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Explain Basic Current Mirror. Discuss Drawback and explain how cascode current Mirror overcomes it.

Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: Medium

cvd • 1.2k  views
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Current mirror Circuits:


Basic current mirror :-

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$I_{Ref}=I_{D_1}=\frac{1}{2}(\frac{W}{L})_1\mu_n C_{ox}(V_{GS}-V_{TH})^2.......(1)$

Similarly,

$I_{out}=I_{D_2}=\frac{1}{2}(\frac{W}{L})_2\mu_n C_{ox}(V_{GS}-V_{TH})^2...........(2)$

$\therefore from \ (1)\ \& \ (2)$

$I_{out}=\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}I_{Ref}$

Thus by adjusting the value of $(\frac{W}{L})$ ratio of both factor,we can get required value of O/P current


Drawbacks of Basic current mirror

  • In Basic CM , we have neglected channel modulation.
  • CLM effects result in significant error.

    $I_{D_1}=\frac{1}{2}\frac{W}{L}\mu_n C_{ox}(V_{GS}-V_{TH})^2(1+\lambda V_{DS_1})$

    $I_{D_2}=\frac{1}{2}\frac{W}{L}\mu_n C_{ox}(V_{GS}-V_{TH})^2(1+\lambda V_{DS_2})$

    $\therefore \frac{I_{D_2}}{I_{D_1}}=\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}\frac{(1+\lambda V_{DS_2})}{(1+\lambda V_{DS_1})}$

    $\therefore$ to suppress effect of CLM, cascode CM is used.


Cascode:-

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$V_b$ is chosen such that $V_x=V_y$ then $I_{out}$ tracks$ I_{Ref}$

enter image description here

Aim : $V_x=V_y$

$\therefore V_b=V_{GS_3}+V_x$

if $\frac{(\frac{W}{L})_3}{(\frac{W}{L})_4}=\frac{(\frac{W}{L})_2}{(\frac{W}{L})_1}$

then $V_{GS_3}=V_{GS_4} \& V_x=V_y$

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