The threshold voltage of an n-channel MOSFET can be increased by
increasing the channel doping concentration
Read the threshold expression and every option answers itself :
\(V_{T}=V_{FB}+2\phi_{F}+\dfrac{\sqrt{2q\varepsilon_{s}N_{A}\left(2\phi_{F}\right)}}{C_{ox}}\qquad C_{ox}=\dfrac{\varepsilon_{ox}}{t_{ox}}\)
which reduces, for the purposes of this question, to
\(V_{T}\propto t_{ox}\sqrt{N_{A}}\)
| Change | Effect on VT | Why |
|---|---|---|
| NA increased | Rises | More depletion charge to balance before inversion |
| NA reduced | Falls | Less depletion charge |
| tox reduced | Falls | Cox rises, so the last term shrinks |
| Channel length reduced | Falls | Short-channel effect — VT roll-off |
So only option 1 raises the threshold. Physically, heavier substrate doping means a wider band of fixed acceptor charge must be depleted before the surface can invert, and the gate must supply the field to balance it — so more gate voltage is required.
Option 4 deserves a note because its answer is not obvious from the simple formula. As the channel shortens, the source and drain depletion regions occupy an appreciable fraction of it, and part of the depletion charge is then controlled by those junctions rather than by the gate. The gate has less work to do, so \(V_{T}\) falls — the threshold roll-off that makes short-channel devices hard to control and drives the need for halo implants and ever-thinner oxides.
The two other handles used in practice :
Body effect. Reverse-biasing the source-substrate junction widens the depletion region and raises the threshold:
\(V_{T}=V_{T0}+\gamma\left(\sqrt{2\phi_{F}+V_{SB}}-\sqrt{2\phi_{F}}\right)\)
Threshold-adjust implantation. A shallow ion implant into the channel shifts \(V_{FB}\), and it is how a CMOS process sets its n- and p-channel thresholds independently and symmetrically — a far more controllable method than adjusting the bulk doping, which would also change the body effect and the junction capacitances.
Hence, the threshold voltage is increased by increasing the channel doping concentration.
Similar Questions
The threshold voltage of an n-channel MOSFET can be increased by
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 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
For the n-channel MOSFET shown in the figure given below, threshold voltage Vth = 2 V. The drain current ID of the MOSFET is 4 mA when the drain resistance RD is 1 kΩ. If the value of RD is increased to 4 kΩ, then the drain current ID will become :
The threshold voltage of a MOSFET can be lowered by
1. using a thinner gate oxide
2. reducing the carrier concentration in the substrate
3. increasing the carrier concentration in the substrateOf these statements :
Important Questions from MOSFET
An n-channel MOSFET is connected as shown in the Figure.
Assume $V_{TH} = 1\text{ V}$, $V_{DD} = 5\text{ V}$, and $\mu C_{ox}\left(\frac{W}{L}\right) = 2\text{ mA V}^{-2}$ and neglect channel length modulation effects.
The gate voltage ($V_G$) of the n-channel MOSFET (in Volt) is __.
(rounded off to two decimal places)
- A semiconductor uses P-type substrates with $N_A = 5 \times 10^{15}cm^{-3}$ in an n-MOS device having depletion region width of 0.415 $\mu$m. What is the depletion charge in magnitude?
- MOSFETs exhibit a non-linear relationship between input voltage and output current due to the device construction. The non-linear relationship is connected through the device parameter 'K'. Calculate 'K' for a specific MOSFET operating with an input voltage of 6 V and output current of 10 mA. The Threshold voltage of the chosen MOSFET is 4 V.
- Given below are two statements: one is labelled as Assertion (a) and the other is labelled as Reason (r).
Assertion (a): In a ratioless logic, the transistors can be of minimum size.
Reason (r): In a ratioless logic, the logic levels are not dependent upon the relative device size. - What are the effects of increasing temperature in the I-V characteristics of the NMOS transistor ?
