A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2 mA is required when the device is biased in the saturation region at VGS = 3 V. Determine the required channel width of the device.
Engineering · College · Tue Nov 03 2020
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To determine the required channel width for the NMOS device to achieve the desired drain current (I_D = 1.2 mA) when biased in the saturation region at V_GS = 3 V, we will use the MOSFET saturation current equation,
\[ I_D = \frac{1}{2} \mu_n C_{ox} \left(\frac{W}{L}\right) (V_{GS} - V_{T_N})^2 \]
where: - \( \mu_n \) is the electron mobility. - \( C_{ox} \) is the capacitance per unit area of the gate oxide. - \( W \) and \( L \) are the channel width and length, respectively. - \( V_{GS} \) is the gate-source voltage. - \( V_{T_N} \) is the threshold voltage. First, we need to calculate the capacitance per unit area \( C_{ox} \), which is given by the equation,
\[ C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}} \]
where: - \( \varepsilon_{ox} \) is the permittivity of the gate oxide material (for SiO2, \( \varepsilon_{ox} \approx 3.45 \times 10^{-11} F/m \)). - \( t_{ox} \) is the thickness of the gate oxide in meters.
Converting \( t_{ox} \) to meters from Angstroms (1 Å = 10^-10 m),
\[ t_{ox} = 200 Å = 200 \times 10^{-10} m = 2 \times 10^{-8} m \]
Now, we can calculate the oxide capacitance per unit area \( C_{ox} \):
\[ C_{ox} = \frac{3.45 \times 10^{-11} F/m}{2 \times 10^{-8} m} = 1.725 \times 10^{-3} F/m^2 \]
Next, we'll rearrange the MOSFET saturation current equation to solve for \( W \):
\[ W = \frac{2I_D}{\mu_n C_{ox} (V_{GS} - V_{T_N})^2} \times L \]
Substituting the given values:
- \( \mu_n = 600 cm^2/V\cdot s = 600 \times 10^{-4} m^2/V\cdot s \) (since 1 cm^2 = 10^-4 m^2) - \( L = 0.8 µm = 0.8 \times 10^{-6} m \) - \( V_{GS} = 3 V \) - \( V_{T_N} = 0.6 V \) - \( I_D = 1.2 mA = 1.2 \times 10^{-3} A \)
\[ W = \frac{2 \times 1.2 \times 10^{-3} A}{600 \times 10^{-4} m^2/V\cdot s \times 1.725 \times 10^{-3} F/m^2 \times (3 V - 0.6 V)^2} \times 0.8 \times 10^{-6} m \]
Calculating the squared term and other values within the equation:
\[ W = \frac{2 \times 1.2 \times 10^{-3}}{600 \times 10^{-4} \times 1.725 \times 10^{-3} \times (2.4)^2} \times 0.8 \times 10^{-6} m \]
\[ W = \frac{2 \times 1.2 \times 10^{-3}}{600 \times 10^{-4} \times 1.725 \times 10^{-3} \times 5.76} \times 0.8 \times 10^{-6} m \]
\[ W \approx \frac{2.4 \times 10^{-3}}{6 \times 1.725 \times 5.76 \times 10^{-5}} \times 0.8 \times 10^{-6} m \]
From here, you can use a calculator to find the numerical value for \( W \). Remember that the result will be in meters, and can be converted to µm if necessary by multiplying by \( 10^6 \).
Extra: MOSFETs (Metal-Oxide-Semiconductor Field-Effect Transistors) are widely used in electronics for their ability to control large currents with minimal input power. The equations used above are part of MOSFET device physics, which describe how the device operates under certain conditions.
Key concepts for school students would include: - Understanding that transistors are like switches that can be turned on and off with a voltage applied to the gate (V_GS). - The threshold voltage (V_T) is the minimum gate-source voltage needed for the transistor to start conducting. - In the saturation region, the drain current (I_D) becomes relatively constant despite increases in the drain-source voltage (V_DS), assuming V_GS is held constant and higher than V_T. - The relationship between channel width (W), channel length (L), and current (I_D) implies that wider channels allow more current to flow, much like a wider hose allows more water to pass through it. - The mobility (\( \mu \)) and permittivity (\( \varepsilon \)) are materials properties that affect how easily electrons can move through the channel and how effectively the gate can control the channel, respectively.