In MOSFET, the carrier velocity between constant mobility regime and the saturation velocity can be described as :
\(\dfrac{\mu_{n}E}{\left[1+\left(\dfrac{\mu_{n}E}{v_{s}}\right)^{n}\right]^{1/n}}\)
The right expression is the one that reduces correctly at both ends — option 1:
\(v=\dfrac{\mu_{n}E}{\left[1+\left(\dfrac{\mu_{n}E}{v_{s}}\right)^{n}\right]^{1/n}}\)
Test the low-field limit. For small E the ratio \(\mu_{n}E/v_{s}\ll1\), so the bracket tends to 1 and
\(v\to\mu_{n}E\)
which is exactly the constant-mobility regime — velocity proportional to field, the ordinary drift relation.
Test the high-field limit. For large E the 1 becomes negligible beside the ratio, so the bracket tends to \(\left(\mu_{n}E/v_{s}\right)^{n}\), whose \(1/n\) power is \(\mu_{n}E/v_{s}\). Then
\(v\to\dfrac{\mu_{n}E}{\mu_{n}E/v_{s}}=v_{s}\)
— the velocity saturates at \(v_{s}\), about \(10^{7}\) cm/s in silicon. A single expression therefore joins the two regimes smoothly, which is what the question asks for.
| Option | Why it fails |
|---|---|
| 1 | ✓ Gives μnE at low field and vs at high field |
| 2 | Numerator lacks μn, and the outer power n is wrong — the low-field limit is E, not μnE |
| 3 | Numerator is a mobility, not a velocity — dimensionally impossible |
| 4 | Minus sign makes the bracket vanish at \(\mu_{n}E=v_{s}\), so v becomes infinite |
Option 4 is the instructive failure. Its only change is the sign inside the bracket, but that is fatal: at the particular field where \(\mu_{n}E=v_{s}\) the denominator becomes zero and the predicted velocity is unbounded — the exact opposite of saturation. Checking a limiting case is what exposes it.
Why velocity saturates at all. At high fields carriers gain energy faster than they can lose it to the lattice by acoustic phonon scattering, until they become energetic enough to emit optical phonons. That channel is very efficient, so any further energy from the field is dumped straight into the lattice and the average velocity stops rising.
The circuit consequence is that a short-channel MOSFET's drain current becomes proportional to \(\left(V_{GS}-V_{T}\right)\) rather than to its square, and the exponent n — about 2 for electrons and 1 for holes — controls how abruptly the transition occurs.
Hence, the correct expression is option 1.
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.
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
