basic&cmos&opamps& - technische universität mü · pdf...
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![Page 1: Basic&CMOS&OPAMPs& - Technische Universität Mü · PDF fileTwo&Stage&CMOS&OPAMP& • Assigning f nd &as&3GBW&and&C n1 /C C ... Telescopic& CascodeAmplifier& • Another&very&importantlimita3on&related&to&](https://reader031.vdocuments.us/reader031/viewer/2022030412/5a9e034b7f8b9a420a8c6ec8/html5/thumbnails/1.jpg)
Basic CMOS OPAMPs
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Amplifiers
Name Input Quan,ty Output Quan,ty
Opera3onal Amplifier (OPAMP) Voltage Voltage
Opera3onal Transconductance Amplifier (OTA)
Voltage Current
Opera3onal Current Amplifier (OCA) Current Current
Current Mode Amplifier Current Voltage
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OPAMPs
• An OPAMP is essen3ally a single pole amplifier. It exchanges gain for bandwidth. All other poles are beyond the GBW.
• Mul3stage amplifiers have several poles and can work properly at one gain. They do not exchange gain for bandwidth.
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Two Stage Miller OPAMP VDD
VSS
Vin- Vin+
VoutQ1
Q4
Q5
Q6
Q7
+
I1
Q8
Q2
Q3
Cc
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Two Stage Miller OPAMP
• The gain can be wriJen simply as,
• The capacitances are, €
A1 = −Gm1R1 = −gm1 r02 r04( )A2 = −Gm2R2 = −gm6 r06 r07( )A = A1A2Rout = r06 r07
€
C1 = Cgd 2 +Cdb2 +Cgd 4 +Cdb4 +Cgs6
C2 = Cdb6 +Cdb7 +Cgd 7 +CL
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Two Stage Miller OPAMP
• From the previous discussion on frequency compensa3on,
• To achieve -‐20 dB/dec down to 0 dB, €
f p1 ≈1
2πR1Gm2R2CC
f p2 ≈Gm2
2πC2
fz ≈Gm2
2πCC
€
GBW = f t = A fp1 =Gm1
2πCC
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Two Stage Miller OPAMP
• This frequency must be lower than fp2 and fz. Thus,
• In prac3ce, C2 is typically equal to CL. C1 is a parasi3c capacitance and should be included into the calcula3ons as a correc3on.
€
Gm1
CC
<Gm2
C2
Gm1 <Gm2
€
GBW =Gm1
2πCC
, fnd =Gm2
2πCL
1
1+Cn1
CC
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Two Stage CMOS OPAMP
• Assigning fnd as 3GBW and Cn1/CC as 0.3, we find a simple rule,
• This expression tells us to use larger current in the second stage.
• Once CC is chosen, the design can be completed easily.
€
Gm2
Gm1≈ 4 CL
CC
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Two Stage Miller OPAMP
• The total phase at f=ft is simply
• The phase of the zero is added, not subtracted because it is a posi3ve zero. €
φtotal = 90° + tan−1 f tf p2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ + tan
−1 f tf z
⎛
⎝ ⎜
⎞
⎠ ⎟
€
PhaseM argin = 90° − tan−1 f tf p2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ − tan
−1 f tf z
⎛
⎝ ⎜
⎞
⎠ ⎟
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Two Stage Miller OPAMP
• This zero is actually due to the feedforward through the capacitance.
• The current passing through the capacitor cancels out the output current of the amplifier, causing a zero.
• To get rid of this zero, we have to make the flow unidirec3onal; that is, cut the feedforward, but keep the feedback.
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Two Stage Miller OPAMP
• There are several solu3ons by using source followers or cascodes, but they are complicated circuits.
• The simplest solu3on is to use a resistance R in series with CC.
• The func3onality of this resistor can be easily understood if we write the current through CC.
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Two Stage Miller OPAMP
• Thus,
• The new loca3on of the zero is at
• By selec3ng R = 1/Gm2, the zero can be eliminated.
