ECE-343 Test 1: Feb 10, 2010
6:00-8:00pm, Closed Book
Name : SOLUTION
CDepl =
CJ0
m
(1 + VR /Vo )
b1 =
X
τi
CDiff = τF gm
τi = Ri Ci
ωT =
ωH ≈
1
b1
gm
Cµ + Cπ
Z1 =
ωT =
gm
1/2
∝ ID or VOV
CGD + CGS
Z
1−A
Z2 =
Z
1 − 1/A
1. (12 pts) The circuit below shows a simple common-source amplifier with a source-degeneration resistor. Use a
small-signal analysis to find the low-frequency gains Ao = vout /vin and A1 = v1 /vin . Give your answer in
terms of Rin , RS , RD , and any applicable small-signal parameters.
Assume that γ = 0, λ = 0, and CDB = CSB = 0
Solution: The small signal model is illustrated on the right above. The controlled current source sets the current
through Rs , and determines v1 :
v1 = (gm vgs )Rs = gm Rs (vin − v1 )
v1 (1 + gm Rs ) = gm Rs vin
A1 =
v1
gm Rs
=
vin
1 + gm Rs
Once v1 is determined, the output voltage is found using
vout = −RD (gm vgs ) = −gm RD (vin − v1 )
Substituting the above value of v1 gives
gm Rs
gm Rs
−gm RD
vout = −gm RD vin −
vin = −gm RD 1 −
vin =
vin
1 + gm Rs
1 + gm Rs
1 + gm Rs
Ao =
vout
−gm RD
=
vin
1 + gm Rs
2.
(a) (8 pts) For the circuit of problem 1, examine the impact of the parasitic capacitors CGD and CGS . Use the
Miller theorem to redraw the circuit showing the equivalent capacitors between the various circuit nodes
and ground. Give your component values in terms of CGD , CGS , Ao = vout /vin and A1 = v1 /vin .
Solution: Miller’s theorem relates the equivalent capacitance to the gain between the nodes that the capacitor is
tied to. Use A1 to deal with CGS , and Ao for CGD .
Cin = CGS (1 − A1 ) + CGD (1 − Ao )
Co = CGD (1 − 1/Ao )
C1 = CGS (1 − 1/A1 )
(b) (8 pts) If RS is chosen sufficiently large, and RD RS , then the gains from problem 1 can be approximated as
Ao =
vout
−RD
≈
vin
RS
A1 =
v1
≈1
vin
Using these approximations, argue that the bandwidth of the amplifier is primarily determined by CGD
(and not CGS ).
Solution: If A1 ≈ 1, then all of the above CGS terms vanish. Essentially, no current will flow through CGS if the
two terminal voltages are equal, so it drops out of the circuit. Only the CGD term remains... and the contribution
to the input capacitance is potentially large. The above reduce to:
RD
Cin ≈ CGD 1 +
RS
RS
Co ≈ CGD 1 +
≈ CGD
RD
C1 ≈ 0
3.
(a) (12 pts) Considering only CGD , and using the approximations given in problem 2b, use the open-circuit
time constant method to give an approximate expression for the amplifier bandwidth.
Solution:
The open circuit time constants are found by scaling each capacitor
by the (DC) resistance measured at the capacitor terminals. At the
input, this resistance is just Rin . The resistance at the output node is
RD in parallel with the resistance seen looking into the drain of the
transistor. In this case ro = ∞ is given, so the result is just RD .
RD
τ1 = Rin Cin = Rin CGD 1 +
RS
RS
τ2 = RD Co = RD CGD 1 +
= (RD + RS )CGD
RD
The -3 dBfrequency is approximately the reciprocal of the sum of all time constants:
ωH ≈
1
=h
τ1 + τ2
Rin 1 +
1
RD
RS
i
(in rad/sec)
+ RD + RS CGD
(b) (3 pts) Assume that gm = 10 mA/V, RD = 5 kΩ, Rin = 100 kΩ, CGD = 1 pF. Evaluate the (approximate) amplifier gain, 3-dB bandwidth (in kHz), and gain-bandwidth product for RS = 500 Ω.
Solution:
Ao =
RD
= 10
RS
1
= 904.6 krad/sec
[(100 kΩ)(11) + (5 kΩ) + (500 Ω)](1 pF)
ωH
= 143.96 kHz
fH =
2π
GBW = 10 · fH = 1.44 MHz
ωH ≈
(c) (3 pts) Repeat problem 3b for RS = 1 kΩ.
