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. Calculate the unloaded voltage gain and output resistance of the circuit in Fig.13 Assume Ignore body effect.

Figure 13: Figure for Question 1

2. Derive the expression for and for the p-channel input differential amplifier.

3. Design the current and of the current mirror load differential amplifier to satisfy the following specification slew rate and a small signal gain of and

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. For a folded cascode current  mirror differential  amplifier  shown in Fig.14 assume for all the transistors .

1. Find the maximum input common mode voltage and the minimum com- mon mode voltage
2. What is the ICMR?
3. Find the small signal gain
4. If a capacitor is connected from the output to ground, what is the frequency for in hertz .

Figure 14: 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. Consider the NMOS and PMOS input stages as shown in Fig. 15. Given that ans the saturation voltages of the current sources is also determine

1. The common input range of the P channel input pair
2. The common input range of the N channel input pair
3. The common input range of the complementary input stage shown in Fig. 16

Figure 15: Figure for Question 5

The common mode input range for the P channel input stage is given by

Thus the ICMR is The common mode input range for the N channel input stage is given by

Thus the ICMR is

Figure 16: Figure for Question 5

For the complementary input stage the ICMR is rail to rail, i.e due to the overlap of the two ICMR’s of the complementary stage.