Introduction / Context:
Threshold voltage engineering in MOSFETs often uses channel implants to shift Vth. The threshold shift relates to the added sheet charge via the oxide capacitance per unit area. The sign of the implant (n or p) determines whether Vth moves positive or negative for an NMOS on a p-type substrate.
Given Data / Assumptions:
- Initial Vth = +1 V; desired Vth = −1 V ⇒ ΔVth = −2 V.
- Oxide thickness t_ox = 5 × 10^-6 cm.
- εr = 3.9, ε0 = 8.85 × 10^-14 F/cm ⇒ C_ox = εr ε0 / t_ox.
- Electronic charge q = 1.6 × 10^-19 C.
- NMOS device; n-type channel implant lowers Vth (toward negative), p-type raises Vth (toward positive).
Concept / Approach:
Threshold shift from a sheet charge density Qs is ΔVth = −Qs / C_ox (sign per convention). The required magnitude of the implant (in charges per cm^2) is |Qs|/q. The polarity required to move from +1 V to −1 V is that which makes Vth more negative for NMOS: n-type (donors) in the channel region.
Step-by-Step Solution:
Compute oxide capacitance: C_ox = (3.9 × 8.85 × 10^-14) / (5 × 10^-6) F/cm^2 ≈ 6.903 × 10^-8 F/cm^2.Required charge: |Qs| = C_ox × |ΔVth| = 6.903 × 10^-8 × 2 ≈ 1.3806 × 10^-7 C/cm^2.Convert to sheet density: N = |Qs| / q ≈ (1.3806 × 10^-7) / (1.6 × 10^-19) ≈ 8.63 × 10^11 cm^-2.Choose implant type: n-type to reduce Vth (make it more negative) in NMOS.
Verification / Alternative check:
Order-of-magnitude sanity: 10^12 cm^-2 level is typical for threshold trims of a few volts with thin oxides in the sub-100 nm range (scaled accordingly).
Why Other Options Are Wrong:
p-type implants push Vth positive (wrong direction).1.02 × 10^12 cm^-2 overestimates the required charge for a 2 V shift with the given C_ox.0.86 × 10^9 cm^-2 is far too small to create a 2 V shift.
Common Pitfalls:
Mixing centimeters and meters; forgetting to divide by q to convert C/cm^2 to charges/cm^2; choosing the wrong implant polarity.
Final Answer:
8.6 × 10^11 cm^-2, n-type implant
Discussion & Comments