Peter Lau

Bill Chun Wai Hung

**Lab 12 Report
– NMOS Common Source Amplifier**

**A. Describe the Set-up**

The Circuit
setup is shown in figure 1.

**Figure 1**

**B1. Describe Inputs – Ideal Case**

Stimuli:

vi:

Sinusoidal

Amplitude
100mv

K = 0.82

VT = 2.9
V

Vdd =
10 V

Av = 20

MOSFET =
NMOS = Model BS170

C1 = C2
= 10uF.

Av= gm(Rd)

Av = 2K(Vgs – Vto)(Rd)
…(1)

Id(Rd)
= 5V

K(Vgs – Vto)^2 (Rd) = 5V … (2)

Solving
(1) and (2)

Av = 2K(Vgs – Vto)(Rd)
… (1)

20 =
2(0.82)(Vgs – 2.9) (Rd) …
(3)

K(Vgs – Vto)^2 (Rd) = 5V … (2)

(0.82)(Vgs – 2.9)^2(Rd) = 5V …(4)

Solving
(3) and (4)

Vgs =
3.40V

Rd =
24.4 ohm

Substitue Vgs = 3.40V

Id = K(Vgs – Vto)^2*
…(5)

Id =
(0.82)(3.4 – 2.9)^2

Id =
205mA.

Vds = Vdd – V(Rd) = 10V – 0.205 x 24.4

Vds =
5.0V

In other
words, the Ideal Results are

Id=205mA.

Rd =
24.4 ohm

Vds
=5.0V

Av =
20.1

**B2. Describe Inputs – Experimental Case**

The
setup of the Experimental Case is the same as that of the Ideal case, expect

R1=6.9k

R2=3.3k

Rd = 36

The
experimental circuit setup is shown in Figure 2.

Figure 2

Therefore

Vgs=Vdd(R2/(R1+R2))

Vgs=10V(3.3k/(6.9k+3.3k))

Vgs =
3.24V

From (3)

20 =
2(0.82)(Vgs – 2.9) (Rd)

(20/(2*0.82)) = (Vgs – 2.9) (Rd)

12.2=
(3.24 – 2.9) (Rd)

Rd =
35.9 ohm, for which, a 36 ohm resistor is used.

From (5)

Id = K(Vgs – Vto)^2
…(5)

Id =
(0.82)(3.24 – 2.9)^2

Id=94.8mA

V(Rd)
= Id x Rd = 0.0948 x 36 = 3.40V, which is not the ideal 5V.

So Vds = Vdd-V(Rd)
= 10V – 3.4V = 6.6V,

for V(Rd) is the voltage across
Rd

In other
words, the Expected Experimental Results are

Id=94.8mA

Rd = 36
ohm

Vds
=6.6V

Av = 20

**C. Describe What you Observe**

The
experimental Vo =Vds = 4.384V.

The
input (Vin) amplitude is
100mA, which is 200mA peak to peak

Amplification
(Av) = Vo/Vi = 22

The
measured Id is

5.616/36ohm
= 0.156A

In other
words, the Actual Experimental Results are

Id=156 mA

Rd = 36
ohm

Vds =4.384V

Av = 22

**D. What you can deduce**

The
Percentage Difference of the Amplification

=|Target
Value – Experimental Value|/Target Value x 100%

=|20 –
22|/20 x 100%

=10%

The
Percentage Difference of the Id

=|Target
Value – Experimental Value|/Target Value x 100%

=|94.8mA
–156mA|/94.8mA x 100%

=64.6%

The
Percentage Difference of the Vds

=|Target
Value – Experimental Value|/Target Value x 100%

=|6.6V –4.384V|/6.6V
x 100%

=33.6%

The Experimental
values are off from the calculated value by the percentages shown above. The
error maybe caused by the non-perfect transistor or non-perfect resistors.

**Q1) Qualitatively,
describe what happens if Rd increase.**

If Rd
Increase, then Vds decreases.
Because Vds decreases, so Vo
decrease, and the amplification decreases.

**Q2)Qualitatively,
describe what happens if R2 decreases.**

If R2
decrease, then Id decreases. Since Id decreases, V(Rd)
decreases, and Vds increases, so the Vo is increased,
so the amplification increased.