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 circuits shown in Fig.  33.  Which of these source degenerated circuits have higher input referred noise voltage, and by what factor?

Figure 33 :  Source Degenerated Circuits

2. Design circuit in Fig. 34 for an input referred thermal noise voltage of and maximum output swing. Assume and

Figure 34: Common Gate Stage

3. The circuit in Fig. 35 is designed with and If an output swing of is required, estimate the dimension of and such that the input referred thermal noise current is minimum .

Figure 35: Common Gate Stage

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. Assume Fig. 36. If the contribution of to the input referred noise voltage must be one-fifth of that of , what is the maximum output voltage swing of the amplifier?

Figure 36 : (a) Common Source Amplifier (b) Source Follower

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 circuit of Fig. 37 - Fig. 38. To decrease overall noise of circuit,

1. Whether transconductance of should be kept high or low in two cases of Fig. 37?

Figure 37 : (a) Common Source Amplifier (b) Inverter

2. Whether should be minimized to reduce noise in Fig. 38?

Figure 38 : Source Degenerated Stage

6. Consider input stage of op-amp shown in Fig. 39,

1. Should transconductance of be decreased to decrease input-referred thermal noise?
2. How sizing of   and should be done to reduce flicker noise?

Figure 37 : (a) Common Source Amplifier (b) Inverter

Solution :

Part (a)

Step 1: Find gains from each noise source to the output node.

Figure 39 : Input stage of CMOS Opamp

Figure 40 : Opamp input MOSFET noise sources

1. Gain from and .
2. Gain from and .
3. Gain from and .

(This gain factor is relatively small compared to the others, so, it could be ignoredfor further calculations.)

Step 2: Using the gain factors, find the output noise.

Step 3:  Find input referred thermal noise voltage by dividing output noise value with gain which results in

where,

Thus, noise contributed by two pair of transistors is inversely proportional to the transconductance . Therefore, should be made as large as possible to minimize thermal noise contrinution.

Part (b)

Step 1: In output noise equation, substitute transconductance with transistor size as parameter.

Substituting

Step 2: Substitute noise sources with spectral density

Thus, from the input-referred flicker noise equation,

1. Taking longer greatly helps due to inverse squared relationship in the second term .
2. The input noise is independent of and therefore it could be made large to maxi- mize signal swing at the output.
3. Taking wider also helps to minimize noise.
4. Taking increases the noise because the second term is domnant  (due to Thus, large decreases the input-referred noise of p-channel drive transistors which are not the dominant noise source, but, increases the input-referred noise of n-channel load transistors, which are dominant noise sources.