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
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 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
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 :
Important Questions from MOSFET
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