Chapter 8
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.
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,
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