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. Use the Miller approximation to calculate the frequency of the small signal voltage gain of a common source stage as shown in Fig. 21, using and the following transistor data. Ignore channel-length modulation.

Figure 21: Figure for Question 1

2. Calculate the frequency of the circuit in Fig. 22 assuming the following parameter values.

Figure 22 : Figure for Question 2

Assume

3. Use zero value time constant method to estimate the small signal dominant pole for the current gain of the cascode current mirror shown in Fig. 23. Assume an input ac current source in parallel with and a zero load impedance with The bias current Compare your answer with the value of the devices. Device parameters are Ignore channel length modulation and substrate effect. Given

Figure 23 : 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. The ac schematics of a common source stage and a common source common gate(cascode) stage are shown below, with and Using the transistor and oper- ating point data as Given Assume

1. Calculate the low frequency small signal voltage fain for each circuit
2. Use the zero value time constant method to calculate and compare the frequencies of teh gains of the two circuits .
3. Estimate the 10% to 90% rise times for each circuit for a small step input and sketch the output voltage waeform over 0 to 300ms for a step input

Figure 24 : 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. Find an expression for the transfer function of an source follower amplifier (Fig.25). Ignore body effect.

Figure 25 : Figure for Question 5

Consider the ac equivalent of the source followe as shown in Fig. 26. Applying at the outout node we can write

Figure 26 : Figure for Question 5

Thus

Applying KVL for the input section we can write

From above two equations we can write the transfer function as

Note : Note that the transfer function has a zero. Thus the estimation of bandwidth cannot be done using the dominant pole approximation. i.e. the zero value time constant technique cannot be used for circuits with zero’s in the transfer function. Such circuits can be identified by inspecting for a direct capacitive path from the input to output. The pole value estimated by the zero value technique will be correct but will not correspond to the bandwidth of the circuit due to the zero