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 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 :
The threshold voltage of an n-channel MOSFET can be increased by
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 ?