• Even if complete matching cannot be achieved, the zero can be pushed to higher frequencies or converted to a nega3ve zero.
€
Vi2
R +1sCC
=Gm2Vi2
€
s =1
CC1Gm2
− R⎛
⎝ ⎜
⎞
⎠ ⎟
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Two Stage Miller OPAMP
• However, too large a value should not be chosen either since the nega3ve zero is beneficial. Hence choose the zero to be less than 3GBW.
• Thus, a range for R can be determined as,
€
1Gm2
< R <13Gm1
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Two Stage Miller OPAMP
• Slew rate is generally due to the first stage and the compensa3on capacitance.
• Slew rate is simply I/CC, where I is the tail current of the first stage.
• Also, from these equa3ons, SR = VOVωt.
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Two Stage Miller OPAMP
• Define a Figure of Merit (FOM) for our OPAMP designs as
• This will be used to evaluate our designs. €
FOM =GBWxCL
IBIAS
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Two Stage Miller OPAMP
• Let us try to design a two stage OPAMP with GBW of 400 MHz and CL = 5 pF.
• We have two equa3ons rela3ng the three unknown variables, Gm1, Gm2, and CC.
• If you start by choosing CC first, its minimum is about 3Cn1 and maximum is CL/{2-‐3}.
• By adjus3ng CC, a minimum power point can be found.
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Two Stage Miller OPAMP
0.0E+00
5.0E-‐06
1.0E-‐05
1.5E-‐05
2.0E-‐05
2.5E-‐05
3.0E-‐05
0 1E-‐12 2E-‐12 3E-‐12 4E-‐12 5E-‐12 6E-‐12
I1
I6
Itot
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Two Stage Miller OPAMP
• This graph was drawn for GBW = 1MHz, CL = 10pF, Cn1 = 0.4pF, VGS – VT = 0.2V, and fnd = 3MHz.
• You can play with these values in the excel chart to obtain your op3mum.
• Note that Cn1 actually changes with the sizing of transistor M6. Thus, this graph is not exact.
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Two Stage Miller OPAMP
• The second alterna3ve is to choose Gm2 first. The absolute minimum for this value is 3(GBW)(2π)CL.
• Now, choose a Gm2 which is 30% larger (corresponding to CC = 3Cn1). Then, the parasi3c Cn1 is determined right away. One can move from here to calculate other variables.
• The third alterna3ve is to choose Gm1 first. This is useful to minimize noise.
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Two Stage Miller OPAMP
• Concentrate the choices in coefficients:
• Then, the GBW is given as
€
CL = αCC ,CC = βCn1 = βCGS6, fnd = γGBW
€
GBW =gm62πCgs6
1αβγ 1+1 β( )
fT6
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Two Stage Miller OPAMP
• Choose α,β,γ 2 3 2 • Calculate fT6 from GBW 6.4 GHz • Calculate L6 for VGS – VT of 0.2V 0.5 µm • Calculate W6 from CL 417 µm – Determine IDS6 2.3 mA – Determine Cn1 0.83 pF
• Calculate CC from CL and α 2.5 pF • Calculate gm1 and IDS1 0.63 mA
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Two Stage Miller OPAMP
• The total current consump3on is 3.56 mA. • The FOM is 561 MHzpF/mA.
• Remember fT in ac3ve region:
• And fT in subthreshold region:
€
fT =1.5 µn
2πL2(VGS −VT )
€
fT =12π
ItkTq
1CD
1L2
IDIM
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Two Stage Miller OPAMP
• For smaller GBW values, the transistors can be biased in weak inversion or in the boundary.
• Rewri3ng the above equa3ons by using the inversion coefficient technique,
€
fTfTH
= i 1− e− i( ) ≈ i
fTH =2µkT q2πL2
for small i
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Two Stage Miller OPAMP
• Let us now do a design for GBW = 1MHz and CL = 5pF.
• The transistors are probably in weak inversion. • Thus, it may be a good idea to use the EKV equa3ons.
