with $V^+ = 10V$ and $V^- = 0$ and the transistor parameters are: $V_{TN} = 1.8 V, 1/2\mu_nC_{ox} = 20 \mu A/V^2$ and $\lambda = 0.01 V^{-1}$. The transistor width to width ratios are: $(W/L)_2 = 6, (W/L)_1 = 12, (W/L)_3 = 3$. Determine (i) $I_{REF}$ (ii)$I_o$ at $V_{DS2} = 2V$

From figure,

$I_{D1} = I_{D3}$ i.e $k_{n1}(V_{GS1}-V_{TN1})^2(1+\lambda V_{DS1}) = k_{n3}(V_{GS3}-V_{TN3})^2(1+\lambda V_{DS3})$

Assuming $\mu_{n} , C_{ox} $ and $V_{TN}$ common for both the transistors,

Therefore,

$V_{GS1} = \sqrt {\dfrac {(W/L)_3}{(W/L)_1}} V_{GS3} + (1 - \sqrt {\dfrac {(W/L)_3}{(W/L)_1}}) V_{TN}$

where $V_{TN}$ is the threshold voltage of both transistors. From the circuit we see that

$V_{GS1} + V_{GS3} = V^+ - V^-$

Therefore,

$V_{GS1} = \dfrac {\sqrt {\dfrac {(W/L)_3}{(W/L)_1}}}{1 +\sqrt {\dfrac {(W/L)_3}{(W/L)_1} }} (V^+ - V^-) + \dfrac{1 - \sqrt {\dfrac {(W/L)_3}{(W/L)_1}}}{1 +\sqrt {\dfrac {(W/L)_3}{(W/L)_1}}} V_{TN} = V_{GS2}$

Substituting, we get

$V_{GS1} = V_{GS2} = 3.93 V$

Therefore, $V_{GS3} = (V^+ - V^-) - V_{GS1} = (10-0) - 3.93 = 6.07V$

Therefore, $I_{REF} = \dfrac {1}{2} \mu_n C_{ox} (\dfrac {W}{L})_3 (V_{GS3}-V_{TN})^2(1+\lambda V_{DS}) = 1.115 mA$

Now, $I_O = \dfrac {1}{2} \mu_n C_{ox} (\dfrac {W}{L})_2 (V_{GS2}-V_{TN})^2(1+\lambda V_{DS2}) = 0.555 mA$