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.
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 ?
