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Magazine Chapter 3: Problem 1
(a) Calculate the drain current in an NMOS transistor with parameters \(V_{T N}=0.4 \mathrm{~V}, k_{n}^{\prime}=120 \mu \mathrm{A} / \mathrm{V}^{2}, W=10 \mu \mathrm{m}, L=0.8 \mu \mathrm{m}\), and with ap- plied voltages of \(V_{D S}=0.1 \mathrm{~V}\) and (i) \(V_{G S}=0\), (ii) \(V_{G S}=1 \mathrm{~V}\), (iii) \(V_{G S}=2 \mathrm{~V}\), and (iv) \(V_{G S}=3 \mathrm{~V}\). (b) Repeat part (a) for \(V_{D S}=4 \mathrm{~V}\).
Short Answer
Expert verified
For \(V_{DS} = 0.1\,\text{V}\): \(0, 67.5, 139.5, 229.5\,\mu\text{A}\). For \(V_{DS} = 4\,\text{V}\): \(0, 270, 1920, 5070\,\mu\text{A}\).
Step by step solution
01
Understand the Problem
We are calculating the drain current of an NMOS transistor using the long-channel model equations. The given parameters are threshold voltage \(V_{TN}=0.4\,\text{V}\), \(k_n' = 120\,\mu\text{A/V}^2\), \(W = 10\,\mu\text{m}\), and \(L = 0.8\,\mu\text{m}\). We need to find the drain current for different values of \(V_{GS}\) and \(V_{DS}\).
02
Determine Transistor Regions
The NMOS transistor can operate in the cutoff, triode, or saturation region. Compare \(V_{GS}\) to \(V_{TN}\) to determine if the transistor is off (cutoff). If \(V_{GS} > V_{TN}\), it is on, and we further compare \(V_{DS}\) to \(V_{GS} - V_{TN}\) to determine if it is in the triode (\(V_{DS} < V_{GS} - V_{TN}\)) or saturation region (\(V_{DS} \geq V_{GS} - V_{TN}\)).
03
Calculate Drain Current for Case (i)
For \(V_{GS} = 0\,\text{V}\), \(V_{GS} < V_{TN} (0.4\,\text{V})\), thus the NMOS is in cutoff and \(I_D = 0\). This applies for both \(V_{DS}=0.1\,\text{V}\) and \(V_{DS}=4\,\text{V}\).
04
Calculate Drain Current for Case (ii)
For \(V_{GS} = 1\,\text{V}\):- For \(V_{DS} = 0.1\,\text{V}\), the transistor is in the triode region since \(0.1\,\text{V} < 1\,\text{V} - 0.4\,\text{V} = 0.6\,\text{V}\). Use the equation \(I_D = k_n' \cdot \frac{W}{L}\cdot \left[(V_{GS} - V_{TN})V_{DS} - \frac{V_{DS}^2}{2}\right]\) to find \(I_D\).- For \(V_{DS} = 4\,\text{V}\): the transistor is in the saturation region because \(4\,\text{V} > 0.6\,\text{V}\). Use \(I_D = \frac{k_n'}{2} \cdot \frac{W}{L} \cdot (V_{GS} - V_{TN})^2\).
05
Perform Calculation for Case (ii 鈥� Triode)
\[I_D = 120 \times \frac{10}{0.8} \times \left[(1 - 0.4) \times 0.1 - \frac{0.1^2}{2}\right] = 120 \times 12.5 \times (0.6 \times 0.1 - 0.005) \]\(I_D = 90 \times (0.055) = 67.5\,\mu\text{A}\).
06
Perform Calculation for Case (ii 鈥� Saturation)
\[I_D = \frac{120}{2} \times \frac{10}{0.8} \times (1 - 0.4)^2 = 60 \times 12.5 \times 0.36 = 270\,\mu\text{A}\].
07
Calculate Drain Current for Case (iii)
For \(V_{GS} = 2\,\text{V}\):- For \(V_{DS} = 0.1\,\text{V}\), the transistor is still in the triode region since \(0.1\,\text{V} < 1.6\,\text{V}\). Use the triode region equation.- For \(V_{DS} = 4\,\text{V}\), it is in saturation. Use the saturation equation.
08
Perform Calculation for Case (iii 鈥� Triode)
\[I_D = 120 \times \frac{10}{0.8} \times [(2 - 0.4) \times 0.1 - \frac{0.1^2}{2}] = 120 \times 12.5 \times (1.6 \times 0.1 - 0.005)\]\(I_D = 90 \times (0.155) = 139.5\,\mu\text{A}\).
