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.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both statements are true, and small size together with process simplicity is exactly why MOS took over digital electronics — option 1.
The size argument. A MOSFET is a self-isolating device: the reverse-biased source and drain junctions keep neighbouring transistors apart, so no separate isolation structure is needed. A bipolar transistor in an integrated circuit requires isolation diffusions or trenches occupying substantial area around every device. The result is that a MOS gate occupies a small fraction of the area of its bipolar equivalent, and packing density — the number of functions per chip — is what a digital process is judged on.
The fabrication argument. A basic NMOS process needs far fewer masking and diffusion steps than a bipolar one, and MOSFETs need no resistors: a transistor itself serves as the load, and in CMOS the complementary device does the job. Every step removed raises yield, and yield decides cost. Fewer steps also means fewer opportunities for misalignment, which is what allows the geometry to be shrunk in the first place.
| MOS | Bipolar | |
|---|---|---|
| Isolation | Self-isolating junctions | Separate structures needed |
| Area per gate | Small | Large |
| Process steps | Fewer | More |
| Static power (CMOS) | ≈ Zero | Milliwatts per gate |
A third reason the assertion does not mention is arguably the decisive one, and it reinforces rather than contradicts the reason given: CMOS dissipates essentially no power when idle, because in either logic state one transistor of the complementary pair is off and no DC path exists from supply to ground. Power is consumed only in switching, \(P=fCV^{2}\). Without that property, packing millions of gates onto one die would be impossible whatever their size — the heat could not be removed.
What MOS gave up was speed, and for a time that kept bipolar ECL in the fastest machines. Scaling eventually removed even that disadvantage, since a smaller MOSFET is also a faster one — the same shrink that improves density improves speed, which is the reason the industry has pursued it for fifty years.
Hence, both (A) and (R) are true and (R) is the correct explanation of (A).
Similar Questions
The threshold voltage of an n-channel MOSFET can be increased by
In MOSFET, the carrier velocity between constant mobility regime and the saturation velocity can be described as :
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
