wide_swing_current_mirror

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+ | {{:ws_cm_small.jpg|}} | ||

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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: | ||

+ | |||

+ | \[V_{B}=2V_{DSSAT}+V_T\] | ||

+ | |||

+ | 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