ECE-343 Test 2: Mar 23, 2011, 6:00-8:00pm, Closed Book
Name : SOLUTION
1. (20 pts) Complete the design of the difference amplifier circuit shown below. Neglect the body effect and
channel-length modulation (γ = 0, λ = 0). Assume that kn0 = 400 µA/V2 and Vtn = 0.4 V.
Design for an overdrive voltage of 0.3 V for all transistors, and formulate your design to give gm1 = gm2 =
1 mA/V.
(a) Determine all MOSFET size ratios
M1
M2
M3
W/L
8.33
8.33
12.77
M4
M5
8.33
8.33
(b) Select a value for RD to maximize the available output voltage swing when vicm = 1.5 V.
RD = 6.33 kΩ
(c) Determine the input common-mode range for your design.
1 V < vicm < 2.45 V (= 3 − (150 µA)RD + 0.4)
Solution:
(a) Select the size of M1 and M2 to give the desired gm at VOV = 0.3 V.
W
VOV =0.3 V, k0 =400 µA/V2
0
gm = 1 mA/V = k
VOV
−−−−−−−−−−−−−−−−−→
L 1,2
W
L
1,2
= 8.33
M4 and M5 are the same size, since they carry the same drain current at the same overdrive voltage as M1
and M2 . The actual drain current is
ID1 = ID2 = ID3 = ID4 =
400 µA/V2
(8.33)(0.3 V)2 = 150 µA
2
Since VG3 = Vt + VOV = 0.7 V, the drain current in M3 is determined by the 10 kΩ resistor, ID3 =
(2.3 V)/(10 kΩ) = 230 µA. Given ID3 and VOV , the required size is determined
400 µA/V2 W
W
230 µA =
(0.3 V)2
−→
= 12.77
2
L 3
L 3
(b) For vicm = 1.5 V, M1 and M2 remain saturated as long as the output voltages remain above 1.5−Vt = 1.1 V.
To get the maximum output swing, bias the output voltages midway between this value and the positive supply
rail, at 2.05 V.
3 − 2.05
RD =
= 6.33 kΩ
150 µA
(c) M4 and M5 become ohmic if vD4 or VD5 drop below VOV = 0.3 V. Since VGS1 = VGS2 = Vt + VOV =
0.7 V, this requires that vicm > 0.3 + 0.7 = 1.0 V.
To keep M1 and M2 pinched off (for the 2.05 V output bias voltage), we need vicm < 2.05 + Vt = 2.45 V.
2. (20 pts) The circuit of problem 1 is reproduced below, except that M4 and M5 have been replaced by non-ideal
current sources, and values of RD are given. (Note that in practice implementing RD = 50 kΩ requires using
an active load... not allowed in problem 1.). Continue to ignore the body effect and channel-length modulation
for M1 and M2 , and assume gm1 = gm2 = 1 mA/V. Also, you may assume that Rs << 100 kΩ. Define the
differential input signal as v1 − v2 , and take vo1 as the (single-ended) output signal.
(a) Draw the differential-mode half-circuit, write down the differential gain expression in terms of the value of RS . Select
RS so that the design has a differential gain of -10.
(b) Draw the common-mode half circuit, and calculate the
common-mode gain and common-mode rejection ratio of
your design.
Solution:
Part a:
Using an approximation 100 kΩ k (Rs /2) ≈ Rs /2,
the gain of the amplifier is
−gm1 (50 kΩ)
vo1
=
vd /2
1 + gm1 (Rs /2)
vo1
1 −(1 mA/V)(50 kΩ)
Ad =
=
vd
2 1 + (1 mA/V)(Rs /2)
Setting the gain to Ad = −10 and solving for Rs
gives
Rs = 3 kΩ
Part b:
The gain of this common-source amplifier is
vo1
−gm1 (50 kΩ)
=
vicm
1 + gm1 (100 kΩ)
−50
=
= −0.495
100
Acm =
The common-mode rejection ratio is
CMRR =
Ad
10
=
= 20.2
Acm
.495
(= 26.1 dB)
3. (10 pts) An implementation of the circuit of problem 2 is shown below, in which the resistance Rs has been
replaced by two transistors M6 and M7 . Note that M6 and M7 share the same gate to source voltages as M1
and M2 , but operate in the ohmic (triode) region. All transistors are fabricated in the same technology as for
problem 1 (kn0 = 400 µA/V2 , Vtn = 0.4 V, γ = 0, λ = 0). Also assume that the design of problem 1 was
successful, so that M1 and M2 are saturated with Vov1 = Vov2 = 0.3 V
(a) At v1 = v2 = 1.5 V, what DC voltages appear at the sources
of M6 and M7 ? What current flows through M6 and M7 ?
