The threshold voltage of a MOSFET can be lowered by 1. using a thinner gate oxide Of these statements :
2. reducing the carrier concentration in the substrate
3. increasing the carrier concentration in the substrate
1 and 2 are correct
Read the threshold-voltage expression and both answers fall straight out :
\(V_{T}=V_{FB}+2\phi_{F}+\dfrac{\sqrt{2q\varepsilon_{s}N_{A}\left(2\phi_{F}\right)}}{C_{ox}}\qquad\text{where}\qquad C_{ox}=\dfrac{\varepsilon_{ox}}{t_{ox}}\)
Statement 1 — a thinner gate oxide lowers VT. Correct. Substituting \(C_{ox}=\varepsilon_{ox}/t_{ox}\) shows the last term is proportional to \(t_{ox}\):
\(V_{T}\propto t_{ox}\sqrt{N_{A}}\)
Physically, a thinner oxide means a larger capacitance, so a smaller gate voltage places the charge needed to invert the surface.
Statement 2 — reducing the substrate doping lowers VT. Correct. The depletion charge that must be supported before inversion begins is \(\sqrt{N_{A}}\), so lighter doping means less charge to balance and a lower threshold.
Statement 3 is therefore the exact opposite of statement 2 and must be wrong: increasing NA raises VT. So 1 and 2 — option 2.
| Change | Effect on VT | Why |
|---|---|---|
| Thinner oxide tox ↓ | Lower | Cox rises |
| Substrate doping NA ↓ | Lower | Less depletion charge |
| Substrate doping NA ↑ | Higher | More depletion charge |
| Reverse body bias VSB ↑ | Higher | Body effect |
Two further handles on VT exist and are used in practice. The first is the body effect: reverse-biasing the source-substrate junction widens the depletion region and raises the threshold as
\(V_{T}=V_{T0}+\gamma\left(\sqrt{2\phi_{F}+V_{SB}}-\sqrt{2\phi_{F}}\right)\)
The second is threshold-adjust implantation — a shallow ion implant into the channel that shifts \(V_{FB}\), which is how CMOS processes set n- and p-channel thresholds independently and symmetrically.
Why a low threshold is not simply better. Lowering VT lets the device switch faster at a lower supply voltage, since drive current scales with \(\left(V_{GS}-V_{T}\right)^{2}\). But it also raises subthreshold leakage exponentially, so a low-VT transistor never turns fully off — the dominant source of static power in modern chips, and the reason processes offer several threshold flavours on the same die.
Hence, the correct statements are 1 and 2.
Similar Questions
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
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:
Approximate oxide capacitance value (Cgd) for saturation operating mode of MOS transistor 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 a MOSFET the drain saturation current is
The threshold voltage of an n-channel MOSFET can be increased by
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 carrier velocity between constant mobility regime and the saturation velocity can be described as :
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 :
Important Questions from MOSFET
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