1. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 KEEE 4426 VLSI WEEK 3 CHAPTER 1 MOS Capacitors (PART 2) CHAPTER 1

2. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3 MOS Calculation

3. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(1) If there is no charge present in the oxide or at the oxide-semiconductor interface, the flat band voltage simply equals the difference between the gate metal work function,  M, and the semiconductor work function,  S. (1.3.1) V FB =  M -  S The work function is the voltage required to extract an electron from the Fermi energy to the vacuum level. This voltage is between three and five Volt for most metals. The actual value of the work function of a metal deposited onto silicon dioxide is not exactly the same as that of the metal in vacuum.

4. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(2) Table 1.3.1Workfunction of selected metals as measured in vacuum and as obtained from a C-V measurement on an MOS structure. The workfunction of a semiconductor,  S, requires some more thought since the Fermi energy varies with the doping type as well as with the doping concentration. This workfunction equals the sum of the electron affinity in the semiconductor, , the difference between the conduction band energy and the intrinsic energy divided by the electronic charge in addition to the bulk potential. This is expressed by the following equation: (1.3.2) (1.3.2)

5. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(3) For MOS structures with a highly doped poly-silicon gate one must also calculate the work function of the gate based on the bulk potential of the poly-silicon. Where N a,poly and N d,poly are the acceptor and donor density of the p-type and n-type poly-silicon gate (1.3.3)

6. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(4) For a pMOS capacitor, which has an n-type substrate with doping density N d, the work function difference equals: (1.3.4) The flat band voltage of real MOS structures is further affected by the presence of charge in the oxide or at the oxide-semiconductor interface. The flat band voltage still corresponds to the voltage, which, when applied to the gate electrode, yields a flat energy band in the semiconductor.

7. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(5) Any charge in the oxide or at the interface affects the flat band voltage. For a charge, Q i, located at the interface between the oxide and the semiconductor, and a charge density, ρ ox, distributed within the oxide, the flat band voltage is given by: (6.3.5) where the second term is the voltage across the oxide due to the charge at the oxide-semiconductor interface and the third term is due to the charge density in the oxide.

8. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.1 Flatband voltage calculation(6) The actual calculation of the flat band voltage is further complicated by the fact that charge can move within the oxide. The charge at the oxide-semiconductor interface due to surface states also depends on the position of the Fermi energy. Since any additional charge affects the flat band voltage and thereby the threshold voltage, great care has to be taken during fabrication to avoid the incorporation of charged ions as well as creation of surface states.

9. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 Example 1.1 Calculate the flat band voltage of a silicon nMOS capacitor with a substrate doping N a = 10 17 cm -3 and an aluminum gate (  M = 4.1 V). Assume there is no fixed charge in the oxide or at the oxide-silicon interface. Solution The flat band voltage equals the work function difference since there is no charge in the oxide or at the oxide-semiconductor interface. The flat band voltages for nMOS and pMOS capacitors with an aluminum or a poly-silicon gate are listed in the table below.

10. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.2 Inversion layer charge(1) The basis assumption as needed for the derivation of the MOSFET models is that the inversion layer charge is proportional with the applied voltage. In addition, the inversion layer charge is zero at and below the threshold voltage as described by: (1.3.6) The linear proportionality can be explained by the fact that a gate voltage variation causes a charge variation in the inversion layer. The proportionality constant between the charge and the applied voltage is therefore expected to be the gate oxide capacitance. This assumption also implies that the inversion layer charge is located exactly at the oxide-semiconductor interface.

11. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.2 Inversion layer charge(2) Because of the energy band gap of the semiconductor separating the electrons from the holes, the electrons can only exist if the p-type semiconductor is first depleted. The voltage at which the electron inversion-layer forms is referred to as the threshold voltage. To justify this assumption we now examine a comparison of a numeric solution with equation (1.3.6) as shown in Figure 1.3.2

12. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.2 Inversion layer charge(3) Fig. 1.3.2 Charge density due to electrons in the inversion layer of an MOS capacitor. Compared are the analytic solution (solid line) and equation (1.3.6) (dotted line) for N a = 10 17 cm -3 and t ox = 20 nm. (1.3.6)

13. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(1) We now derive the MOS parameters at threshold with the aid of Figure 1.3.3. To simplify the analysis we make the following assumptions: 1) we assume that we can use the full depletion approximation 2) we assume that the inversion layer charge is zero below the threshold voltage. Beyond the threshold voltage we assume that the inversion layer charge changes linearly with the applied gate voltage

14. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(2) Fig. 1.3.3 Electrostatic analysis of an MOS structure. Shown are (a) the charge density, (b) the electric field, (c) the potential and (d) the energy band diagram for an nMOS structure biased in depletion.

15. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(3) The derivation starts by examining the charge per unit area in the depletion layer, Q d. As can be seen in Figure 1.3.3 (a), this charge is given by: (1.3.7) (1.3.8) Where x d is the depletion layer width and N a is the acceptor density in the substrate. Integration of the charge density then yields the electric field distribution shown in Fig. 1.3.3 (b). The electric field in the semiconductor at the interface,E s, and the field in the oxide equal, E ox Fig. 1.3.3

16. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(4) The electric field changes abruptly at the oxide-semiconductor interface due to the difference in the dielectric constant. At a silicon/SiO 2 interface the field in the oxide is about three times larger since the dielectric constant of the oxide (  ox = 3.9  0 ) is about one third that of silicon (  s = 11.9  0 ). The electric field in the semiconductor changes linearly due to the constant doping density and is zero at the edge of the depletion region, based on the full depletion approximation. Fig. 1.3.3

17. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(5) The potential shown in Figure 1.3.3 (c) is obtained by integrating the electric field. The potential at the surface,  s, equals: (1.3.9) The calculated field and potential is only valid in depletion. In accumulation, there is no depletion region and the full depletion approximation does not apply Fig. 1.3.3

18. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(6) In inversion, there is an additional charge in the inversion layer, Q inv. This charge increases gradually as the gate voltage is increased. However, this charge is only significant once the electron density at the surface exceeds the hole density in the substrate, N a. We therefore define the threshold voltage as the gate voltage for which the electron density at the surface equals N a. This corresponds to the situation where the total potential across the surface equals twice the bulk potential,  F. (1.3.10)

19. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(7) The depletion layer in depletion is therefore restricted to this potential range: (1.3.11) For a surface potential larger than twice the bulk potential, the inversion layer charge increases exponentially with the surface potential. Consequently, an increased gate voltage yields an increased voltage across the oxide while the surface potential remains almost constant. We will therefore assume that the surface potential and the depletion layer width at threshold equal those in inversion. The corresponding expressions for the depletion layer charge at threshold, Q d,T, and the depletion layer width at threshold, x d,T, are: (1.3.12) (1.3.13)

20. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(8) Beyond threshold, the total charge in the semiconductor has to balance the charge on the gate electrode, Q M, or: where we define the charge in the inversion layer as a quantity, which needs to determined but should be consistent with our basic assumption. This leads to the following expression for the gate voltage, V G : In depletion, the inversion layer charge is zero so that the gate voltage becomes: (1.3.14) (1.3.15) (1.3.16)

21. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 1.3.3 Full depletion analysis(9) while in inversion this expression becomes: the third term in (1.3.17) states our basic assumption, namely that any change in gate voltage beyond the threshold requires a change of the inversion layer charge. From the second equality in equation (1.3.17), we then obtain the threshold voltage or : (1.3.17) (1.3.18)

22. Norhayati Soin 06 KEEE 4426 WEEK 3/2 13/01/2006 Example 1.2 Calculate the threshold voltage of a silicon nMOS capacitor with a substrate doping N a = 10 17 cm -3, a 20 nm thick oxide (e ox = 3.9 e 0 ) and an aluminum gate (F M = 4.1 V). Assume there is no fixed charge in the oxide or at the oxide-silicon interface. Solution The threshold voltage equals: Where the flatband voltage was already calculated in example 1.1. The threshold voltage voltages for nMOS and pMOS capacitors with an aluminum or a poly-silicon gate are listed in the table below.