Monkeybad wrote on Aug 2nd, 2007, 7:04pm:Why not? The current source will become more ideal with large L.
Besides, I'm wondering that why the large L will affect the Vdsat and the input common mode voltage.
If you keep the Vgs of M1 the same, then you can still have the same Vdsat of the M1 for the saturation.
But now the current of M1 is lower when you increase the L, so you can increase the W to keep the
same current. Isn't it? (It's just my thought and correct me if I'm wrong.)
BEST REGARD
Yes, you are absolutely right. The drain-source current is proportional to W/L and the output resistance to 1/L, so by increasing W and L with the same factor you can achieve a higher output resistance. But a larger transistor also has higher capacitances, so the effective output capacitance of your current source will increase. If you do not expect to have any voltage swing at the output node of the current source, then the extra capacitance will not matter. But if you expect large voltage swings at high frequenies, it might actually degrade the output impedance. A larger transistor also takes up more area which might be valuable. Summarizing you can say that, increasing W and L with the same factor for the output transistor in this current source results in higher output resistance, larger area and larger output capacitance.