Write down the threshold expression and every option can be tested against it :

\(V_{T}=V_{FB}+2\phi_{F}+\dfrac{\sqrt{2q\varepsilon_{s}N_{A}\left(2\phi_{F}\right)}}{C_{ox}},\qquad C_{ox}=\dfrac{\varepsilon_{ox}}{t_{ox}}\)

which condenses, for the purposes of this question, to

\(V_{T}\propto t_{ox}\sqrt{N_{A}}\)

So only option 1 raises the threshold. Physically, heavier substrate doping means a thicker sheet of ionised acceptors must be depleted before the surface can invert, and the gate must supply the field that balances that charge — hence more gate voltage is needed to turn the device on.

Option 3 deserves a moment because a thinner oxide brings the gate closer to the channel, which sounds like it should demand more control voltage, not less. In fact proximity increases the capacitance per unit area, so the same gate voltage terminates more field lines on the channel — the gate becomes more effective, and the threshold falls. This is one of the reasons oxides were thinned as processes scaled.

Option 4 is the least obvious. In a short channel the source and drain depletion regions occupy a significant fraction of the channel, so part of the depletion charge is supported by those junctions rather than by the gate. With less work left to do, the gate reaches inversion at a lower voltage — the V_T roll-off that makes very short devices hard to control and drives the use of halo implants.

Two practical handles on the threshold are worth knowing alongside these. The body effect raises it when the source-substrate junction is reverse biased:

\(V_{T}=V_{T0}+\gamma\left(\sqrt{2\phi_{F}+V_{SB}}-\sqrt{2\phi_{F}}\right)\)

and a shallow threshold-adjust implant into the channel shifts \(V_{FB}\), which is how a CMOS process sets its n- and p-channel thresholds independently — a far more controllable method than changing the bulk doping, which would also alter the body effect and the junction capacitances.

Hence, the threshold voltage is increased by increasing the channel dopant concentration.