# 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:
$$2I_{DS1} = 2I_{DS2}$$
$$\boxed{I_{DS1} = I_{DS2}}$$