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.Consider a variable delay inverter with delay where is a control voltage to adjust the delay. If this inverter is used to make a three stage ring oscillator, find the oscillation frequency with respect to the control voltage considering

2. Consider a three stage ring oscillator shown in Fig. 41. The varactor capacitance is given by

1. Derive an expression for small signal open loop gain of the circuit in terms of .
2. From the above expression derive the frequency of oscillation.

Figure 41 : Figure for Question 2

3. A three stage ring oscillator is made with inverters that have asymmetric delay for the rising and falling edge, due to difference in channel mobilities of and devices .

Given and Will this circuit oscillate. If yes calculate the frequency of oscillation and the duty cycle of the oscillator output .

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 small signal equivalent of an oscillator as shown in Fig. 42. The inductor has a series resistance of . Estimate the maximum permissible value of in terms of and for sustained oscillations.

Figure 42 : 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 oscillator shown in Fig. 43. Capacitance of the varactor is given by

Figure 43 : Figure for Question 5

1. Estimate the gain
2. If a sinusoidal signal overrides over the control voltage, determine the frequency components at the output. Given

6. Consider the oscillator shown in Fig. 44. Treating the transistors to be ideal switches (the output voltage has only fundamental harmonic), sketch the spectrum of the current through the inductor, capacitor and resistor.

Figure 44 : Figure for Question 6

Given sinusoidal output voltage, fundamental component of the current through inductor and capacitor are equal and out of phase. The current through the transistors is square. This square current will have all odd harmonics. The tank together presents open circuit for the fundamental frequency. Hence this current flows through the resistor. The higher harmonics flow through the capacitor as it presents a low impedance as compared to the inductor and resistor. The bias current for the transistor flows only through the inductor as it has zero DC resistance.

Fig. 45 shows the sketch spectrum of the currents through the inductor, capacitor and resistor.

Figure 46 : Solution to Question 6 : Spectra of currents through R L and C