# The Analog Dog

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wide_swing_current_mirror

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 — wide_swing_current_mirror [2015/05/06 20:27] (current)104.228.198.109 created 2015/05/06 20:27 created 2015/05/06 20:27 created Line 1: Line 1: + {{:​ws_cm_small.jpg|}} + + + 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: + + $V_{B}=2V_{DSSAT}+V_T$ + + We can write the current equation for M5 as: + + $I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( V_B-V_T \right )^2$ + + + $I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( 2V_{DSSAT}+V_T-V_T \right )^2$ + + $I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}\left ( 2\left ( V_{GS1}-V_T \right ) \right )^2$ + + $I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}4\left ( \left ( V_{GS1}-V_T \right ) \right )^2$ + + + $I_{D1}=\frac{1}{2}\mu C_{ox}\frac{W_{1}}{L_{1}}\left ( V_{GS1}-V_T \right )^2=I_{REF}=I_{D5}=\frac{1}{2}\mu C_{ox}\frac{W_{5}}{L_{5}}4\left ( V_{GS1}-V_T \right )^2$ + + $\frac{W}{4L}=\frac{W_{5}}{L_{5}}$ + + So, M5 must be 4X smaller than the other devices to enable the widest possible swing.