EE I15B
Frequencyresponse
Log plots
Ifltr yto p lot
=
onaline a rs c a le f o rv a lu e s o f xb e t we e n 0 a n d l0 , I g e t a f ai r l y r e a +."
sonableplot. If I try to extendthe rangeto include 100, 1000,or one million, the plot quickly
stops conveying any useful information (try it sometime).
So what if I plot the 1ogof f(x) vs. the log of x? Now I find that I can actually draw a curve over
many orders of magnitude of values of x and f(x) and still get useful information about it. Better
yet, all I really need to draw is straight lines.
/(x)
The Bell - a unit of relative power (that'sa.joke-gert?)
One Bell is one order of magnitude of power gain. The Bell is a relative unit, in the sensethat
you need to specify a referencepower (or a ratio of powers) in order to calculatehow many Bells
somethingis. For example,for soundpressure,the unit of power commonlyusedis the minimum
power audible by the human ear, which correspondsto a pressurewave of roughly 0.00002
N / m2. Conversationalspeech,at roughly 5 Bells (or 50 deci-Belts)SPL (soundpressurelevel)
has then 5 orders of magnitude more power than the minimum audible sound. A loud concert
might be somewherearound 10 Bells, or 10 ordersof magnitudemore soundpower than the minimum audiblesound! 10 ordersof magnitude= 10,000,000,000.No wonderyour earsring a little
after the concert!
To calculatehow many Bells you have,take the log (base10) of the ratio of two powers:
.
/P\
togro[
If you find that this unit is inconvenientlylarge,you might want to multiply by 10
t,, re.f/]
"
and talk about tenths of Bells, or deci-Bells, which is usually contractedto decibels or dB.
Probablythe most common use of Bells or decibelsis to representthe power gain of a linear system, in which casethe ratio of powers is the output power over the input power. For a truly linear
system,this ratio is independentof the magnitudeof the input power.
Another common referencepower level is the milli-Watt. Often power in RF circuits is written as
somenumber of dBm. or deci-Bellsrelativeto a milli-Watt.
Sometimesthe referencepower will be somemid-band power level in a given circuit, as in "the
power is 3 dB down from it's DC value".
Becausewe usually work with voltagesand currentsrather than power levels, we can replace the
powers in the equation above with the expressionfor power as a function of voltage to get
roe,ofio
J = rosr.l
Y)
--\r ref/
'"lri"/
= togro(
#\'=
zrogls(Jl)
't \ v ref/
vVrrf)
r)
Which is theformulato usewhencalculatingBells basedon voltage. To calculatedeci-Bells
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basedon voltage,you needto multiply by 10 of course,whichis wherewe get thefamiliar forv
mula2Ologf,.
v ref
Who caresabouttransferfunctions?
Given a linear systemwith a transfer function H (s) , if the input of the systemis driven with a
sine wave, the output of the systemwill be a sinusoidwith somedifferent magnitude and phase.
The wonderful thing about the transferfunction is that it tells you what the magnitude and phase
of that output sinusoidwill be.
Inparticular,if v,n = sin (cor), thenvou,= lH(i@)lsin(or + /.H (.irrl)). This is suchasimple property to use, and such a powerful thing to be able to do, that we often plot lH (7co)| and
lH (jul) as a function of crl so that we can quickly look up what's going to happento a sine
wave at a given frequency. If you plot the log of the magnitudevs log frequency,we call these
plots Bode plots. Why do we want to plot the log of magnitudeand frequency,but plot phaseon
a linear scale?
Voltagedividers
Theseare easy:
R.
The thing to rememberis that this works with one or more capacitors
ffi;.
too.
complex numbers- magnitude,phase,and graphicalcalculation
Given a transfer function
H ( s) = a
(s-Z )
(s-Z )...(s-Z*)
= o f I G - 2, )
^ (s - P ,) (s - P z) ... (s - P n )
'fl (s- Pr)
the magnitude and phasecan be conveniently broken down in terms of the magnitude and phase
of the individual terms of the form s - x, which has a nice physical meaningin the complex
plane. s - "r is the vector from x to s . If we needto know the vector'smagnitude,we measureits
length. If we need to know its angle,we measurethat relative to the positive horizontal axis.