09
Perform Calculation for Case (iii 鈥� Saturation)
\[I_D = \frac{120}{2} \times \frac{10}{0.8} \times (2 - 0.4)^2 = 60 \times 12.5 \times 2.56 = 1920\,\mu\text{A}\].
10
Calculate Drain Current for Case (iv)
For \(V_{GS} = 3\,\text{V}\):- For \(V_{DS} = 0.1\,\text{V}\), the transistor is in the triode region since \(0.1\,\text{V} < 2.6\,\text{V}\). Use the triode region equation.- For \(V_{DS} = 4\,\text{V}\), use the saturation equation as it is clearly in saturation.
11
Perform Calculation for Case (iv 鈥� Triode)
\[I_D = 120 \times \frac{10}{0.8} \times [(3 - 0.4) \times 0.1 - \frac{0.1^2}{2}] = 120 \times 12.5 \times (2.6 \times 0.1 - 0.005)\]\(I_D = 90 \times (0.255) = 229.5\,\mu\text{A}\).
12
Perform Calculation for Case (iv 鈥� Saturation)
\[I_D = \frac{120}{2} \times \frac{10}{0.8} \times (3 - 0.4)^2 = 60 \times 12.5 \times 6.76 = 5070\,\mu\text{A}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Drain Current Calculation
Drain current, denoted as \( I_D \), is a key electrical current flowing from the drain to the source of an NMOS transistor. To calculate \( I_D \) effectively, it is essential to identify the transistor's operational region鈥攚hether it is in cutoff, triode, or saturation鈥攂ecause each region uses different equations.
- In the **triode region**, the formula for \( I_D \) is given by \( I_D = k_n' \cdot \frac{W}{L} \cdot \left[ (V_{GS} - V_{TN})V_{DS} - \frac{V_{DS}^2}{2} \right] \), where \( k_n' \) represents the process transconductance parameter, \( W \) is the channel width, and \( L \) is the channel length.
- Conversely, in the **saturation region**, \( I_D \) is calculated using \( I_D = \frac{k_n'}{2} \cdot \frac{W}{L} \cdot (V_{GS} - V_{TN})^2 \). This occurs when the drain-source voltage exceeds the overdrive voltage.
MOSFET Operation Regions
Understanding MOSFET operation regions is fundamental to predicting the behavior of a transistor under different biasing conditions. The NMOS transistor can operate in three primary regions:
- **Cutoff Region:** Here, the transistor acts as an open switch. This happens when the gate-source voltage \( V_{GS} \) is less than the threshold voltage \( V_{TN} \). In this state, \( I_D = 0 \) since no charge carriers are formed to conduct current.
- **Triode Region:** Also known as the linear region. The transistor behaves like a resistor, and current flows through it from the drain to the source. This occurs when \( V_{GS} > V_{TN} \) but \( V_{DS} < V_{GS} - V_{TN} \).
- **Saturation Region:** In this region, the transistor behaves like a current source. \( V_{GS} > V_{TN} \) and \( V_{DS} \geq V_{GS} - V_{TN} \) signify that the channel is pinched off, but current remains constant and independent of \( V_{DS} \).
Threshold Voltage
The threshold voltage, \( V_{TN} \), is the minimum gate-to-source voltage that is necessary to create a conductive channel between the drain and source terminals of an NMOS transistor. When \( V_{GS} \) is less than \( V_{TN} \), the MOSFET is in the cutoff region and effectively stops conducting.
This parameter is influenced by several factors, including:
This parameter is influenced by several factors, including:
- The **doping levels** of the semiconductor's substrate, which affect how easily charge carriers can form a channel.
- **Oxide thickness**, because a thinner layer of silicon dioxide can produce higher electric fields, modifying the effective \( V_{TN} \).
- **Temp茅rature variations**, as threshold voltage can decrease with increasing temperature due to changes in carrier mobility.
Long-Channel Model
The long-channel model is a simplification used in analyzing MOSFET behavior for transistors with channels that are significantly longer than the sum of source and drain depletion widths. This model provides a straightforward approach by assuming uniform electric fields across the device channel.
With this model:
With this model:
- It predicts that the drain current \( I_D \) increases linearly in the triode region and becomes constant upon entering saturation.
- The model is best suited to older technology nodes where the channel length is not aggressively minimized to near the de Broglie wavelength of carriers.
- For transistors with very short channels, short-channel effects like velocity saturation and drain-induced barrier lowering (DIBL) become significant, which are not accounted for in this model.
