EE 321 Analog Electronics, Fall 2013
Homework #13 solution
4.86. Figure P4.86 shows a scheme for coupling and amplifying a high-frequency
pulse signal. The circuit utilizes two MOSFETs whose bias details are not shown
and a 50-Ω coaxial cable. Transistor Q1 operates as a CS amplifier and Q2 as a
CG amplifier. For proper operation, transistor Q2 is required to present a 50-Ω
resistance to the cable. This situation is known as “proper termination” of the
cable and ensures that there will be no signal reflection coming back on the
cable. When the cable is properly terminated, its input resistance is 50 Ω. What
m ust gm2 be? If Q1 is biased at the same point as Q2 , what is the amplitude
of the current pulses in the drain of Q1 ? What is the amplitude of the voltage
pulse at the drain of Q1 ? What value of RD is required to provide 1 − V pulses
at the drain of Q2 ?
The input resistance of Q2 is the inverse of the transconductance gain, so
1
= 0.02 Ω−1
Rin2
Since Q1 is biased at the same point as Q2 , gm1 = gm2 . The amplitude of the current pulses
at the drain of Q1 is then
gm2 =
Id1 = Vg1 gm1 = 0.02 × 5 × 10−3 = 0.1 mA
This is also the drain current of Q2 , so the relationship between the output voltage amplitude,
Vo , and the drain current pulse amplitude is
Id1 RD = Vo
The needed resistance to get Vo = 1 V is then
RD =
1
Vo
=
= 10 kΩ
Id1
0.1 × 10−3
1
4.87. The MOSFET in the circuit of Fig. P4.87 has Vt = 1 V, kn′ W
= 0.8 mA/V2 ,
L
and VA = 40 V.
(a) Find the values of RS , RD , and RG so that ID = 0.1 mA, the largest possible
value of RD is used while a maximum signal swing at the drain of ±1 V is
possible, and the input resistance at the gate is 10 MΩ.
(b) Find the values of gm and ro at the bias point.
(c) If terminal Z is grounded, terminal X is connected to a signal source having
a resistance of 1 MΩ, and terminal Y is connected to a load resistance of
40 kΩ, find the voltage gain from signal source to load.
(d) If terminal Y is grounded, find the voltage gain from X to Z with Z opencircuited. What is the output resistance of the source follower?
(e) If terminal X is grounded and terminal Z is connected to a current source
delivering a signal current of 10 µA and having a resistance of 100 kΩ, find
the voltage signal that can be measured at Y . For simplicity, neglect the
effect of ro .
(a) We see right away that RG = 10 MΩ, and that VG = 0 V. Since we want a ±1 V voltage
swing while using the largest value of RD , we need the drain to be biased 1 V from
saturation, so VD = VG − Vt + 1 V = 0 V. We find RD from
RD =
VDD − VD
5
=
= 50 kΩ
ID
0.1 × 10−3
2
To find RS we must first find VS , and for that we find VGS from
ID =
kn′ W
(VGS − Vt )
2 L
VGS =
s
ID
′
kn
W
2 L
=
s
0.1
0.8
2
+ Vt
+1
=1.5 V
Thus, VS = −1.5 V, and then
RS =
−1.5 + 5
VS − VSS
=
= 35 kΩ
ID
0.1 × 10−3
(b) The value of gm ,
2 × 0.1 × 10−3
2ID
=
= 0.4 × 10−3 Ω−1
gm =
VGS − Vt
1.5 − 1
Also,
ro =
40
VA
=
= 400 kΩ
ID
0.1 × 10−3
(c) This is a common-source amplifier, and the overall voltage gain is (following the book’s
suggestion of ignoring ro )
RG
gm (RD ||RL )
RG + Rsig 1 + gm RS
0.4 × 10−3 × (50 × 103 ||40 × 103 )
10 × 106
×
=
10 × 106 + 1 × 106
1 + 0.4 × 10−3 × 35 × 103
=0.54
Gv =
(d) This is a source follower. The open-loop gain is
Avo =
ro
= 0.99
ro + g1m
and the output resistance at the source is
Rout = ro ||
1
1
= 400 × 103 ||
= 2484 Ω
gm
0.4 × 10−3
3
(e) I assume that current source is connected in parallel with the 100 kΩ resistor, as this
is the usual way for this type of connection. In that case, it is equivalent to the voltage
source with vsig = isig Rsig = 1 V and resistance Rsig = 100 kΩ. In that case the voltage
signal at Y is
gm RD
1 + gm Rsig
0.4 × 10−3 × 50 × 103
=1 ×
1 + 0.4 × 10−3 × 100 × 103
=0.49 V
vo =vsig Gv = vsig
4.88.
(a) The NMOS transistor in the source-follower circuit of Fig P4.88(a) has
gm = 5 mAV and a large ro. Find the open-circuit voltage gain and the
output resistance.
(b) The NMOS in the common-gate amplifier of Fig P4.88(b) has gm = 5 mA/V
and a large ro . Find the input resistance and the voltage gain.
(c) If the output of the source follower in (a) is connected to the input of the
common-gate amplifier in (b), use the results of (a) and (b) to obtain the
overall voltage gain vo /vi .
(a) This is a source follower. The open circuit voltage gain is
Avo = 1
and the output resistance is
Rout =
1
1
=
= 200 Ω
gm
5 × 10−3
4
(b) This is a common-gate amplifier. The input resistance is
Rin =
1
1
=
= 200 Ω
gm
5 × 10−3
The voltage gain is
Av = gm (RD ||RL ) = 5 × 10−3 5 × 103 ||2 × 103 = 7.14
(c) The overall gain is the cascade of the two amplifiers,
Rib
Avb
Roa + Rib
200
=1 ×
7.14 = 3.57
200 + 200
Gv =Avoa
5