1

Ve311 Homework #6

Note:

(1) Please use A4 size papers.

(2) Please use the SPICE model in page 3 for simulation and calculation.

1. [Common-Source with Diode-Connected Load]

(a) [30%] Assume λ = 0 and γ = 0. For VDD = 5 V, VIN = 0.9 V,

(Wdrawn/Ldrawn)2 = 5 µm / 2 µm and (Wdrawn/Ldrawn)1 = x µm / 2 µm, what is

the value of x to obtain a voltage gain Av = –5? What is the range of VIN

for M1 to stay in the saturation region?

(b) [10%] Using the design and biasing condition in (a), plot VOUT as a

function of VIN (from 0 V to 5 V) in Pspice. Compare the handcalculation results in (a) with the simulation results here. Note: the slope

of the VOUT versus VIN curve at VIN = 0.9 V is the Av

.

(c) [10%] Using the design and biasing conditions in (a), plot Vout as a

function of time (from 0 to 0.1 second) in Pspice, when Vin = 0.9 +

0.001 × sin(2π100t) (V). Compare the hand-calculation results in (a)

with the simulation results here. Note: the small-signal voltage amplitude

of Vout divided by the small-signal voltage amplitude of Vin is the Av

.

2

2. [Common-Source with Source Degradation]

(a) [30%] Assume λ = 0 and γ = 0. For VDD = 5 V, VIN = 1.2 V,

(Wdrawn/Ldrawn)1 = 200 µm / 2 µm, RD = 100 kΩ and RS = 20 kΩ, what is

the voltage gain Av? Does the voltage gain approach –RD / RS as expected?

(b) [10%] Using the design and biasing condition in (a), plot VOUT as a

function of VIN (from 0 V to 5 V) in Pspice. Compare the handcalculation results in (a) with the simulation results here. Note: the slope

of the VOUT versus VIN curve at VIN = 1.2 V is the Av

.

(c) [10%] Using the design and biasing conditions in (a), plot Vout as a

function of time (from 0 to 0.1 second) in Pspice, when Vin = 1.2 +

0.001 × sin(2π100t) (V). Compare the hand-calculation results in (a)

with the simulation results here. Note: the small-signal voltage amplitude

of Vout divided by the small-signal voltage amplitude of Vin is the Av

.

3

Vacuum permittivity (𝛜𝐨) = 𝟖. 𝟖𝟓 × 𝟏𝟎−𝟏𝟐 (F / m)

Silicon oxide dielectric constant (𝛜𝐫

) = 𝟑. 𝟗

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