Question: Explain qualitative analysis of differential pair.
1

Subject: CMOS VLSI Design

Topic: Differential Amplifiers

Difficulty: Medium

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modified 7 months ago by gravatar for Ankit Pandey Ankit Pandey70 written 18 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 270
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Qualitative analysis of basic differential pair:

  • Assume $V_{in_{1}}-V_{in_{2}}$ varies from $-\infty \,\, to + \infty$

enter image description here

Fig:Diffrential mode characteristics(DM behaviour)

  • $V_{in}$ is more -ve than $V_{in_{2}},\,M_1$ - OFF & $M_2$- ON & $I_{D_{2}}=I_{SS}$
    $\therefore\,\,V_{out_{2}}=V_{DD}-R_D\,I_{SS}$
    $\therefore\,\,V_{out_{1}}=V_{DD} $

  • As $V_{in}$ increases, $M_1$ is gradually turned ON, drawing a fraction of current from $I_{SS}$ & hence $V_{out_{1}}$ decreases and $I_{D_{2}}$ decreases and $V_{D_{2}}$ increases.

enter image description here

  • $V_{in_{1}}=V_{in_{2}}, V_{out_{1}}=V_{out_{2}}=V_{DD}-R_D\,I_{SS}/2$

  • $V_{in_{2}}$ is more positive than $V_{in_{2}}$, $M_1$ carries more current.

    $\therefore\,\, V_{out_{1}}$ drops below $V_{out_{2}}$.
    $\therefore\,\, V_{out_{1}}=V_{DD}-I_D\,R_D=V_{DD}-I_{SS}\,R_D$
    $\therefore\,\, V_{out_{2}}=V_{DD}$


Two important attributes:

(a) minimum & maximum levels at the output are well defined ($V_{DD}\,\,and \,\,V_{DD}-I_{SS}\,R_D$).
(b) $V_{out_{1}}-V_{out_{2}}$ is independent of input common mode level.


Common Mode Characteristics: CM behaviour

$I_{SS}$ implemented by an NMOS.

enter image description here

  • If $V_{in,cm}=0,\,\,M_1\,\,and\,\,M_2$ -> OFF , yeilding $I_{DS}$ =0
    with $I_{D_{1}}=I_{D_{2}}=0$, the circuit is incapable of signal amplification
    $\therefore\,\,V_{out_{1}}=V_{out_{2}}=V_{DD}$.

  • If $V_{in,cm}$ becomes positive, $M_1$ and $M_2$ - ON,
    if $V_{in,cm} \geq V_{Th}, \,\,I_{D_{1}}$ and $I_{D_{2}}$ continues to increases .
    $\therefore \,\, I_{D_{1}}$ and $I_{D_{2}}$ increases -> $V_{out_{1}}$ and $V_{out_{2}}$ decreases.

  • For sufficiently high $V_{in,cm}$ the $V_{DS_{3}}$ exceeds $V_{GS_{3}}-V_{Th_{3}}$, allowing $M_3$ to operate in saturation.
    $\therefore$ Total current through $M_1$ and $M_2$ will remain constant. So beyond this, even if $V_{in,cm}$ increases, $V_{out_{1}}$ and $V_{out_{2}}$ will remain constant i.e

    $V_{out_{1}}=V_{out_{2}}=V_{DD}-\frac{I_{SS}}{2}R_D$

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modified 18 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 270 written 18 months ago by gravatar for dukare030296hemant dukare030296hemant90
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