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 Cascode amplifier (Fig. 7 ) for a gain A_V > −50V /V , V out_max of 1.3V, V out_min of 0.4V. The slew rate with 5pF load should be at least 10V /µsec. Supply voltage is 1.8V and maximum power dissipation is 200µW .

Figure 7: Figure for Question 1

2. Calculate the transconductance and output resistance of the cascode circuit shown in Fig. 8 . Assume both transistors are in saturation and g_m = 2mA/V and r₀ = 20kΩ. Ignore body effect. Repeat the calculation including body effect and observe the percentage of error.

Figure 8: Figure for Question 2

3. Calculate the output impedance of the circuit shown in Fig. 9 and state the advantages and disadvantages of the circuit.

Figure 9: 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. Find the output resistance for the active cascode circuit shown in Fig.  10 excluding resistor R. What is the maximum allowable output swing?

Figure 10: 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. Determine input resistance, transconductance, output resistance and maximum open circuit voltage gain of cascode if I_D = 300µA, W/L = 200.

Figure 11: Figure for Question 5

6. Consider the cascode amplifier shown in Fig. 12. Given g_m1 = 1mA/V , g_m2 = 0.5mA/V , calculate the gain of the amplifier and the swing on node X. Ignore channel length modulation and body effect.

Figure. 6 Circuit for Q6

The gain of the amplifier is given by

Hence the gain is approximately 10.  The net impedance from node to ground is which is about 2kΩ. The current through is given by . Hence the swing at node is given by

Note that here the cascode transistor reduces the swing on node thus providing shielding between the output and the  input.This  shielding  property  is  important  in high gain and large signal amplifiers.