79 lines
1.3 KiB
Markdown
79 lines
1.3 KiB
Markdown
# VLSI Exam 1 Section 1 - Aidan Sharpe
|
|
|
|
## Problem 1
|
|
Aluminum is a suitable material when access to polysilicon is restricted. Restrictions may be in terms of access or the cost of the material. Academic institutions, for example, would not have access to polysilicon.
|
|
|
|
## Problem 2
|
|
```python
|
|
L_n = 0.6E-6
|
|
L_p = 0.6E-6
|
|
|
|
K_n = 122E-6
|
|
K_p = 61E-6
|
|
|
|
C_ox = 33E-4
|
|
|
|
V_tn = 0.7
|
|
V_tp = -0.7
|
|
|
|
V_gs = 5
|
|
V_ds = 5
|
|
|
|
V_dsat = V_gs - V_tn
|
|
|
|
I_ds = 2.82E-3
|
|
```
|
|
|
|
### 2a
|
|
```python
|
|
w_n = (2*I_ds*L_n) / (K_n * V_dsat**2)
|
|
```
|
|
$W_n = 1.5[\mu\text{m}]$
|
|
|
|
### 2b
|
|
```python
|
|
w_p = (2*I_ds*L_p) / (K_p * V_dsat**2)
|
|
```
|
|
$W_p = 3[\mu\text{m}]$
|
|
|
|
|
|
### 2c
|
|
Inverter input capacitance: 3C
|
|
|
|
Fanout capacitance 9C
|
|
|
|
Total load capacitance: 12C + 22$\mu$F
|
|
```python
|
|
C_g = C_ox*w_n*L_n
|
|
C_equiv = 12*C_g + 20E-15
|
|
beta = K_n*w_n/L_n
|
|
I_sat = beta * (V_gs - V_tn)**2 / 2
|
|
t_sat = V_tn*C_equiv / I_sat
|
|
R = 10E3
|
|
tau = R*C_equiv
|
|
t_lin = -tau*log(0.5/4.3)
|
|
t = t_lin + t_sat
|
|
```
|
|
|
|
$t_\text{sat} = 13.81[\text{ps}]$
|
|
|
|
$t_\text{lin} = 1.19[\text{ns}]$
|
|
|
|
$t = 1.21[\text{ns}]$
|
|
|
|
### 2d
|
|
|
|
|
|
|
|
|
|
|
|
### 2e
|
|
|
|
Since there is only one capacitor node, the Elmore delay and the RC delay are the same.
|
|
|
|
For both:
|
|
```python
|
|
t_pd = R*C_equiv
|
|
```
|
|
$t_{pd} = 556.43[\text{ps}]$
|