To get the magnitudeof H (s) , we seethat it is just the producflquotientof a bunch of lengths:
flls - z,l
ln(,)t=|"^|ffi
So we just needto measureall of the vectorsfrom zerosto 's', and divide that productby the
product of all of the lengths of the vectors from all of the poles to 's'.
Similarly for angles,if you representthe complex numbersas exponentials,you can quickly convince yourself (seebelow if you needa little help in the argument)that the angle of H(s) is just the
sum/difference of a bunch of angles:
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EE 115B
zH (s)= Llg -z)
-Ez(s- P; )
The long and short of it is that you can draw your poles and zeros,and think abouthow all of
theselengthsand angleschangeas you move aroundin the complexplane. Usually you will be
particularly interestedin sliding 's' up and down the positivecomplex axis. (If you can't figure
out why, go back up and look at 'Who caresabouttransferfunctions?')
The capacitor- a frequencydependentresistor?
One very useful way to think about capacitorsis just as frequencydependentresistors. If you see
an RC circuit and want to know how it will perform at a given frequency,just replace the capacitor (mentally!) with a resistorof value I^.
o lpF capacitorgives you lMe
at (D = 106 radl
sec, and IkQ at ot = 109 radlsec. (Quick quiz - what are thosefrequenciesin Hertz?) If the
capacitor is in parallel or serieswith a l0 kO resistor,somethinginteresting is going to happen
between those two frequencies.
Is this the right way to think about capacitors? In somecasesyes - but don't forget that you're
neglecting the phaseshift (that's really supposedto be a 7conot an crrin the formula for the
impedanceof a capacitor).
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Complexnumbers
A complex number of the form a + bj canalsobe representedas an exponentialof the form ,rio
.
In the first form, a andb specifythepositionof a point in a two dimensionalspacein a familiar
Cartesianway. In the secondform, thepositionof thepoint is specifiedasa radiusr and.anangle
from thehorizontalaxis, 0. You canusegeomeffyto figureout whattherelationshipis between
thepair a, b andthepair r, 0.
One advantageof the exponential form of a complex number is that the magnitudeand phase
can
be read directly from the representation. The magnitudeis r and the phaseis 0.
This feature makes it convenient to work with transferfunctions using exponential notation. If
we representa transfer function as
H( s) =o
^ ffi=o^ W" .(s - rr) (s - p ) ... (s - p,)
( s _ p;)
andthensubstitutein an exponentialrepresentation
for eachof the (s - Z,) and (s - p,) terms,
we get
l1 ( s) = a^
jor,
j!rjop
n
' fp r€
" ...rp ,e
,p ru
flrrejo''
ffr, ejo''
now remember that products of exponentialsturn into an exponentialof a sum, and we see
that
H (s) = o fI",
Et",)
*fr,,,
"t(Lu',-
The magnitude of the transfer function at a given point s is just the product of the
distancesfrom
the zerosto s divided by the distancesto all of the poles (anddon't forget about a^). Thephase
is the sum of all of the phasesof vectorsdrawn from zerosto s minus the phasesof vectors
drawn
from poles to s.
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EE115B
Frequencyresponse,
part II
Most amplifiers can be broken down into a seriesof stages. Each stagehas its own
relationship
betweengain and frequency,which we call the frequencyresponseoithr stage. The
overalt gain
of the amplifier as a function of frequency is just the product of the frequenc! depenoent
gain of
each stage. In many cases,the stagescan be modeledindependentof each other.
We will seethat
in some cases(notably Miller compensation)the gain of one stagehas a sffong
effect on the
fequencyresponseof anotherstage.
Singlestagegain and frequencyresponse
Most gain stagescan be modeled by a voltage controlled current sourceor transconductance
(B*), an output resistance(,
(C
ou), and an output capacitance
"il.
The low frequency gain of the amptfier is independentof the capacitorvalue,
and is just -g ou,.
^r
At some higher frequency,the capacitor will begin to "steal" a signifcant fraction
of the current
coming out of the transconductor.This will meanthat lesscurrent is flowing in
the resistor,hence
the output voltage will decrease,and the gain of the systemwill drop ,".orJingly.