(b) Find the size ratio W/L for M6 and M7 which would be
appropriate to implement the circuit of problem 2 with RS =
3 kΩ.
Solution:
(a) The gate to source voltages of M1 and M2 are given as Vt + Vov = 0.4 + 0.3 = 0.7 V. So the source voltages
are
VS6 = Vs7 = 1.5 V − 0.7 V = 0.8 V
Since there’s know voltage drop across M6 and M7 , no current flows in these transistors (M6 and M7 are
ohmic).
(b) To implement Rs = 3 kΩ, select rDS = 1.5 kΩ for each of M6 and M7 .
rDS =
1
k 0 (W/L)(VGS
− Vt )
= 1.5 kΩ
These transistors are biased at the same gate-to-source voltages as M1 and M2 (VGS − Vt = VOV = 0.3 V),
so set
1
W
=
= 5.55
L 6,7
(400 µA/V2 )(1.5 kΩ)(0.3 V)
4. For the two-stage amplifier shown below, neglect the body effect, and assume that
W
0
Vt = 1 V
kn
= 2 mA/V2
(VA )M 1 = (VA )Q2 = 40 V
L M1
βQ2 = 200
Cgs = 2 pF
Cgd = 1 pF
Cdb = 0
Csb = 0
Cµ = 0.8 pF
Cπ = 9.5 pF
The circuit should look familiar. It’s identical to a circuit from Test 1, except the the capacitor values have been
specified. Hopefully, you were able to show on Test 1 that by selecting R1 = 3 kΩ, you obtain (at midband)
ID1 = 103 µA
vout
= −18.5 V/V
vin
VOV 1 = 0.316 V
v1
= 0.649
vin
IC2 = 1 mA
vout
= −28.51 V/V
v1
Note: In the original exam, the value of VOV 1 was incorrectly given as 0.613 V instead of 0.316 V. The solution
here uses the corrected value (shown in red).
(a) (10 pts) Use the Miller theorem to write down the (midband) amplifier input impedance Rin , and calculate
the amplifier gain vout /vsig .
Solution: The feedback resistor RF has been replaced by the Miller equivalent resistances in the schematic above
(in red). The input impedance can be seen by inspection:
Rin =
The system gain is
vout
vin vout
=
·
=
vsig
vsig vin
10 MΩ
= 512.8 kΩ
19.5
512.8
512.8 + 100
(18.5) = 15.54 V/V
(b) (20 pts) Estimate the lower 3-dB frequency of the amplifier.
Solution: Identify the “pole” frequencies for each bypass capacitor (use ro2 = 40 V/1 mA = 40 kΩ):
1
= 16.3 rad/sec
(0.1 µF)(100 kΩ + 512 kΩ)
1
ω2 =
= 263 rad/sec
(1 µF)(1 kΩ + 9.5 MΩ k 3 kΩ k ro2 )
ω1 =
f1 = ω1 /(2π) = 2.6 Hz
f2 = ω1 /(2π) = 41.9 Hz
The approximate lower 3-dB cutoff frequency is the sum of the pole frequencies:
fL ≈ f1 + f2 = 44.5 Hz
(c) (20 pts) Estimate the upper 3-dB frequency of the amplifier.
Solution: To approximate fH , evaluate the time constants for the parasitic capacitors at nodes À, Á, and Â. At
each node, the Miller theorem is used to find the equivalent capacitance between the node and ground, and the time
constant is evaluated by finding the equivalent impedance looking into the node.
Node À:
C1 = Cgd + Cgs (1 − .649) = 1.7 pF
τ1 = (1.7 pF)(100 kΩ k 512.8 kΩ) = 142.2 ns
Node Á:
C2 = Cgs (1 − 1/.694) + Cπ + Cµ (1 + 28.51) = 32 pF
τ2 = (32 pF)(rπ k 6.8 kΩ k 1/gm1 )
The small signal parameters are
rπ =
β
200 · 25.8 mV
=
= 5.16 kΩ
gm2
1 mA
Substituting gives the time constant
gm1 =
2ID1
2(103 µA)
=
= 651.9 µA/V.
