Biasing NMOS using current mirror

  • Thread starter salil87
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
es1
324
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Not sure what "bias an NMOS" means. Could you be more specific?
 
  • #3
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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 wont change. Not able to put the pieces together :-(. A Brief Explanation or a good doc will help me out :-). Thanks.

Thanks
Salil
 
  • #4
798
1
Not entirely sure if I understand your wording correctly. Are you wondering how there is a voltage gain given a constant current?
 
  • #5
26
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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 cant frame my question well. My notes are not giving good explaination. 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
 
  • #6
es1
324
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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.
 
  • #7
26
0
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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