How Does the H Parameter Model Calculate Current and Voltage Amplification?

In summary, the problem is that the author is not able to correctly calculate voltage amplification. Av is defined as Vout/Vs, but they are not correctly calculating Vout/Vs. They need to take into account Rs when calculating voltage amplification.
  • #1
null void
102
1

Homework Statement


M.aspx?v=8c69668658a4757aa666.jpg


calculate current and voltage amplification

The Attempt at a Solution


Conversion to h parameter model,
M.aspx?v=8c696686596370aeb29b.jpg


and this is my working step,
https://www.filesanywhere.com/fs/v.aspx?v=8b6b628b5e646faf6e9b
 
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  • #2
there are something i can't understand, in my calculation, i use equation 3 to find Ai, i get -132... But if i use equation 6 instead of 3, i get 132..., same number but different sign. I suppose there is something went wrong in my equation 6...
 
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  • #3
Your eqn(2) doesn't correctly show how hfeib divides between the two resistances.

I suggest that, in future, after you create something as a .docx file that you take a screenshot of your effort and post it as a jpeg instead of the .docx file. This will allow those who rely on a basic tablet to still follow the thread, using software add-ons no more sophisticated than a jpeg viewer.

A further suggestion: I think it's a good idea to informatively label dimensionless parameters, e.g.,
hfe = ... amps/amp
hre = ... volts/volt

It helps you to stay correctly focussed, and makes checking easier (especially by those of us who may have forgotten some of these. :wink:
 
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  • #4
null void said:
there are something i can't understand, in my calculation, i use equation 3 to find Ai, i get -132... But if i use equation 6 instead of 3, i get 132..., same number but different sign. I suppose there is something went wrong in my equation 6...

Your equation 2 is Iout = hfe Ib + hoe Vout

Equation 3 is Iout = -hfe Ib () What are these parentheses?

Why is there a minus sign in front of hfe in equation 3 but not in equation 2?

Shouldn't equation 2 be Iout = hfe Ib + hoe Vout + Vout/Rc?

Also, I don't see anywhere where you have taken Rs into account. Rs has a very large effect on Av.

I would think that Av would be defined as Vout/Vs. Is your problem defining it as Vout/Vin, which is the same as Vout/Vb?
 
  • #5
yeah sorry messed up a few thing up there, this is my second attempts
M.aspx?v=8c6966865b9ea57bab6b.jpg


so is this the correct ways?

Also, I don't see anywhere where you have taken Rs into account. Rs has a very large effect on Av.

I would think that Av would be defined as Vout/Vs. Is your problem defining it as Vout/Vin, which is the same as Vout/Vb?

I think the arrow in the diagram is telling us the Vin is Vb
 
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  • #6
This is for the voltage amplification, but i can't really figure out how to include the Rs in the calculation
M.aspx?v=8c6966865c66b676a497.jpg


A further suggestion: I think it's a good idea to informatively label dimensionless parameters, e.g.,
hfe = ... amps/amp
hre = ... volts/volt

Do you mean by putting amps/amp behind the term like a unit?
 
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  • #7
For your Iout/Iin expression, the algebra looks right, though there's a transcription error: you show RB hie where it should be RB + hie. I haven't checked your arithmetic, and haven't checked your voltage equation.

You may need a sign change. The arrow direction for iout is indicating iout = - vout / RC
 
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  • #8
If VB is correctly denoted as Vin, then as you observed earlier, Rs does not enter into your calculations.
 
  • #9
In post #5 and #6, you have used a value of .87k for hie, but in post #1 its value was given as .86k; which is correct?

null void said:
...i can't really figure out how to include the Rs in the calculation

An easy way is to calculate the input resistance at the Vin node, and use that impedance plus the 1k Rs to form a voltage divider. The voltage division ratio from the Vs node to the Vin node can be multiplied times the Av you already calculated to get the overall voltage gain from Vs to Vout.
 
  • #10
You may need a sign change. The arrow direction for iout is indicating iout = - vout / RC
Oh yeah, I get what u mean.

hah sorry hie it is actually 0.86k ohms

An easy way is to calculate the input resistance at the Vin node, and use that impedance plus the 1k Rs to form a voltage divider. The voltage division ratio from the Vs node to the Vin node can be multiplied times the Av you already calculated to get the overall voltage gain from Vs to Vout.

I find it is a bit tricky to find the input impedance even tough Rs is not included because of the HreVout, it causes great pain for me to kill off Ib , Iin and Vb.
 
  • #11
null void said:
I find it is a bit tricky to find the input impedance even tough Rs is not included because of the HreVout, it causes great pain for me to kill off Ib , Iin and Vb.

In post #1, you said your problem was to "calculate current and voltage amplification", but you didn't say whether voltage amplification, Av, was defined as Vout/Vs or Vout/Vin.

Have you decided which it is?

Do you still need to calculate Vout/Vs?
 
  • #12
NascentOxygen said:
You may need a sign change. The arrow direction for iout is indicating iout = - vout / RC

from the

-hfeIb = Vout (hoe + 1/Rc)

i should change it to

hfeIb = Vout (hoe + 1/Rc)

then at the end, my Ai,

Ai = RB/ [ (Rchoe + 1)( RB + hie )/hfe + hreRc ]

With this change, i get positive 132.4439 amplification
 

What is the "H parameter model" in scientific research?

The "H parameter model" is a mathematical model used in scientific research to describe and analyze the behavior of complex systems. It was developed by physicist Hermann Haken and is based on the principles of self-organization and non-equilibrium thermodynamics.

How does the "H parameter model" differ from other mathematical models?

Unlike other mathematical models, the "H parameter model" takes into account the interactions and feedback between different components of a system, rather than treating them as isolated entities. This allows for a more comprehensive understanding of complex systems, particularly in the fields of biology, chemistry, and physics.

What are some practical applications of the "H parameter model"?

The "H parameter model" has been used to study a wide range of phenomena, including chemical reactions, neural networks, and ecological systems. It has also been applied in fields such as economics, sociology, and engineering to better understand and predict the behavior of complex systems.

What are the limitations of the "H parameter model"?

While the "H parameter model" has proven to be a useful tool in studying complex systems, it also has its limitations. It is a highly theoretical model and may not always accurately reflect real-world systems. Additionally, it requires a significant amount of computational power and data to accurately simulate and analyze complex systems.

How can the "H parameter model" benefit scientific research?

The "H parameter model" can provide scientists with a more holistic understanding of complex systems and their behavior. It can also help identify patterns and relationships within these systems, leading to new insights and discoveries. Additionally, the model can aid in predicting the behavior of complex systems under different conditions, allowing for more informed decision-making in various fields of study.

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