To find the frequency at which the capacitor has the samenmgnitudeimpedanceas
the resisioi:
lll
lM)=
rou,
,oc;rr=
Now solve for rrr to get crr. =
r outcey
If we were to replace the capacitorwith a resistor of equalmagnitudeimpedance,
the voltage gain
of the amplifier would decreaseby a factor of two. Becausethe capacitorhas
a complex impedance,the actual decreasein gain at frequency or, is only
you
J1. (Why?
should be able to show
this both graphically and atgebraically).
For frequencieshigher than cDc,most of the current from the ffansconductor
goes into the capacitor, and the resistor plays an increasingly small part in the circuit, the load
impedancelooks very
muchli,."
&,
andtheoutputvoltage
becomes
, =
" ffi,
,
Another frequency of interest is the frequ ency atwhich the amplifier gives gain
a
of 1. The unity
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m 1158
gain frequencyCI)tis thefrequencyat which amplifier
stopsgiving voltage and.
gain,and begins
to attenuatethe signal' This occurswhen , =
!-.
v rff
Thesethreeexpressions
for theDC gain,
corner frequency,and unity gain frequency are very general
- they apply to MoS amplifiers, bipolar and vacuum tube-amplifiers,evenhydraulic
,yrt.Lr. They are basedon the assumptionthat
the stageis linear, which is always trusfor "smal'I"
signals- where n'small,,essentiallymeanssigare small enough to make the amplifier looklinear (so
we've gor some circuiar logic
ffi}]rt:"
The relationships can be summarizedgraphically
using a plot of the log of the magnitude of the
gain versus the log of the input frequency.
log (g
^r
or,)
Let's plug in somenumbers:if we have a g
, an outputresistanceof one Megaohm,
* of 1
and a 1 pF capacitgl-we get a DC gain of -1000, ^N'f
a corner frequency of one million radians per
second' or around 160kHz, and a unity gain frequency
of one billion rad/sec,or around 160 MHz.
It is worth remembering how the three components
affect the shapeof the transfer function as
well' Increasing the output resistanceror, increases
the low frequencygain, but decreasesthe
corner frequency' so the high frequencyperformance
remainsunchanged.Increasingthe effective
capacitanc" c
decreasesboth the corner frequency and the unity
gain frequency,but has no
"fi
effect on low frequency gain. Increasingthe
hansconductanceg, increasesboth the low frequency gain' and the unity gain frequency,
but doesnot affect the corner frequency.
Notice that the unity gain frequency is equal to
the corner frequency times the Dc gain,
01 = (8
o.
h
general,
the
gain
'
at frequenciesabovethe cornerfrequencyis given
or)
by
^r
8.,-foY'
Ao'"
A ( rrl) = =8 ^r our@,=
= GB
r + (D /(D c
(D
(D
(D
where GB is the gain-bandwidth product of the amplifier.
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EE1158
IncreasinErou,
DecreasingCrf
Increasingg,
Feedbackand stability
Feedbackimproves linearity, reducesoutput impedance,reducesinput offset, increasesbandwidth (sort of), allows gain to be set accurately,and all sorts of other wonderful stuff. Unfortunately, it also introduces the possibility of instability
vtw
vou,
With a little algebrayou can show that the gain from input to output in this figure is
t/,
A
_=O UT
V,*
1+A B
All of the quantities in this equation can be frequencydependent,and the relationship still holds,
and is often written G (s) = r ,O,(f). . If A is very large (or A(s) hasa largemagnitude),you
1 +A (s ) a
p
can approximatethe gain
. fo get a largegain, you needa small B, which may seemcoun", p|
terintuitive, but one way to look at it is that the amplifier needsto set V
o* so that V* and V _ are
(approximately)equal.
If P issmall,ittakesmoreofVour toproducethedesiredeffect
atV-,
resulting in large gain.
From the equation for the gain, it is clear that somethinginterestingwill happenf AA = - | .
The equation indicates that the amplifier will provide infinite gain in this case. The unfortunate
reality of this is that the amplifier will oscillate,becomeunstable,rtng,buzz,pop, and in some
cases,melt.. It will certainlynot provide you with the nice cleansignalsthat you would like to
see.
