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wide_swing_current_mirror [2015/05/06 20:27] (current)
104.228.198.109 created
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 +{{:​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.
  
wide_swing_current_mirror.txt ยท Last modified: 2015/05/06 20:27 by 104.228.198.109