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) \(\dfrac{g_{m}}{2\pi\left(C'_{ca}+C'_{par}\right)}\) Choose the most appropriate answer from the options given below :
(B) \(\dfrac{g_{m}}{2\pi C_{gs}}\)
(C) \(\dfrac{g_{m}}{2\pi C_{gd}}\)
(D) \(\dfrac{g_{m}}{2\pi\left(ZLC_{ox}+C'_{par}\right)}\)
(A) and (D) Only
Statements (A) and (D) are the same expression written two ways — option 4 — and recognising that is the whole point of the question.
The cut-off frequency is the frequency at which the device's current gain falls to unity:
\(f_{T}=\dfrac{g_{m}}{2\pi C_{in}}\)
where \(C_{in}\) is the total capacitance the gate must drive.
Why (A) and (D) are identical. The channel capacitance is the oxide capacitance per unit area multiplied by the channel area:
\(C'_{ca}=Z\times L\times C_{ox}\)
Substituting that into (A) gives (D) exactly. The question defines \(Z\times L\) as the channel area in its own preamble, which is the hint that the two are meant to be recognised as one.
| Statement | Capacitance used | Verdict |
|---|---|---|
| (A) | Channel + parasitic | ✓ Complete |
| (B) | Cgs alone | ✗ Ignores the parasitics |
| (C) | Cgd alone | ✗ The wrong capacitance entirely |
| (D) | ZLCox + parasitic | ✓ The same as (A) |
Why the parasitic term cannot be dropped, which rules out (B). The intrinsic channel capacitance is only part of what the gate drives. Overlap capacitance where the gate extends over the source and drain diffusions, fringing capacitance from the gate edges, and the capacitance of the interconnect all add to it — and as devices are scaled down the intrinsic part shrinks while the overlap and fringing parts do not shrink as fast, so the parasitics come to dominate. An expression that omits them overestimates \(f_{T}\) badly for a real microwave device.
Why (C) is wrong in principle. \(C_{gd}\) is the feedback capacitance from output back to input. It matters greatly — through the Miller effect it is multiplied by the gain when referred to the input — but it is not the input capacitance, and it appears in the maximum oscillation frequency \(f_{max}\) rather than in \(f_{T}\).
How to raise fT : shorten the channel, which raises \(g_{m}\) and lowers \(C_{ca}\) together, giving roughly \(f_{T}\propto1/L^{2}\) in the long-channel limit — which is why the drive to shorter gates has been as much about speed as about density.
Hence, the correct statements are (A) and (D).
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 carrier velocity between constant mobility regime and the saturation velocity can be described as :
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