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Input impedance darlington tor

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The Electrician : I see that it is only problem when we consider the network with transistor alone. However, in addition to the active matrix by transistor, we also have the passive network by components around so it is not problem.

I tried to calculate input impedance with active matrix singular and it works fine. However, with adding Rg (gate resistance as you mentioned) to get the final result of input impedance I have to take the limit for Rg approaching infinity.


upload_2018-2-3_18-57-49.png


upload_2018-2-3_19-1-31.png
 
I tried to calculate input impedance with active matrix singular and it works fine. However, with adding Rg (gate resistance as you mentioned) to get the final result of input impedance I have to take the limit for Rg approaching infinity.

If, in fact, Rg is not infinite then the expression you obtained is correct. If there is no Rg, then it is to be expected that you will have to take the limit, because if you leave it out of the passive matrix and solve the system, you immediately get the result that your Y matrix is singular:

anhnha.png


Using a finite value for Rg is a workaround for the problem of obtaining Z when the Y matrix is singular. Using a finite Rg allows us to invert Y and then taking the limit with Rg becoming infinite after obtaining your result for Zin gives the correct result.
 

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Actually I was not clear with the designation.

Rg: small-signal internal gate resistance (intentionally added)
RG: external feedback resistance which is drawn in the self-bias circuit

So without adding Rg into the transistor matrix (as post #22) or including Rg (as in the image below), the result is same.
upload_2018-2-4_4-2-20.png


However, I just realized that the fact not adding internal Rg only work with this special circuit, i.e. self-bias circuit.

If RG is infinity meaning no feedback path from drain to gate then the problem of singular arises as you mentioned above.

upload_2018-2-4_4-7-35.png
 
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