Common Data : V_DD = 1.8V , L_min  = 0.18µm, µ_0N = 350cm²/V s, µ_0P  = 100cm²/V s, |V_TH | = 0.55, t_ox = 3.8nm, λ = 0.07V⁻¹, χ = 0.1.

For the curved surface of the cylinder, the unit vector is which gives  Parameterize ,Since we are confined to the first octant The flux through the slant surface is .  The top and the bottom caps are in the  directions, the contribution from these two give zero by symmetry.  There are two more surfaces if we consider the first octant, they are the positive x-z plane and the positive y-z plane., the normal to the former being in the direction of  while that for the latter is along  directions. The flux from the former is , while that from the latter is .  Adding up all the contributions, the total flux from the closed surface is zero. This is consistent with the fact that the divergence of the field is zero.

1. Design a simple MOS current mirror of the type shown in Fig. 17 to meet the following constraints.

1. Transistor must operate in the active region for values of to within from ground .
2. The output current must be
3. The output current must change less than 5% for change in output voltage of 0.4V

Figure 17: Figure for Question 1

Make and identical.  You are to minimize the total device area within given constraints. The device area is the total gate area Assume

2. Design a voltage reference using the Beta-multiplier.  Using spice simulation show the temperature performance of your design, by plotting reference voltage vs temperature.

3. Consider the circuit of Fig. 18 assuming and

1. Determine such that
2. If deviates from the value calculated by what is the mismatch between and
3. If the circuit fed by cascode current source changes by by how much does change?

Figure 18: Figure for Question 3

The divergence of the field is 3. The flux, therefore, is  3 times the volume of the cone which is The flux is thus  The  direct calculation of the flux involves two surfaces, the slant surface and the cap, as shown in the figure. The cap is in the xy plane and has an outward normal . The flux (because on the cap z=1 and the cap is a disk of unit radius).  Thus it remains to be shown that the flux from the slanted surface vanishes. At any height z, the section parallel to the cap is a circle of radius z. Since, the height and the radius of the cap are 1 each, the semi angle of the cone is 45⁰.

Thus the normal to the slanted surface has a component  along the z direction and  in the x-y plane. The component in the xy plane can be parameterized by the azimuthal angle  and we can write  . The area element can be written as  , the factor  appears because the length element is along the slant. Thus the contribution from the slanted surface is  . Using  , this integral can be evaluated and shown to be zero.

4. Consider the circuit of Fig. 19 with and

1. Calculate the deviation of from if is less than
2. Determine the of the amplifier .

Figure 19: Figure for Question 4

This problem is to be attempted similar to the problem  5 of the tutorial, i.e., by closing the cap and subtracting the contribution due to the cap. The divergence being 3, the  flux from the closed cone is 3 times the volume of the cone which gives   The contribution from the top face (which is a disk of  radius 2 ) is . Thus the net flux is zero.  (You can also try to get this result directly as done in problem 4, where we showed that the flux from the curved surface is zero).

5. Calculate the quiscent curent and temperature coefficient of the circuit shown in Fig.20 where and The resistor has a temperature coefficient of

Figure 20 : Figure for Question 5

We have

 

Hence

The temperature coefficient is given by