Biasing NMOS using current mirror

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Discussion Overview

The discussion revolves around the use of a current mirror to bias an NMOS transistor using a constant current model. Participants seek clarification on the operation of the current mirror, its role in setting the drain current (Id) of the NMOS, and how this relates to signal transfer and voltage gain.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Salil expresses confusion about how a current mirror biases an NMOS and questions where the input is applied.
  • Salil asks how the current can remain constant while still allowing for signal transfer, indicating uncertainty about the relationship between Id and input signals.
  • Another participant seeks clarification on whether Salil is asking about achieving voltage gain with a constant current.
  • Salil acknowledges confusion about the workings of the NMOS when biased with a constant current and requests further explanation.
  • A participant shares a link to slides that explain the current mirror concept, suggesting that the relationship between input and output currents can be derived from basic principles.
  • Salil expresses gratitude for the information but still feels uncertain about how the NMOS operates under constant current biasing.

Areas of Agreement / Disagreement

The discussion reflects a lack of consensus, with participants expressing varying levels of understanding and confusion about the concepts involved. Some seek clarification while others provide resources, but no definitive answers are reached.

Contextual Notes

Participants have not fully resolved the assumptions regarding the behavior of the NMOS under constant current biasing, nor have they clarified the implications for signal transfer and voltage gain.

salil87
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Hi

I need a brief explanation on how a Current Mirror is used to Bias an NMOS using Constant Current Model. My Doubts are :-
* Where the input is given ?
* If the current remains constant how does it work ?
* If the current does not remain constant then why say constant current ?
Bit Confused. Little help needed. :)

Thanks
Salil
 
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Not sure what "bias an NMOS" means. Could you be more specific?
 
Thanks for replying es1. By Biasing I mean setting the Qpoint of the NMOS using the current mirror i.e. setting Id constant. But I am not getting how the signal would be transferred from the input to the output. If Id remains constant what is the quantity that changes with the input signal? Coz i think since Id is set using constant current source ... it won't change. Not able to put the pieces together :-(. A Brief Explanation or a good doc will help me out :-). Thanks.

Thanks
Salil
 
Not entirely sure if I understand your wording correctly. Are you wondering how there is a voltage gain given a constant current?
 
sandy.bridge said:
Not entirely sure if I understand your wording correctly. Are you wondering how there is a voltage gain given a constant current?

Sorry for that :-(. But i guess i m a bit confused about how it works. May be that's why i can't frame my question well. My notes are not giving good explanation. I am just not getting the main idea how it works. If you have time sandy.bridge , could you pls explain me how there is voltage gain ? that would definitely work for me.

Thanks
Salil
 
Slides 2-4 explain the mirror concept pretty well I think.
http://www.aicdesign.org/SCNOTES/2010notes/Lect2UP160_(100325).pdf

On slide 4 is should be pretty trivial to find iI when it is set by a resistor from a constant source in the standard most basic configuration. It's just iI=(Vdd-Vgs)/Rref because the gate and drain are shorted.

And on slide 4 you can see that if the transistors are matched (in this case that means they have the same W & L) then iO=iI.
 
es1 said:
Slides 2-4 explain the mirror concept pretty well I think.
http://www.aicdesign.org/SCNOTES/2010notes/Lect2UP160_(100325).pdf

On slide 4 is should be pretty trivial to find iI when it is set by a resistor from a constant source in the standard most basic configuration. It's just iI=(Vdd-Vgs)/Rref because the gate and drain are shorted.

And on slide 4 you can see that if the transistors are matched (in this case that means they have the same W & L) then iO=iI.

Wow, Thanks. I got how the current mirror works. Phew! But still confused how the NMOS works when it is biased using constant current. :-)

Thanks
Salil
 

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