## Chapter 8

## Q. 8.1

Given V_{DD} = 3 V and using I_{REF} = 100 μA, design the circuit of Fig. 8.1 to obtain an output current whose nominal value is 100 μA. Find R if Q_{1} and Q_{2} are matched and have channel lengths of 1 μm, channel widths of 10 μm, V_{t} = 0.7 V, and k_{n}^{′} = 200 μA/V². What is the lowest possible value of V_{O}? 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 V_{O}.

## Step-by-Step

## 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,

V_{OV} = 0.316 V

and

V_{GS} = V_{t} + V_{OV} = 0.7 + .316 \simeq 1 V

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

V_{Omin} = V_{OV} \simeq 0.3 V

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

V_{A} = 20 × 1 = 20 V

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

The output current will be 100 μA at V_{O} = V_{GS} = 1 V. If V_{O} changes by +1 V, the corresponding change in I_{O} will be

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