Solution:
Ao =
RD
=5
RS
1
= 1650 krad/sec
[(100 kΩ)(6) + (5 kΩ) + (1 kΩ)](1 pF)
ωH
fH =
= 262.6 kHz
2π
GBW = 5 · fH = 1.31 MHz
ωH ≈
4. The circuit below shows a folded cascode CMOS amplifier utilizing a simple current source M2 , supplying
current 2I, and a cascoded current source (M4 , M5 ) sinking current I. All transistors are saturated. Assume,
for simplicity, that all transistors have equal small-signal parameters gm and r0 , and that gm r0 1. Neglect
the body effect, and neglect all parasitic capacitors.
(a) (18 pts) Give approximate expressions for all of the resistances indicated: Ro1 , Ro2 , Ro3 , Ro4 , Ro5 , Rin3
and Rout . Give your answers in terms of gm and ro .
Solution: Start with the easy cases:
Ro1 = Ro2 = Ro5 = ro
The impedance looking into the drains of M4 and M3 have an additional term given by the resistance attached at
their source terminals, scaled up by their intrinsic gains:
Ro4 = ro4 + (1 + gm4 ro4 )ro5 ≈ gm ro2
Ro3 = ro3 + (1 + gm3 ro3 )(Ro1 k Ro2 ) = ro + (1 + gm ro )(ro /2) ≈
gm ro2
2
The resistance into the source of M3 is 1/gm3 plus and additional term given by the resistance attached to the drain
terminal, scaled down by the intrinsic gain of M3 :
Rin3 =
1
gm3
+
Ro4
1
gm ro2
1
≈
+
=
+ ro ≈ ro
1 + gm3 ro3
gm
gm ro
gm
Finally,
Rout = Ro3 k Ro4 ≈
gm ro2
2
gm ro2
2
+ gm ro2
gm ro2
=
gm ro2
3
(b) (12 pts) Re-evaluate Rin3 for the case when the output node is shorted to ground, and use current division
to find the short-circuit transconductance Gm . Show that Gm ≈ gm1 .
Solution: For the output shorted, the resistance attached to the drain of M3 changes to 0 (instead of Ro4 ), so
Rin3 =
1
gm3
If it helps, redraw the small-signal model for the case in which the output is shorted to ground as shown below.
(With some practice, you should get to the point where you can work directly from the original circuit diagram.)
The controlled current source output from M1 is divided between Ro1 k Ro2 = ro /2 and Rin3 = 1/gm . The
output current is the current through Rin3 :
ro /2
gm ro
iout = gm1 vin
vin
= gm1
ro /2 + 1/gm3
gm ro + 2
gm ro
iout = gm1
Gm =
vin short
gm ro + 2
≈ gm1
(for gm ro Large)
(c) (6 pts) Evaluate the voltage gain vout /vin for the case gm = 2 mA/V, and r0 = 10 kΩ.
Solution: Here, the intrinsic gain is gm ro = 20. The amplifier voltage gain is
vout
= −Gm Rout
vin
2 2
ro
−(20)2
−gm
=
= −133.3
≈
3
3
5. (18 pts) Using VDD = 1.8 V and a pair of MOSFETs, design the current sink (shown below) to provide an output
current iO of 200 µA nominal value. To simplify matters, assume that the nominal value of the output current
is obtained at vO ≈ VGS . It is further required that the circuit operate for vo in the range of 0.20 V to VDD
and that the change in iO over this range be limited to 3% of the nominal value of iO . Find the required value
of R and the device dimensions of M2 . For the fabrication-process technology utilized, µn Cox = 250 µA/V2 ,
VA0 = 20 V/µm, and Vt = 0.6 V.
Solution: The required current sink characteristic is shown on the right above. The length of M2 is found using the
required output impedance.
ro2 ≥
1.8 − 0.2
1.6 V
=
= 266.6 kΩ
6 µA
6 µA
(20 V/µm)L2
≥ 266.6 kΩ
200 µA
L2 ≥ 2.66 µm.
(I’ll use the minimum required device length L2 = 2.66 µm.) Since the current source must operate at vo = 0.2 V,
the overdrive voltage of M2 (and M1 ) is set to 0.2 V, giving VG = 0.8 V for both MOSFETS. The width of M2 is
set to give the desired 200 µA output current:
250 µA/V2 W2
200 µA =
(0.2 V)2
2
L2
W2
= 40
L2
W2 = 40L2 = 40(2.66 µm) = 106.4 µm.
The current through M1 will be one fourth of that in M2 because of the different value of W1 /L1 . So the design
should set IREF = 50 µA. Using the known gate voltage of VG = 0.8 V, the value of R is determined:
R=
1.8 − 0.8 V
= 20 kΩ.
50 µA