Current Mirror using MOSFETs

Discussion in 'Homework Help' started by BrentM, Nov 7, 2012.

  1. BrentM

    Thread Starter Member

    Oct 26, 2012
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    Hi, I am making a current mirror for one of my classes using NMOS transistors. The required reference, Id, current is roughly 1mA, and the desired load current is 100μA. Basically, in order to lower the current to the load I need to add a resistor going from the source of the second MOSFET to ground. I have attached a schematic of the design. Here is the work I have done so far:

    MOSFET parameters: Kn = 111e-6, W/L = 3, Vt = 2 V

    since 1mA reference current(Id) is desired, and we know that N1 is in saturation,

    Id1 = (1/2)*(Kn)*(W/L)*(Vgs - Vt)^2

    plugging in our values

    1mA = (1/2)*(111e-6)*(3)*(Vgs - 2)^2

    Vgs = 4.582 V

    therefore the resistor, Rd:

    Rd = (Vdd - Vgs) / Id = (12 V - 4.852 V) / (0.001) = 7.4kΩ

    I know that my desired current, Is, is 100μA and is one tenth of the reference current, Id:

    Is = Id / 10

    I am having trouble calculating a value for Rs to get 100uA. Is there some way I should be manipulating the Id equation that I used above? Or should I be using simple voltage division and ohms law? Thanks for any help.


    Brent
     
  2. Ron H

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    Apr 14, 2005
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    Have you been told that you must add a resistor? The normal way of scaling currents is to scale transistor geometry (W and L).
     
  3. BrentM

    Thread Starter Member

    Oct 26, 2012
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    I was told that I had to add a resistor, because the MOSFETs are on an IC with equal geometry.
     
  4. Ron H

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    It' pretty straightforward. Solve the Id equation for Vgs2 when Id=100uA. Getting Rs is simple once you have Vgs2.
    Come back with your method and answer.
     
  5. BrentM

    Thread Starter Member

    Oct 26, 2012
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    Ok that seems easy enough. So once I determine Vgs2, I can use ohms laws like this:

    (Vgs2 - Vss) / (100uA) = Rs


    Does that sound right?
     
  6. BrentM

    Thread Starter Member

    Oct 26, 2012
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    Here is what I came up with:

    @ 100uA, Vgs2 = 2.775 V and Vss = 0 V

    Rs = (2.775 - 0) / (100uA) = 27.75k
     
  7. Jony130

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    Feb 17, 2009
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    No your answer is wrong. Voltage across Rs resistor is not equal to 2.775V.
    Look at diagram and use KVL to find voltage across Rs.

    PS. All so I get Vgs1 = 4.45072V and Vgs2 = 2.77498V and Rs = 16.7574KΩ = 16KΩ
     
    Last edited: Nov 8, 2012
  8. BrentM

    Thread Starter Member

    Oct 26, 2012
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    using KVL I get Vd1 + Vss - (100uA)(Rs) = Vgs2?
    so the gate voltage at the second transistor, Vg2, is 2.775V.
    Vss = 0V


    Am I on the right track?
     
  9. Jony130

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    Feb 17, 2009
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    No, you need to use this KVL loop
    [​IMG]
     
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  10. Ron H

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    That's what I get also.
     
  11. BrentM

    Thread Starter Member

    Oct 26, 2012
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    so KVL would,

    Vgs1 = Vgs2 + Vrs?
     
  12. BrentM

    Thread Starter Member

    Oct 26, 2012
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    I tried using the resistor value you came up with and I did not get a load current of 100uA. However when I used a value of 1.4k I get the desired load current.


    Can you explain your process, I am clearly missing something.


    Brent
     
  13. Ron H

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    Apr 14, 2005
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    That's correct.

    Rs=(Vgs1-Vgs2)/100uA
     
  14. Jony130

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    Feb 17, 2009
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    You build this circuit is simulator? For Rload = ?
     
  15. BrentM

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    Oct 26, 2012
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    I used a load of 1k.
     
  16. Ron H

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    Show us how you tested the 16k value, and how you got the 1.4k result.
     
  17. BrentM

    Thread Starter Member

    Oct 26, 2012
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    I used MicroCap 10 to simulate the circuit. I have attached a schematic. V1 and V2 were both 12 VDC sources, and R2 is the load resistor(1k). When I simulate the current through resistor, R3, it's right around 45uA. However, if I change R3 to 1.4k, the current through it is right around 100uA.
     
  18. Ron H

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    Do both of your transistors have the parameters you specified in your first post? If not, then your simulation will not give you the correct results.
     
  19. BrentM

    Thread Starter Member

    Oct 26, 2012
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    Good point. I'll check it out. Thanks for all of your help Ron and Jony. I really appreciate it.
     
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