VLSI homework 1, and IoT reading assignment 1 done

7th-Semester-Fall-2024/VLSI/homework/homework-1/homework-1.md

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+# VLSI Homework 1 - Aidan Sharpe
+
+## Problem 1
+```python
+import numpy as np
+import matplotlib.pyplot as plt
+
+e_0 = 8.85E-14
+k_ox = 3.9
+e_ox = e_0 * k_ox
+
+mu = 350                # 350 cm^2 / (V*s)
+t_ox = 100E-8           # 100A
+
+C_ox = e_ox / t_ox
+W = 1.2
+L = 0.6
+
+beta = mu * C_ox * W/L
+
+V_g_range = np.arange(6)
+V_t = 0.7
+V_dd = np.linspace(0, 5, 50)
+
+def I_ds(V_gs, V_ds, V_t, beta):
+    if V_gs < V_t:
+        return np.zeros_like(V_ds)
+
+    V_dsat = V_gs - V_t
+
+    return np.where(V_ds < V_dsat,
+                    beta * (V_gs - V_t - V_ds/2) * V_ds, 
+                    beta * (V_gs - V_t)**2 / 2)
+
+for V_s in (0, 2):
+    for V_g in V_g_range:
+        V_gs = V_g - V_s
+        V_ds = V_dd - V_s
+        plt.plot(V_ds,
+                 1000*I_ds(V_gs, V_ds, V_t, beta),
+                 label="$V_{gs}$ = "+str(V_gs))
+
+    plt.xlabel("$V_{ds}$[V]")
+    plt.ylabel("$I_{ds}$[mA]")
+    plt.legend()
+    plt.show()
+```
+![](Figure_1.png)
+
+## Problem 2
+Show that a MOSFET with channel length $2L$, when operating in the linear region, has the same current $I_{DS1}$ as two MOSFETs in series each with channel length $L$. Assume the same gate voltage $V_{DD}$ for all MOSFETs in the design and the same supply voltage $V_{DS}$ for both designs.
+
+### Known
+The current through a MOSFET in the linear region is:
+$$I_{DS} = \beta\left(V_{GS} - V_t - \frac{V_{DS}}{2}\right)V_{DS}$$
+The parameter $\beta$ is defined as:
+$$\beta = \mu C_\text{ox} \frac{W}{L}$$
+
+### Defining Variables
+The factor $\beta_1$ for the singular MOSFET is $\mu C_\text{ox} \frac{W}{2L}$, and the factor $\beta_2$ for each of the MOSFETs in series is $\mu C_\text{ox} \frac{W}{L}$. Therefore, $\beta_2 = 2\beta_1$.
+
+The current $I_{DS1}$ through the singular MOSFET is:
+$$I_{DS1} = \beta_1\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS}$$
+The current $I_{DS2}$ through both MOSFETs in series is both:
+$$I_{DS2} = \beta_2\left(V_{DD} - V_1 - V_t - \frac{V_{DS} - V_1}{2}\right)(V_{DS} - V_1)$$
+$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
+
+### Expanding $I_{DS2}$
+The first equation for $I_{DS2}$ can be rewritten as the difference of two terms:
+$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)(V_{DS} - V_1)$$
+$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_{DS}}{2} - \frac{V_1}{2}\right)V_1$$
+
+The term $-\frac{V_1 V_{DS}}{2}$ can be pulled out of the first term, and its complement can be pulled out of the second term. Therefore, $I_{DS2}$ can be simplified to the difference of two terms each in the form of the current through a MOSFET in the linear region.
+$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
+
+### Reducing $I_{DS2}$
+Notice that the second term is the inverse of the second equation for $I_{DS2}$ under "Defining Variables". Setting them equal:
+$$I_{DS2} = \beta_2\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} - \beta_2\left( V_{DD} - V_t - \frac{V_1}{2}\right)V_1 = \beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
+
+Therefore:
+$$\beta_2\left(V_{DD} -V_t - \frac{V_{DS}}{2}\right)V_{DS} = 2\beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
+
+Substituting $\beta_2$ on the left for $2\beta_1$ gives:
+$$2\beta_1\left(V_{DD} - V_t - \frac{V_{DS}}{2}\right)V_{DS} = 2\beta_2\left(V_{DD} - V_t - \frac{V_1}{2}\right)V_1$$
+
+Therefore:
+$$2I_{DS1} = 2I_{DS2}$$
+$$\boxed{I_{DS1} = I_{DS2}}$$

7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_1.py

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+import numpy as np
+import matplotlib.pyplot as plt
+
+e_0 = 8.85E-14
+k_ox = 3.9
+e_ox = e_0 * k_ox
+
+mu = 350                # 350 cm^2 / (V*s)
+t_ox = 100E-8           # 100A
+
+C_ox = e_ox / t_ox
+W = 1.2
+L = 0.6
+
+beta = mu * C_ox * W/L
+
+V_g_range = np.arange(6)
+V_t = 0.7
+V_dd = np.linspace(0, 5, 50)
+
+def I_ds(V_gs, V_ds, V_t, beta):
+    if V_gs < V_t:
+        return np.zeros_like(V_ds)
+
+    V_dsat = V_gs - V_t
+
+    return np.where(V_ds < V_dsat,
+                    beta * (V_gs - V_t - V_ds/2) * V_ds, 
+                    beta * (V_gs - V_t)**2 / 2)
+
+for V_s in (0, 2):
+    for V_g in V_g_range:
+        V_gs = V_g - V_s
+        V_ds = V_dd - V_s
+        plt.plot(V_ds,
+                 1000*I_ds(V_gs, V_ds, V_t, beta),
+                 label="$V_{gs}$ = "+str(V_gs))
+
+    plt.xlabel("$V_{ds}$[V]")
+    plt.ylabel("$I_{ds}$[mA]")
+    plt.legend()
+    plt.show()

7th-Semester-Fall-2024/VLSI/homework/homework-1/problem_2.py

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+import numpy as np
+import matplotlib.pyplot as plt
+
+e_0 = 8.85E-14
+k_ox = 3.9
+e_ox = e_0 * k_ox
+
+mu = 350                # 350 cm^2 / (V*s)
+t_ox = 100E-8           # 100A
+
+C_ox = e_ox / t_ox
+W = 1.2
+L = 0.6
+
+beta = mu * C_ox * W/L
+
+V_t = 0.7
+
+def I_ds(V_gs, V_ds, V_t, beta):
+    if V_gs < V_t:
+        return np.zeros_like(V_ds)
+
+    V_dsat = V_gs - V_t
+
+    return np.where(V_ds < V_dsat, beta * (V_gs - V_t - V_ds/2) * V_ds, beta * (V_gs - V_t)**2 / 2)
+
+V_g = 0.8
+V_d1 = 1
+V_s1 =