Is BJT Impedance Proportional to Ic?

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The discussion centers on the relationship between BJT impedance and collector current (Ic). It clarifies that the transconductance (gm) is defined as gm = Ic/Vt, indicating that gm is proportional to Ic. The impedance of the current source is indeed represented as 1/gm, leading to the conclusion that the current source's impedance is inversely related to the transconductance. Additionally, the input resistance of a diode-connected BJT is expressed as (rbe||1/gm). This reinforces the concept that understanding these relationships is crucial for analyzing BJT behavior in circuits.
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In the equivalent diagram of BJT, Vbe and rbe represent the base voltage and resistance.
gmVbe is a current source, which represents the current gain.

Since gm = Ic/Vt

Vt/Ic = 1/gm
Does this mean the impedance of the current source is 1/gm ?
I know this sounds a bit weird. It's because of a homework problem posted here -
https://www.physicsforums.com/showthread.php?t=569541
(the input resistance of the diode connected BJT is (rbe||1/gm).
 
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likephysics said:
In the equivalent diagram of BJT, Vbe and rbe represent the base voltage and resistance.
gmVbe is a current source, which represents the current gain.

Since gm = Ic/Vt

Vt/Ic = 1/gm
Does this mean the impedance of the current source is 1/gm ?
I know this sounds a bit weird. It's because of a homework problem posted here -
https://www.physicsforums.com/showthread.php?t=569541
(the input resistance of the diode connected BJT is (rbe||1/gm).

You're confusing up things. gm means trans-conductance. it is derived as gm = dIc/dVbe = Ic/Vt. This Vt is thermal equivalent of voltage and is constant for a particular temperature.
It only means gm is proportional to Ic. And it does mean the current source has 1/gm impedance.
 

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