To avoid this unstablestateof affairs, let's first seehow it can happen.
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EE 1158
A(s) equal to negative one implies that the magnitudeof A is one, and the phaseof A is 180
degrees. Is this the only combination to worry about? What if the magnitude of A is 2 when the
phaseis 180 degrees? The answer,as you might expect,is that if the magnitudeA is greater than
or equal to 1 when the phaseof A is 180 degrees,the systemwill be unstable. [rt's call the
fequency at which the amplifier gives a phaseshift of 180 degreesor80. From the figure we can
seethat a signal arriving at the positive input of the amplifier with frequency crlrro will be ampli("rrrso) , and effectively multiplied by -1 (due to the 180 degreesof
lA
|
phaseshift) and sent into the negativeinput of the op-amp. The two minus signs will cancel, and
the processwill repeat. Each time around the loop the signal will be multiplied by
(",rrao) ,
lA
|
(7olrr/ is greaterthan 1, the output signal will increaserapidly
and inverted twice. Clearly, if
lA
|
fied in magnitude by
with time until something non-linear happens(like hitting the supply rails of the amplifier).
In summary, to design a stable amplifier, you needto make sure that you design the frequency
responseso that the magnitude of the transferfunction is less than one by the time the phasehas
reached180.
Miller compensation
Thetwo stageCMOSop-ampbelowhasa primarypoleassociated
with eachstage.Bothpoles
Tt
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areof theform, @p= +,
r
Thecapacitance
interestinghappens.
butin bothcasessomething
ourLeff
seenat the output of the differential amplifier is much larger than the apparentvalues of the
capacitorsC*+ C ,. This is due to the Miller multiplicationof the compensationcapacitorC, .
When a capacitoris placed acrossa gain stageit's effective capacitancechanges. Consider the
figure below, in which Vo = -AV,. The chargeon the capacitoris given by
Vo
Vi
but sincewe know whatVo is oncewe know V,,we canjust write
Q = C (V i-V),
= C (1+A )V i = Crf y ' , wh e reCrf f = (1 + A )C. S o t h e e f f e c t iv e c a p a c Q = C ( Vi--A V )
itance is equal to the actual capacitor value times the gain of the stageit's across.
In the figure below, this means that Crrr, = g m.r out2C* andthat the pole of the differential
11
amplifier is at cl)or =-=rorttc ,1t
r o rt l(8 ^ 7 t ' o u , zC)* '
If this were the only pole of the op-amp, the correspondingunity gain would be
(8^zr o''t) (8 nr o''z)
= y.
As we will see,it turnsout that this generallyis
01 = A o@ p I r o u tL(B
C
^ 7 t' o u t2 )
,
Cx
the unity gain frequency of a well designedMiller compensatedop-amp.
The pole frequency of the gain stageis also affectedby the compensationcapacitor,but not
becauseof Miller multiplication. There is no Miller multiplication since the gain looking from
the output of the stageback toward the input is small. At higher frequencies,however, the compensationcapacitor does effectively couple the output signal back to the gate of transistor M7 ,
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m 1158
resultingin an effectiveoutputresistance
of r uutL
out2= +
for frequencies
aboverrrol.
I m'7
The capacitanceseenat the output node is the sum of the load capacitanceC, and the seriescomb in a ti on ofC* andCr.If CxrrCr,th e n t h e o u t p u t c a p a c it a n c e is ju
s tu t 2 = CL + Ct . G e n e r Co
ally the load capacitancewill be higher than the internal capacitanceC1, so the secondpole
frequency is approxim ately r;-r, = y
'
vL
There is one additional effect of the compensationcapacitor. In addition to supplying a high-frequency coupling from the output to the gate of l:N.{7,italso allows signalsto favel from V, to V
o
without going through the inverting path through M7. This resultsin a right half plane zero
8
,n 7
-fi.
located
at az =
Noise
There are many sourcesof noise in analog circuits. There is externalnoise, which may come
from electrostatic or magnetic interference(you've all seen 60Hz on an oscilloscope). This type
of noise can be minimized by using differential signals,running signalsin twisted pair wires, or
co-ax cable. Unfortunately for the analog circuit designer,there is a more fundamentaltype of
noise in electrical circuits - thermal noise.