Vov1
.316 V
τ2 = 32 ns
Node Â:
C3 = Cµ (1 + 1/28.51) = 0.83 pF
τ3 = (0.83 pF)(1 kΩ k 3 kΩ k 9.5 MΩ k ro2 ) = 611 ps
The time constants are combined to approximate fH :
ωH ≈
1
1
= 5.72 Mrad/sec
=
τ1 + τ2 + τ3
174.8 ns
fH =
ωH
≈ 910 kHz
2π
iC = αiE = βiB = Is e
vBE /VT
vCE
1+
VA
W
1 2
(vGS − Vt )vDS − vDS
L
2
0
k W
=
(vGS − Vt )2 (1 + λvDS )
2 L
VT =
iD = k 0
k 0 = µn Cox
iD
λ=
IC
gm =
VT
β
rπ =
gm
VA + |VCE |
VA
r0 =
≈
IC
IC
α
re =
gm
kT
≈ 25.8 mV at T = 300 K
q
1
1
λ0
=
VA
L
α=
β
β+1
W
W
(VGS − Vt ) = k 0 VOV
rDS
L
L
hp
i
p
Vt = Vt0 + γ
2φf + VSB − 2φf
= k0
gm = k 0
p
W
2ID
VOV =
= 2k 0 (W/L)ID
L
VOV
VA + |VDS |
VA
r0 =
≈
ID
ID
gmb
!
γ
p
=
gm
2 2φf + |VSB |
|
{z
}
χ
−gm (RC k ro )
1 + gm (RC k ro )RE /RC
Avo =
large
r0
−gm RC
≈
1 + gm RE
Ais = β
large
g m RE
≈
R ≈ rπ + (β + 1)RE
−RC
RE
Rin = rπ + (β + 1)RE
Rout = RC k {ro + (1 + gm ro )(RE k rπ )}
RE k ro
≈1
re + RE k ro
Ais = β + 1
Avo =
R≈
rπ + RB
RB
= re +
β+1
β+1
Rin = rπ + (β + 1)(RE k ro )
Rout = re k RE k ro ≈ re
R = r0 + (1 + gm r0 )(RE k rπ )
Avo = gm (ro k RC )
With a base resistor, RB is added to rπ , and gm
is scaled by rπ /(rπ + RB ):
Ais = α ≈ 1
RC
gm ro
= RC k ro
Rin = re +
Rout
r0 RC
≈
re
1
RC
k r0 +
gm
1 + gm r0
r0 RC
1
RC
1
≈ rπ k
+
≈ rπ k
= re
gm
gm r0
gm
R = rπ k
Avo =
−gm (RD k ro )
1 + gm (1 + χ)(RD k ro )RS /RD
large
r0
−gm RD
1 + gm (1 + χ)RS
Ais = ∞
≈
large
g m RE
≈
−RD
(1 + χ)RS
With a base resistor, RB is added to rπ , and gm
is scaled by rπ /(rπ + RB ):
Rin = ∞
0
Use gm
= gm + gmb = (1 + χ)gm :
Rout = RD k {ro + (1 + gm (1 + χ)r0 )RS }
0
R = r0 + (1 + gm
r0 )RS
gm (RS k ro )
1
≈
1 + gm (1 + χ)(RS k ro )
1+χ
Ais = ∞
Avo =
Rin = ∞
Rout =
1
1
k RS k ro ≈
gm (1 + χ)
gm (1 + χ)
0
Use gm
= gm + gmb = (1 + χ)gm :
1
RD
k r0 +
0
0 r
gm
1 + gm
0
1
RD r0 RD 1
≈ 0 + 0
≈
0
gm
gm r0
gm
R=
Avo = gm (1 + χ)(RD k ro )
Ais = 1
1
RD
+
gm (1 + χ) gm (1 + χ)ro
r0 RD
1
≈
gm (1 + χ)
= RD k ro
Rin =
Rout
CDepl =
a1 =
CJ0
m
(1 + VR /Vo )
X 1
X 1
=
≈ ωL
τi
Ri Ci
CDiff = τF gm
b1 =
X
ωT =
τi =
X
gm
Cµ + Cπ
ωT =
gm
1/2
∝ ID or VOV
CGD + CGS
1
ωH
Z1 =
Z
1−A
Ri Ci ≈
Z2 =
Z
1 − 1/A