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 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 :
The threshold voltage of an n-channel MOSFET can be increased by
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 ?
