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The task at hand is to find the size of M5 which allows the output voltage to drop as low as 2VDSSAT.

Assume M1-M4 have the same $\frac{W}{L}$

In general: \[V_{DSSAT}=V_{GS}-V_T\] or \[V_{DSAT}=V_{G}-V_T\]

Therefore the gate of M2/M4/M5 or $V_{B}$ must be equal to:


We can write the current equation for M5 as:

\[I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( V_B-V_T \right )^2\]

\[I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( 2V_{DSSAT}+V_T-V_T \right )^2\]

\[I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( 2\left ( V_{GS1}-V_T \right ) \right )^2\]

\[I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}4\left ( \left ( V_{GS1}-V_T \right ) \right )^2\]

\[I_{D1}=\frac{1}{2}\mu C_{ox}\frac{W_{1}}{L_{1}}\left ( V_{GS1}-V_T \right )^2=I_{REF}=I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}4\left ( V_{GS1}-V_T \right )^2\]


So, M5 must be 4X smaller than the other devices to enable the widest possible swing.

wide_swing_current_mirror.txt · Last modified: 2015/05/06 20:27 by