Blocks for SPAM Receiver
Antenna
XO
Lab 2
Decoding
BNC to
ANT
LNA
Mixer
(SA 612)
BNC to
Speaker
PLL
(LM 565)
IF Amp
Lab 1
Lab 3
RF Amplifier from Antenna (and “LNA”)
Tuned Load
(can be integral
with “match”)
Match to
Antenna…
M
TL
Match to
Next Stage (i.e.
Multiplier)
M
O
I
What does “Matched” mean?
Handout (developed by SPAM
“Alumni”) shows that conjugate
gives maximum power transfer…
TL
M
50Ω +jX
M
O
I
1.5KΩ +jY
1
Small-Signal View of Amplifier
ZI
ZI
vs
Cfb
f
vout
+
Vin Rin Cin
--
gmVin
ZL
ZL
f
At low frequencies (neglecting resonance effects) :
Vin = voltage divider (at Rin)
Look Ahead:
If ω0 for two Z(ω)
Vout = -gm Vin RL
are NOT the same…problems
At high frequencies:
We need to consider both Z(ω)’s and C’s…
ZI
ZL
f
Reminders about ac Parameters
For IC=5mA, β=80, τF=30ps (and other parameters per SPICE deck)
gm=qIC/kT=0.193 and rπ=β/gm=414Ω
Parameter
Value
Cπ=CJE(VBE) +gmτF
C
β
80
= 1.3pF f(VBE) + 5.8pF
τ
30ps
C
~7.1pF
F
JE
VBE
CJE
CJC
1.3pF
1.0pF
C
Cµ=CJC(VBC) ~CJC=1pF
CJC
VBC
1pF
Rin~200Ω for CE:
0.193
414Ω
Ω
Hybrid-π
R1||R2=
4KΩ
rπ=
414Ω
7.1pF
2
A quick comment about fT
Reminder about fT
i (s) = β(s)
i (s)
ic
c
b
at..ω T ..β (s) = 1
ib
(a review from EE113 notes…)
β
β (s) =
ω
(from data sheet)
@20mA
gm=0.772
100
Cπ+Cµ=(0.772)/(2π 5x109)
T
=
1+ s r π
(c + c )
π
g
µ
<=1/2πf3dB
m
cπ + c µ =2πfT
β
(I.e. fT)
= 24.6pF
1
Cπ+Cµ=CJE + CJC + gm τF
(now check out the SPICE deck…)
f3dB
fT= βf3dB
f
About Today’s Lecture:
•More on Feedback (HO#11)
(including example with re)
•Smith Chart (HO#10 + #12)
•Examples on Matching (HO#10)
9(Radmanesh examples…HO#8!!)
9Last Year’s MT…(no HO#…
solutions in another week)
3
A Simple (Practice) Biasing Example
+9V
R1
M
Assume IC=5mA
IR1=IR2=0.5mA (a bit big but…)
and X =3V
TL
M
X
I
X - 0.7V
re
RE
R2
O
Turning to the IC biasing
(3V - 0.7V)/RE = 5mA
which in turn gives re+RE=460Ω
9V/(R1+R2) = 0.5mA
thus R1+R2 = 18KΩ
and given X =3V
R2 = 6KΩ, R1 = 12KΩ
R1||R2=4KΩ
Note: for “fun” you might
think about finding R’s for
3mA and 1mA =IC
Now, moving on the the small-signal (ac) modeling...
(we’ll continue to consider the matching issues, per the last few
lectures, after we address the basic small-signal issues)
Putting these numbers together…
RI=50Ω
ZI
vs
1pF
+
vin
--
375Ω
7.1pF
vout
0.193v
in
gmRequiv=0.193 x 103=193
Cmiller=Cfb(1+gmRequiv)=194pF
Cin=7.1pF
vin=vs 375/(375+50)=0.88 vs
vout= -(194)(0.88)vs=-171vs
ZL Requiv=1KΩ
7.1pF +
Cmiller
375Ω
Large voltage gain is not always good…
1/2πRinCT~2.1 MHz!