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Two Stage Miller OPAMP
• Choose α, β, and γ 2 3 2 • Calculate minimum fT6 16 MHz • Choose a channel length, L6 0.5 µm
– Calculate fTH6 2 GHz • Calculate inversion coefficient 0.008 • Calculate W6 from CL 417 µm
– Calculate IDST6 0.33 mA – Calculate IDS6 2.7 µA – Calculate Cn1 0.83 pF
• Calculate CC 2.5 pF • Calculate gm1 and IDS1 1.6 µA
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OPAMP Specifica3ons
• Introductory analysis – DC currents and voltages on all nodes – Small signal parameters of all transistors
• DC Analysis – CM input voltage range vs supply voltage – Output voltage range vs supply voltage – Maximum output current (sink and source)
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OPAMP Specifica3ons
• AC and transient analysis – AC resistance and capacitance on all nodes – Gain vs frequency – GBW vs biasing current – SR vs load capacitance – Output voltage range vs frequency – SeJling 3me – Input impedance vs frequency – Output impedance vs frequency
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OPAMP Specifica3ons
• Specifica3ons related to offset and noise – Offset voltage vs CM input voltage – CMRR vs frequency
– Input bias current and offset – Equivalent input noise voltage vs frequency – Equivalent input noise current vs frequency – Noise op3miza3on for capaci3ve/induc3ve sources
– PSRR vs frequency – Distor3on
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OPAMP Specifica3ons
• Other second order effects – Stability for induc3ve loads – Switching the biasing transistors – Switching or ramping the supply voltages – Different supply voltages, temperatures, …
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Common Mode Input Voltage Range
• The maximum input voltage is VDD – VGS1 – VDS7
• The minimum input voltage is VSS + VGS3 + VDS1 – VGS1
• We can go closer to the nega3ve supply voltage.
• The opposite is true for an NMOS input circuit.
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Output Voltage Range
• If no resis3ve loads are present, the output can go rail to rail.
• Otherwise, there is a resis3ve divider between the load and the output resistors.
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Slew Rate Revisited
• The worst slewing condi3on occurs when an ideal square wave is applied to the OPAMP.
• The square wave is converted to a triangular wave with VOUT,max = SR/4fmax.
• As discussed above
€
SRGBW
=4πIDS1gm1
IDS1gm1
=VGS1 −VT
2IDS1gm1
=nkTq
ICE1gm1
=kTq
Strong inversion
Weak inversion
BJT
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Slew Rate
• Actually, there are two types of slew rate, external and internal.
• SR is also related to seJling 3me. €
SRint =IBCC
SRext =IDS7CL
€
ttot = tslew + t0.1 =VOUTSR
+7
2πBWln(1000) ≈ 7
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Output Impedance
• In the open-‐loop configura3on, the output impedance is
• However, this impedance starts dropping at the dominant pole and drops un3l
• At fnd, there is a second pole with CL. • Thus, the output impedance is not as large as expected.
• Furthermore, the OPAMP is typically used with feedback.
€
Rout = r06 r07
€
Rout = 1gm6
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Noise Behavior
• The first stage is the dominant noise source all the way to GBW.
• It is enough to calculate this noise. • The noise density is given by
• Using the concept of noise bandwidth, the integrated noise is given by
€
viN2 = 4kT 4 3
gmΔf
€
4kT3CC
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A Few Comments on the Input Stage
• Choose p-‐channel over n-‐channel for input differen3al amplifier due to – Lower noise – BeJer slew rate – BeJer GBW
• Noise op3miza3on can be performed on the first stage by changing the NMOS mirror dimensions as well.
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Telescopic Cascode Amplifier VDD
VBIAS1
VIN+VIN-
VBIAS2
VBIAS3
Q1
Q4
Q5 Q6
Q7 Q8
Q9
Q3
Q2
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Telescopic Cascode Amplifier
• Provides more gain at low frequencies. • GBW does not change.
• You can actually use gain boos3ng to the cascodes to increase the gain further. However, the GBW will not change.
