The threshold voltage of an n-channel MOSFET can be increased by
increasing the channel dopant concentration
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
\(V_{T}\propto t_{ox}\sqrt{N_{A}}\)
| Change | Effect on VT | Reason |
|---|---|---|
| NA increased | Rises | More fixed depletion charge to balance before inversion |
| NA reduced | Falls | Less depletion charge |
| tox reduced | Falls | Cox rises, shrinking the last term |
| Channel length reduced | Falls | Short-channel roll-off |
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 VT 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.
Similar Questions
Assertion (A) : MOS ICs based on MOSFET structure find wide applications in digital field.
Reason (R) : MOS ICs have small size and are easy to fabricate.
In MOSFET, the carrier velocity between constant mobility regime and the saturation velocity can be described as :
If Z × L is the total channel area and C'par is the total input parasitic capacitance, then for microwave performance the cut off frequency can be defined as :
(A) \(\dfrac{g_{m}}{2\pi\left(C'_{ca}+C'_{par}\right)}\)
(B) \(\dfrac{g_{m}}{2\pi C_{gs}}\)
(C) \(\dfrac{g_{m}}{2\pi C_{gd}}\)
(D) \(\dfrac{g_{m}}{2\pi\left(ZLC_{ox}+C'_{par}\right)}\)Choose the most appropriate answer from the options given below :
Consider the following statements :
(A) The origin of the punch through phenomena is the lowering of the barrier near the source.
(B) For a long channel device, a drain bias can change the effective channel length, but the barrier at the source end remains constant.
(C) For a short channel device, this barrier is no longer fixed.
(D) The lowering of the source barrier do not cause any injection of extra carriers.
(E) The punch through condition normally occurs inside the bulk region of the semiconductor.Choose the most appropriate answer from the options given below :
In MOSFET, the linear region current is :
(A) \(\dfrac{\mu_{n}C_{ox}w}{L}\left(V_{gs}-V_{th}-\dfrac{V_{ds}}{2}\right)V_{ds}\)
(B) \(\dfrac{\mu_{n}C_{ox}w}{L}\left(V_{gs}-V_{th}\right)V_{ds}\)
(C) \(\dfrac{\mu_{n}C_{ox}w}{2L}\left(V_{gs}-V_{th}\right)V_{ds}^{2}\)
(D) \(\dfrac{\mu_{n}C_{ox}w}{L}\left(V_{gs}-V_{th}\right)^{2}\)Choose the most appropriate answer from the options given below :
Approximate oxide capacitance value (Cgd) for saturation operating mode of MOS transistor is:
For an n-channel MOS transistor with $\mu_n$ = 600 cm2/Vs, Cox = 7 x 10-8 F/cm2, W = 40 $\mu_m$, L = 4$\mu_m$ and VTO=1.0 V, the value of K parameter is:
In MOS
A. The substrate fermi potential ϕF is negative in NMOS
B. The substrate fermi potential ϕF is positive in NMOS
C. The substrate bias voltage VSB is positive in NMOS, negative in PMOS
D. The substrate bias voltage VSB is negative in NMOS, positive in PMOS.
Choose the correct answer from the options given below:
In enhancement mode MOSFET the saturation (drain) current is given by
(a) \(K\dfrac{W}{L}(V_{gs}-V_{th})^{2}\)
(b) \(K\dfrac{W}{L}(V_{gs}-V_{th})(1+\lambda V_{ds})\)
(c) \(K\dfrac{W}{L}(V_{gs}-V_{th})^{2}(1+\lambda V_{ds})\)
(d) \(K\dfrac{W}{L}(V_{gs}-V_{th})^{2}(1-\lambda V_{ds})\)
Out of these
In a MOSFET the drain saturation current is
Important Questions from MOSFET
The O in a MOSFET stands for _______ layer which provides _______ to the device.
Which industry does aluminium smelting belong to?
In an N-channel MOSFET, the drain current ID increases as _______.
Consider an ideal long channel nMOSFET (enhancement-mode) with gate length 10 µm and width 100 µm. The product of electron mobility (µn) and oxide capacitance per unit area (COX) is µn COX = 1 mA/V2 . The threshold voltage of the transistor is 1 V. For a gate-to-source voltage VGS = [2 − sin (2t)] V and drain-to-source voltage VDS = 1 V (substrate connected to the source), the maximum value of the drain-to-source current is ________.
Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are
