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:
42 x 10-5 A/V2
To solve for the value of the transconductance parameter \( K \) for the given n-channel MOSFET, we use the formula:
\(K = \frac{\mu_n \cdot C_{ox} \cdot W}{2L}\)
where:
- \(\mu_n\) = electron mobility = 600 cm2/Vs
- \(C_{ox}\) = oxide capacitance per unit area = \(7 \times 10^{-8}\) F/cm2
- \(W\) = width = 40 μm = \(40 \times 10^{-4}\) cm
- \(L\) = channel length = 4 μm = \(4 \times 10^{-4}\) cm
Substituting the known values into the formula:
\(K = \frac{600 \cdot (7 \times 10^{-8}) \cdot (40 \times 10^{-4})}{2 \times (4 \times 10^{-4})}\)
First, calculate the numerator:
\(600 \cdot 7 \cdot 40 = 168000\)
Therefore, the numerator is \(168000 \times 10^{-8} \times 10^{-4} = 168000 \times 10^{-12}\)
Calculate the denominator:
\(2 \cdot 4 \times 10^{-4} = 8 \times 10^{-4}\)
Now divide the numerator by the denominator:
\(K = \frac{168000 \times 10^{-12}}{8 \times 10^{-4}} = \frac{168000}{8} \times 10^{-8} = 21000 \times 10^{-8}\)
Simplifying further, we get:
\(21000 \times 10^{-8} = 2.1 \times 10^{-4} = 42 \times 10^{-5} \text{ A/V}^2\)
Thus, the value of the \(K\) parameter is 42 x 10-5 A/V2.
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
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 :
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
Which industry does aluminium smelting belong to?
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
A switched mode power supply operating at 20 kHz to 100 kHz range uses as the main switching element is __________.
Which semiconductor power device out of the following, is not a current triggering device?
A CMOS amplifier when compared to an N–channel MOSFET, has the advantage of
