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## Q. 8.1

Given VDD = 3 V and using IREF = 100 μA, design the circuit of Fig. 8.1 to obtain an output current whose nominal value is 100 μA. Find R if Q1 and Q2 are matched and have channel lengths of 1 μm, channel widths of 10 μm, Vt = 0.7 V, and $k_{n}^{′}$ = 200 μA/V². What is the lowest possible value of VO? Assuming that for this process technology, the Early voltage $V_{A}^{′}$ = 20 V/μm, find the output resistance of the current source. Also, find the change in output current resulting from a +1-V change in VO. ## Verified Solution

$I_{D1} = I_{REF} = \frac{1}{2} k_{n}^{′} \left(\frac{W}{L}\right)_{1} V_{OV}^{2}$

$100 = \frac{1}{2} × 200 × 10 V_{OV}^{2}$

Thus,

VOV = 0.316 V

and

VGS = Vt + VOV = 0.7 + .316 $\simeq$ 1 V

$R = \frac{V_{DD} – V_{GS}}{I_{REF}} = \frac{3 – 1}{0.1 mA} = 20 k\Omega$

VOmin = VOV $\simeq$ 0.3 V

For the transistors used, L = 1 μm. Thus,

VA = 20 × 1 = 20 V

$r_{o2} = \frac{20 V}{100 \mu A} = 0.2 M\Omega$

The output current will be 100 μA at VO = VGS = 1 V. If VO changes by +1 V, the corresponding change in IO will be

$\Delta I_{O} = \frac{\Delta V_{O}}{r_{o2}} = \frac{1 V}{0.2 M\Omega } = 5 μA$