Bandwidth and Low Noise are key metrics…
4
How do we get modest gain and BW!
Gain-BandWidth-Product (GBP)…trade-off gain for BW*
400
w. tuned load;
peaked at Requiv.
40
10
f3dB
10f3dB 40f3dB
Comment about Noise (as in LNA):
Resistor (Thermal or Johnson) Noise-v2=4kT R [BW]
Reduced BW->less total noise
(more about noise soon…)
To do this we can use
FEEDBACK**…
Acl=Aol/(1+Aolf)
Where “ol” is “open-loop” and
“cl” is “closed-loop”
* Assumption: Single-pole roll-off
** For the full story…see Brodersen Notes (UCB :)
FB Configurations and “bottom-line” Results-Gain, Resistances and BW
CBig
RF
“Shunt-Shunt”
feedback
(i.e. the resistor
“shunts” both
input and output)
Gain Expressions:
(extreme limiting cases…see supplemental notes for
derivations)
TL
M
RF
I
re
“Series-Series”
feedback
(i.e. the resistor is in series
with both input and output)
M
O
Av(shunt-shunt)~-RF/RS
Av(series-series)~-RL/re
Impedances (either Output or Input) :
Series…Rin|o-l [1+T]…T=Aolf
Shunt…Rin|o-l/[1+T]…(same T)
Bandwidth:
ω3dB|c-l=ω3dB|o-l[1+T]
What is T? (“loop gain”)…
(now let’s see what this really means!)
5
Series-Series--Nodal Analysis Results…
iT
vm +
g m vi
+
vi
RiT
+
ve
−
v o −gm RL
=
v m 1+ gm re
≈
“transconductance gain” io/vin
RoT
−
-
Comment: Natural Units are
io
f=vfb/io=re
re
vm
≈ RiT (1+ gmre )
iT
gm
io =
⋅v
(1+ gmre ) m
…
A
v
g to
goin
vTest
≈ RoT (1+ gmre )
io
−gm RL −RL
=
gm re
re
All three terms are
“scaled” by:
(1+ gm re )
(if 1 << than the second
term in the denominator)
Approximately
An Example (continued)
RI=50Ω
1pF
ZI
vs
375Ω
vout
7.1pF
0.193v
in
ZL Requiv=342Ω
Lower Requiv
re=4.56Ω
RI=50Ω
1pF
ZI
vs
(motivation to
follow:)
375Ω(1+gmre)
1+0.88
R1||R2=4KΩ || 705Ω
vout
0.193 V’in ZL Requiv=342Ω
(1+gmre)
7.1pF
(1+gmre) 0.103 1+g’mRequiv.
1+35.2
3.78pF + Cmiller(re)=36.2pF
Rin=600Ω
Cin=40pF
600Ω
40pF
0.103
Voltage Gain (w.fb):
-35.2
(-Requiv/re=75)
6
Shunt-Shunt--Nodal
Analysis Results…
R
F
ii
R iT
Comment: Natural Units are
g m vi
+
R oT
vi
−
“transresistance gain” vo/iin
f=ifb/vo=1/RF
+
v x gm vi
RiT
RF
vo
RoT RF
−
Rout | open−loop = RoT || RF
Rin | open−loop = RF || RiT
(
Open-loop vo = −gm RF
The “closed-loop”
expressions will then be
modified by this expression 1 + g m ( R F || R oT )( R F || R iT ) =
RF
for 1+T
1 + T
|| RiT )(RoT || RF )⋅ ii
a=vo/ii
af=T
Gain Expression (Shunt-Shunt)
Closed-loop
vo
=
ii
− g m (R F || R oT )(R iT || R F )
g ⋅ (R F || R oT ) ⋅ (R iT || R F )
1+ m
RF
vo
≈ − RF
ii
ii =
Approximately
(if 1 << than the second
term in the denominator)
vs
RS
∴ Av =
Av ≈
vo vo 1
=
v s ii RS
−RF
RS
7