I was investigating a MOS current mirror on the photo however i could not understand how the equation (3) was obtained from equation (2). I put a question mark also there on the photo.
My second question is that the main advantage of connecting base and collector in a current mirror. What are the advantages of using current mirrors? If you share your idea i will be happy.
(There must be a connection between drain and gate of M1. I forgot to draw it.)
Answer
First you need a connection between drain and gate for you current mirror to work properly.
To answer your question, the step from (2) to (3) involves an approximation. The term \$\lambda Vds\$ is usually quite small, and therefore $$ \frac{1}{1+\lambda V_{ds1}} \approx 1 - \lambda V_{ds1}$$ in the next step (after multiplication) the the term \$ \lambda^2 V_{ds2} V_{ds1} \$ is neglected which results in the approximation $$ (1 + \lambda V_{ds2}) (1 - \lambda V_{ds1}) \approx 1 + \lambda (V_{ds2} - V_{ds1}) $$
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