All resistorsgeneratenoise. Elecffons in a resistor are always bouncing around - they have thermal energy. In the absenceof an externally applied electric field, their motions are random. At
any given time, some of them will be poking their headsout one end of the resistor or the other.
This generatesa voltage which is random, and this voltage is what we call the thermal noise voltage of the resistor. Nyquist (I think) showedthat the amount of power in this noise voltage was
4kT for every Hz of bandwidth. Since the amount of noise power is the sameall acrossthe spectrum (4kT in each Hz) this is often called "white noise" (Q: what does "white" have to do with
evenly distributed power over frequency?). kT is a fairly small amount of energy (about
4 x tA-zr Joulesat room temperature),so we're not talking aboutmuch power here. If we look at
all of the thermal energy from DC to a gigaHertz,we only get about four picoWatts.Equating the
power that Nyquist tells us about with the formula for power in a resistor,we come up with
tt2
V
=
kTA,f and if we want to know what the noise voltage will be, we can solve for V to get
y = J kTRLf
For a lkC) resistor,
thisworksoutto about4 nanoVolts
timesthesquare
rootof thebandwidth
of
interest,or 4 nV/ J-Hz. Soin a 1 MHz bandwidth,
wecouldexpectto seeonaverage
about4
microVolts of noise acrossthe resistor.
Active devices (like MOSFETs and BJTs) generatenoise becausethey have internal resistances.
Ideal passivecomponentssuch as inductors and capacitorsdo not generatenoise, but real devices
will generally have some parasitic resistancewhich will generatenoise.
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EE I15B
Questions
1 . Since the gain of a multistage amplifier is the product of the gains of the individual stages,
why is it that I can talk about adding the magnitudeand phaseof the transfer functions? (you
may want to use both graphical or analytical explanationsto answerthis question)
2. Derive the formula for the output impedanceof a cascodestage(r ou,* g.r2).
5.
Derive the formula for the impedanceat the sourceof the cascodetransistor (r ou,= ;| I tn t
om
cascodeamplifier.
4 . If I invent a new type of MOS ransistor with a drain current given by
forthetransconducI p(V ps,Vss) = X (VZs+BVf,5)[ f * 1,ffi,), a"riu"expressions
tance and output impedance(S^ and ro).
5 . Design a folded cascodeCMOS amplifier with a DC gain of 10,000and a unity gain frequency of 200 MHz. Assume that the load will be capacitive,with no more than 10pF of
capacitance.How would your design changeif you were told to minimize power dissipation
at all costs? What would the minimum power dissipationbe? What if you were told to minimize die area?(Assume that die areais directly related to the sum of W times L for all
devices). What if you were told to designfor a power supply of lessthan 5 volts. How low
could you go?
6 . Design a two stageMiller compensatedCMOS amplifier which will be able to track input signals with better than 0.lVo accuracywhen usedin unity gain feedbackwith a 1k resistive load.
How would your design changeunder the sameconstraintsof power, die area,and supply
voltage above?
7. What is the fastestop-amp you can design which operatesfrom a 5 volt supply, allows input
and output voltage swings to within 1 volt of eachrail, and dissipatesless than 100pW of
power?
8 . Compare the performanceof a CMOS two stageMiller compensatedop-amp drawing 1mA in
the input stageand 5 mA in the output stagewith the performanceof a similar bipolar operating with the samebias currents. In particular, if we assumea Vo, of 250mV for the CMOS
devices, how do the transconductancesand output resistancesof the individual devices compare? What size capacitor is neededin eachcaseto make the amplifiers unity gain stable?
What is the DC gain? Unity gain bandwidth? How do theseanswerschangeif we increaseof
decrease Vo, for the CMOS devices?
9 . Explain why unity gain feedbackis the worst form of passivefeedback("worst" from a stability perspective).For an amplifier with 8.3 = Bp, and C* = I0Ct, what is the minimum
feedback gain (or maximum feedbackfactor B ) for which the systemwill have 90 degreesof
phasemargin?
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