Finding current and voltage amplification via load line graph plotting

In summary, the transistor in a common emitter configuration has a load resistance of 2k ohms. The base current is suppose to be in micro Amp, but this was not included in the data table. TheAttempt at a Solution attempted to solve for the current amplification by using the load line equation with the given input resistance and quiescent operating point. The resulting current gain is 31.25.
  • #1
null void
102
1

Homework Statement


This is a question refer to a transistor in a common emitter configuration.

And this is the table with data of Vce and Ic at certain range of Ib stored in microsoft excel, and the graph is plotted within the file attached(in another tab)
https://www.filesanywhere.com/fs/v.aspx?v=8b6a6b8f595e70a970a0

The question also stated load resistance is 2k Ω

The base current is suppose to be in micro Amp, forgot to include this in the data table..

The Attempt at a Solution



no other pictorial information about how circuit, so i suppose this common emitter, have a load resistance of 2k Ω before collector and a Base resistor

so i come out with my load line equation,

Ic = -Vce / 2k + 4m
I induced this load line in my attachment,

So let's say i choose my quiscent operating point at
Ib = 40 uA, and my output current and voltage peak to peak is :

Vpp = 5.8 - 3.1 = 2.7V
Ipp = 2.45 - 1.2 = 1.25 m A

so my current amplification is
Ai = 1.25m / 40u = 31.25



and if the input resistance is 1.5k Ω,



my voltage amplification is,

Av = (Iout)(Rload) / (Iin)(Rinput)
=Ai x (Rload)/(Rinput)
= 31.25(2/1.5)
= 41.667
 
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  • #2
null void said:

Homework Statement


This is a question refer to a transistor in a common emitter configuration.

And this is the table with data of Vce and Ic at certain range of Ib stored in microsoft excel, and the graph is plotted within the file attached(in another tab)
https://www.filesanywhere.com/fs/v.aspx?v=8b6a6b8f595e70a970a0

The question also stated load resistance is 2k Ω

The base current is suppose to be in micro Amp, forgot to include this in the data table..

The Attempt at a Solution



no other pictorial information about how circuit, so i suppose this common emitter, have a load resistance of 2k Ω before collector and a Base resistor

so i come out with my load line equation,

Ic = -Vce / 2k + 4m
I induced this load line in my attachment,

So let's say i choose my quiscent operating point at
Ib = 40 uA, and my output current and voltage peak to peak is :

Vpp = 5.8 - 3.1 = 2.7V
Ipp = 2.45 - 1.2 = 1.25 m A

so my current amplification is
Ai = 1.25m / 40u = 31.25
Isn't your Ib varying from the quiescent point to produce the change in Ic (A signal superimposed upon the quiescent operating point)?
and if the input resistance is 1.5k Ω,
Presumably this is a given value?
my voltage amplification is,

Av = (Iout)(Rload) / (Iin)(Rinput)
=Ai x (Rload)/(Rinput)
= 31.25(2/1.5)
= 41.667
 
  • #3
Isn't your Ib varying from the quiescent point to produce the change in Ic (A signal superimposed upon the quiescent operating point)?

yes

And the given value is the output resistance(2k ohms), input resistance(1.5k ohms)
And a table of Ic and Vce value at different Ib(20u, 40u, 60u, 80).

In the attachment file i posted in the my first post, the right most column is the value i compute with
Ic = -Vce / 2k + 4m, is not actually the value given in the question.
 
  • #4
null void said:
yes
So shouldn't the actual ΔIb be used to calculate the current gain rather than the fixed quiescent current?
And the given value is the output resistance(2k ohms), input resistance(1.5k ohms)
And a table of Ic and Vce value at different Ib(20u, 40u, 60u, 80).

In the attachment file i posted in the my first post, the right most column is the value i compute with
Ic = -Vce / 2k + 4m, is not actually the value given in the question.

Okay. But your table claims 68 μA for one of the Ib values... Is that a typo? should it in fact be 60 μA? If it is, it would explain your use of 40 μA in the current gain calculation.
 
  • #5
yeah my bad, it is 60u...my typo, and the question also stated the Ib swing at peak to peak value of 40u, forgot to include this.

Vcc = 8v
 
  • #6
null void said:
yeah my bad, it is 60u...my typo, and the question also stated the Ib swing at peak to peak value of 40u, forgot to include this.

Vcc = 8v
Okay. So then it looks like your calculations are fine.
 
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  • #7
Thanks you very much
 

1. How do I plot a load line graph for current and voltage amplification?

To plot a load line graph for current and voltage amplification, you will need to know the equation for the load line, which is given by V = I*R, where V is the voltage, I is the current, and R is the resistance. Using this equation, plot points for different values of I and V on a graph and connect them to create the load line.

2. What is the purpose of a load line graph in current and voltage amplification?

A load line graph is used to analyze the operating point of a circuit and determine the voltage and current amplification. It helps to visually understand the relationship between voltage and current in a circuit and can be used to troubleshoot and optimize circuit design.

3. How do I find the current amplification on a load line graph?

To find the current amplification on a load line graph, you will need to determine the slope of the load line. The slope represents the resistance of the circuit, and the current amplification is equal to the inverse of the slope.

4. How do I find the voltage amplification on a load line graph?

The voltage amplification can be found by determining the intersection point of the load line and the characteristic curve of the circuit. The voltage amplification is equal to the ratio of the voltage at the intersection point to the input voltage.

5. Are there any limitations to using load line graphs for finding current and voltage amplification?

Load line graphs are a simplified representation of a circuit and may not accurately reflect the behavior of the circuit in real-world conditions. Additionally, they assume linear behavior of the circuit, which may not always be the case. Therefore, load line graphs should be used as a general guide and not as an exact measurement of current and voltage amplification.

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