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Lecture 270 – High Speed Op Amps (3/28/10) Page 270-1
INCREASING THE GB OF OP AMPSWhat is the Influence of GB on the Frequency Response?The unity-gainbandwidth represents a limit in the trade-off between closed loop voltagegain and the closed-loop -3dB frequency.Example of a gain of -10 voltage amplifier:
0dB
20dB
|Avd(0)| dB
Magnitude
log10(ω)GBωA ω-3dB
Op amp frequency response
Amplifier with a gain of -10
Fig. 7.2-1
What defines the GB?We know that
GB = gmC
where gm is the transconductance that converts the input voltage to current and C is thecapacitor that causes the dominant pole.This relationship assumes that all higher-order poles are greater than GB.
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-3
Higher-Order PolesFor reasonable phase margin, the smallest higher-order pole should be 2-3 times largerthan GB if all other higher-order poles are larger than 10GB.
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-GB-10GB
Dominantpole
Smallest non-dominant pole
Larger non-dominant poles
GB
10GB0dB
Av(0) dB
GBAv(0)
log10ω
If the higher-order poles are not greater than 10GB, then the distance from GB to thesmallest non-dominant pole should be increased for reasonable phase margin.
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-5
Increasing the GB of a Two-Stage Op Amp1.) Use the nulling zero to cancel the closest pole beyond the dominant pole.2.) The maximum GB would be equal to the magnitude of the second closest pole beyond
the dominant pole.3.) Adjust the dominant pole so that 2.2GB (second closest pole beyond the dominant
pole)Illustration which assumes that p2 is the next closest pole beyond the dominant pole:
0dB
|Avd(0)| dB
Magnitude
log10(ω)
Fig. 7.2-3
-40dB/dec
-p1-p2 = z1-p4-p3
|p1| |p2|
|p4||p3|
-60dB/dec-80dB/dec
Before cancellingp2 by z1 and increasing p1
jωσ
|p1|
GB
-p1New Old
GBIncrease
OldGBNew
Old New
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-6
Example 270-1 - Increasing the GB of the Op Amp Designed in Ex. 230-1Use the two-stage op amp designed
in Example 230-1 and apply the aboveapproach to increase the gainbandwidthas much as possible. Use the capacitorvalues in the table shown along with Cox= 6fF/μm2.Solution1.) First find the values of p2, p3, and p4.
a.) From Ex. 230-2, we see thatp2 = -95x106 rads/sec.
b.) p3 was found in Ex. 6.3-1 asp3 = -1.25x109 rads/sec. (alsothere is a zero at -2.5x109
Example 270-1 - Continued(c.) To find p4, we must find CI which is the output capacitance of the first stage of theop amp. CI consists of the following capacitors,
CI = Cbd2 + Cbd4 + Cgs6 + Cgd2 + Cgd4
For Cbd2 the width is 1.5μm L1+L2+L3=3μm AS/AD=4.5μm2 and PS/PD=9μm.For Cbd4 the width is 15μm L1+L2+L3=3μm AS/AD=45μm2 and PS/PD=36μm.From Table 3.2-1:
Therefore, CI = 6.9fF + 37.8fF + 273.7fF + 0.33fF + 3.3fF = 322fF. Although Cbd2 andCbd4 will be reduced with a reverse bias, let us use these values to provide a margin.Thus let CI be 322fF.
In Ex. 230-2, Rz was 4.564k which gives p4 = - 0.680x109 rads/sec.
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-8
Example 270-1 - ContinuedTherefore, the roots are:
jω
σ x 109
-1-2-3
z3 = -2.5G p3 = -1.25G p4 = -0.68G
p2 = -0.095G New GB
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When p2 is cancelled, the next smaller pole is p4 which will define the new GB. 2.)Using the nulling zero, z1, to cancel p2, gives p4 as the next smallest pole.For 60° phase margin GB = |p4|/2.2 if the next smallest pole is more than 10GB.
GB = 0.680x109/2.2 = 0.309x109 rads/sec. or 49.2MHz.This value of GB is designed from the relationship that GB = gm1/Cc. Assuming gm1 is
constant, then Cc = gm1/GB = (94.25x10-6)/(0.309x109) = 307fF. It might be useful toincrease gm1 in order to keep Cc above the surrounding parasitic capacitors (Cgd6 =18.7fF). The success of this method assumes that there are no other roots with amagnitude smaller than 10GB.
The result of this example is to increase the GB from 5MHz to 49MHz.
