Taking into account the finite output resistance of the transistors, we can write the load and reference currents as follows:

$I_O=K_{n2}(V_{GS}-V_{TN2})^2(1+\lambda_2V_{DS2})$

and

$I_{REF}=K_{n1}(V_{GS}-V_{TN1})^2(1+\lambda_1V_{DS1})$

Since transistors in the current mirror are processed on the same integrated circuit, all physical parameters, such as Vtn, $\mu$, Cox and $\lambda$ are essentially identical for …