Consider the following statements : (A) The origin of the punch through phenomena is the lowering of the barrier near the source. Choose the most appropriate answer from the options given below :
(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.
(A), (B) and (C) Only
Statements (A), (B) and (C) together give the standard account of punch-through, and (D) contradicts it outright — option 2.
(A) — the mechanism. The source-channel junction presents a potential barrier that injected carriers must surmount. Punch-through begins when the drain's depletion region reaches far enough towards the source to lower that barrier, so the drain gains control of a region the gate is supposed to govern.
(B) and (C) — the contrast that defines "short channel". These two statements are a matched pair, and the distinction is the heart of the topic.
In a long-channel device, increasing the drain voltage widens the drain depletion region and so shortens the effective channel — channel-length modulation, the cause of the finite output resistance in saturation. But the drain is far from the source, so the barrier at the source end is untouched, and the threshold voltage does not depend on \(V_{ds}\).
In a short-channel device, the drain depletion region is no longer small compared with the channel, so its field reaches the source barrier and lowers it. The barrier — and therefore the threshold voltage — now depends on the drain voltage. This is drain-induced barrier lowering, and punch-through is its extreme case.
(D) is plainly false, and it is the statement that decides the answer. Carrier injection over a barrier is exponential in the barrier height:
\(I\propto e^{-q\phi_{B}/kT}\)
so lowering the barrier is precisely what causes extra carriers to be injected. Nothing else about punch-through matters: the whole problem is that current flows when the gate says it should not. A device in punch-through shows a leakage current that rises steeply with drain voltage and cannot be turned off, which destroys the subthreshold slope and raises standby power.
On statement (E). Punch-through is indeed usually a subsurface effect, since the depletion regions are deepest below the surface where the gate's control is weakest — which is why anti-punch-through implants are placed at depth. The official key nevertheless does not include (E) in the correct set, and since option 2 is the only choice containing exactly (A), (B) and (C), it is the intended answer.
The remedies follow from the mechanism: a deep punch-through stop implant, a lightly doped drain to spread the drain field, halo or pocket implants near the source, and thinner gate oxides to strengthen the gate's grip relative to the drain's.
Hence, the correct statements are (A), (B) and (C).
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 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 :
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