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-9
Example 270-2 - Increasing the GB of the Folded Cascode Op Amp of Ex. 240-4Use the folded-cascode op amp designed in
Example 240-4 and apply the above approach toincrease the gainbandwidth as much as possible.Assume that the drain/source areas are equal to2μm times the width of the transistor and that allvoltage dependent capacitors are at zero voltage.Solution
The poles of the folded cascode op amp are:
pA -1
RACA (the pole at the source of M6 )
pB -1
RBCB (the pole at the source of M7)
p6 -gm10
C6 (the pole at the drain of M6)
p8 -gm8rds8gm10
C8 (the pole at the source of M8)
p9 -gm9C9 (the pole at the source of M9)
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VPB1
M4 M5
RAI6
VPB2 RB
I4 I5
VDD
I7M6 M7
VNB2
M8 M9
M10 M11
+−
vIN
vOU
VNB1
I1 I2
M1 M2
M3I3
CL
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-10
The value of CB is the same as CA and gm6 is assumed to be the same as gm7 giving pB =pA = -1.456x109 rads/sec.3.) For the pole, p6, the capacitance connected to this node is
C6= Cbd6 + Cgd6 + Cgs10 + Cgs11+ Cbd8 + Cgd8
The various capacitors above are found asCbd6 = (560x10-6)(80x10-6·2x10-6) + (350x10-12)(2·82x10-6) = 147fFCgs10 = Cgs11 = (220x10-12·10x10-6) + (0.67)(10μm·0.5μm·6fF/μm2) = 22.2fFCbd8 = (770x10-6)(10x10-6·2x10-6) + (380x10-12)(2·12x10-6) = 24.5fFCgd8 = (220x10-12)(10x10-6) = 2.2fF and Cgd6 = Cgd5 =17.6fF
The smallest of these poles is pA or pB. Since p6 is not much larger than pA or pB, wewill find the new GB by dividing pA or pB by 4 (which is guess rather than 2.2) to get364x106 rads/sec. Thus the new GB will be 364/2 or 58MHz.Checking our guess gives a phase margin of,
PM = 90° - 2tan-1(0.364/1.456) - tan-1(0.364/3.38) = 56° which is okayThe magnitude of the dominant pole is given as
Ex. 270-3 - Design of a Voltage Amplifier for CascadingDesign the previous voltage amplifier for a gain of Ao = 10 and a power dissipation of nomore than 1mW. The design should permit Ao to be well defined. What is the -3dB forthis amplifier and what would be the -3dB for a cascade of three identical amplifiers?
SolutionTo enhance the accuracy of the gain, we replace M3and M4 with NMOS transistors to avoid thevariation of the transconductance parameter. Thisassumes a p-well technology to avoid bulk effects.The gain of 10 requires,
W1L1 (I3+I5) = 100
W3L3 I3
If VDD = 2.5V, then 2(I3+I5)·2.5V = 1000μW.
Therefore, I3+I5 = 200μA. Let I3 = 20μA and W1/L1 = 10W3/L3.
Choose W1/L1 = 5μm/0.5μm which gives W3/L3 = 0.5μm/0.5μm. M5 and M6 aredesigned to give I5 = 180μA and M7 is designed to give I7 = 400μA.
The dominant pole is gm3/Cout.
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VDD
Vout +−
Vin+
−
Μ1 Μ2Μ3 Μ4Μ5
VNB1
VPB1 VPB1
Μ6
Μ7
I7
I3 I4I5 I6
I1 I2
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-21
CASCADED AMPLIFIERS USING CURRENT FEEDBACK AMPLIFIERSAdvantages of Using Current FeedbackWhy current feedback?• Higher GB• Less voltage swing more dynamic rangeWhat is a current amplifier?
Requirements:io = Ai(i1-i2)
Ri1 = Ri2 = 0
Ro =
Ideal source and load requirements:Rsource =
RLoad = 0
Ri2
i1
i2io
+
-
CurrentAmplifier
Ri1
Ro
Fig. 7.2-8A
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-25
Bandwidth Advantage of a Current Feedback AmplifierConsider the inverting voltage amplifiershown using a current amplifier withnegative current feedback:
The output current, io, of the currentamplifier can be written as
io = Ai(s)(i1-i2) = -Ai(s)(iin + io)The closed-loop current gain, io/iin, can befound as
ioiin =
-Ai(s)1+Ai(s)
However, vout = ioR2 and vin = iinR1. Solving for the voltage gain, vout/vin gives
voutvin =
ioR2iinR1 =
-R2R1
Ai(s)1+Ai(s)
If Ai(s) = Ao
(s/ A) + 1 , then
vinvout
+- +
-i1
i2 io
io
vout
CurrentAmplifier
R1R2iin
VoltageBufferFig. 7.2-9
voutvin =
-R2R1
Ao1+Ao
A(1+Ao)s + A(1+Ao) Av(0) =
-R2AoR1(1+Ao) and -3dB = A(1+Ao)
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-26
Current Feedback AmplifierIn a current mirror implementation of the current amplifier, it is difficult to make the inputresistance sufficiently small compared to R1.
This problem can be solved using a transconductance input stage shown in the followingblock diagram:
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GMAi
RF
VinVout
+
−
VoutVin =
-GMRFAi1 +Ai
Lecture 270 – High Speed Op Amps (3/28/10) Page 270-28
SUMMARY• Increasing the GB of an op amp requires that the magnitude of all non-dominant poles
are much greater than GB from the origin of the complex frequency plane• The practical limit of GB for an op amp is approximately 5-10 times less than the
magnitude of the smallest non-dominant pole ( 100MHz)• To achieve high values of GB it is necessary to eliminate the non-dominant poles
(which come from parasitics) or increase the magnitude of the non-dominant poles• The best way to achieve high-bandwidth amplifiers is to cascade high-bandwidth
voltage amplifiers• If the gain of the high-bandwidth voltage amplifiers is well defined, then it is not
necessary to use negative feedback around the amplifier• Amplifiers with well defined gains are obtainable with a -3dB bandwidth of 100MHz