

#### berkeman

Mentor
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

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

#### berkeman

Mentor

"Input/Output Impedance"

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