High Swing Bias Design for a Cascode Transistor

[Could not find the bibliography file(s)

High Swing Bias

A high swing bias scheme for a cascode transistor is shown in Figure 1.

Figure.1 High swing bias generation circuit for cascode transistor

  • Transistors M1-M5 are in saturation and Mx1-Mxn are in triode region
  • Square law model of MOST is considered for analysis

Consider the general case where M3 and M4 have different aspect ratios. But M3 & Mx and M4 & M5 have same aspect ratios hence same \beta. In such a case from square law model the channel current is given by,

(1)   \begin{eqnarray*} I_o &=& \beta_3 V_{ov3}^2 =\beta_4 V_{ov4}^2 \\     &=& {\beta_3 \over n}\left( 2(V_{gsx}-V_{Tx})V_{dsx}-V_{dsx}^2\left) \end{eqnarray*}

From Eq.1, V_{ov4} = \sqrt{\beta_3 \over \beta_4} V_{ov3}

By KVL at node X,

(2)   \begin{eqnarray*} V_{gsx} &=& V_{dsx} + V_{gs5}\\         &=& V_{ds3} + V_{gs4} \end{eqnarray*}

If M4 and M5 are matched, V_{gs4}=V_{gs5} \Rightarrow V_{ov4}=V_{ov5} and V_{dsx}=V_{ds3}.

From Eq.2

(3)   \begin{equation*} V_{gsx}-V_{Tx} = V_{dsx}+V_{T5}+V_{ov5} - V_{Tx}=V_{dsx} + V_{ov5} + V_{\Delta T} \end{equation*}

Substituting Eq.3 in Eq.1, we get

(4)   \begin{eqnarray*} I_o = {\beta_3 \over n}\left(2(V_{dsx} + V_{ov5} + V_{\Delta T})V_{dsx}-V_{dsx}^2\right) =\beta_3 V_{ov3}^2   \\ \Rightarrow {1\over n}\left(2(V_{dsx} + \sqrt{\beta_3 \over \beta_4} V_{ov3} + V_{\Delta T})V_{dsx}-V_{dsx}^2\right)= =\beta_3 V_{ov3}^2   \\ \end{eqnarray*}

    \[ \Rightarrow V_{dsx}^2 + 2 \gamma V_{ov3} - nV_{ov3}^2 =0 \]

Solving for V_{dsx}, we get,

    \[ V_{dsx}=(\sqrt{\gamma^2 +n } - \gamma) V_{ov3} \]

where \gamma = \sqrt{\beta_3 \over \beta_4}  + {V_{\Delta T} \over V_{ov3}}

  • Mx1-Mxn devices replace quarter-size device
  • Less sensitive to body effect[?][?]


Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.