€
GBW =gm12πCL
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Telescopic Cascode Amplifier
• This is a single stage amplifier. No major issues about stability.
• The output and input swings are quite small.
• This circuit is more an OTA than an OPAMP due to its high output resistance.
• High output resistance is not a big problem when driving capaci3ve loads.
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Telescopic Cascode Amplifier
• The maximum output swing for a standard configura3on is given by
• For our case, it is increased slightly. • The input swing is also limited. €
Vswing = 2 VDD − 2Vov,n + 2Vov,p +Vcs( )[ ]
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Telescopic Cascode Amplifier
• Another very important limita3on related to the above is the voltage mismatch between input and outputs. Imagine unity gain F/B between gate of M2 and drain of M6.
• For the new circuit, both transistors have to remain in ac3ve region. Thus, VBIAS3 must be chosen very carefully. Also, the swing is very limited.
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Folded Cascode OTA
VIN-
VIN+VBIAS1
VBIAS2
VBIAS3
VBIAS4
VOUT
VDD
Q1
Q4
Q6
Q8
Q9
Q10 Q11
Q2
Q5
Q7
Q3
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Folded Cascode OTA
• This circuit is symmetrical since M1 and M2 see the same impedance.
• The output is the only high resistance point; thus no compensa3on is necessary.
• The input is again high-‐swing. • The output swing is slightly higher (by one current source voltage) than the telescopic cascode.
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Folded Cascode OTA -‐ DC
• Choose bias current through M9 as 100 µA as an example.
• M1 and M2 each conduct 50 µA. • Choose the currents through M10 and M11 as 100 µA. Then, the rest of the transistors will conduct 50 µA.
• It is not a necessity to equate the currents through M9 and M10-‐11. However, good choice for symmetry.
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Folded Cascode OTA
• The power consump3on is thus twice the telescopic OTA.
• The Gain and GBW expressions are exactly the same.
• Why bother with the folded cascode rather than the telescopic cascode? Same performance at twice the power?
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Folded Cascode OTA
• The first non-‐dominant pole comes from the drains of M1 and M2. They form together one single non-‐dominant pole at approximately fT3/3.
• The other non-‐dominant poles arising from M5-‐M6-‐M7-‐M8 are followed immediately by zeros and are not discernible.
• Hence, this OTA has only one important non-‐dominant pole and is quite easy to design.
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Folded Cascode OTA
• Choose Vov1,2 as 0.2V and Vov10,11 as 0.5V. • Then, inputs can go all the way down to VSS.
• Hence, high input swing can be achieved. • Input and output voltage levels can easily be matched.
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A Small Comparison
• GBW = 100MHz, CL = 2pF, VGS – VT = 0.2V.
Type ITOT Swing
2-‐stage Miller 1.1 large
Telescopic 0.25 small
Folded Cascode 0.5 average
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Symmetrical CMOS OTA
VDD
Vout
VBIAS
1 : BB : 1
Q1
Q3 Q6
Q8Q9
Q2
Q4Q5
Q7
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Symmetrical CMOS OTA
• Although this has 3 current mirrors as opposed to only one in a simple single stage amplifier, the performance is in essence the same.
• However, this amplifier is the best you can achieve in terms of symmetry.
• Furthermore, B can be used to obtain more gain.
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Symmetrical CMOS OTA
• The gain is now gm1RoutB. • The BW is again given by
• Thus, GBW is
• How large can we make B?
€
BW =1
2πRoutCL
€
GBW = B gm12πCL
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Symmetrical CMOS OTA
• The two nodes drain of M1 and drain of M2 cause a single pole. The capacitance at this node is given by
• The pole at the drains of M5 and M7 is closely followed by a zero and can be ignored.
• Thus, the maximum of B can be found by equa3ng fnd to 3GBW. It is typically 3…5.
€
C = 1+ B( )Cgs4 +Cdb4 +Cdb2 ≈ 3+ B( )Cgs4
fnd =gm42πC
≈fT 43+ B
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Symmetrical CMOS OTA
• A symmetrical OTA can be built easily with cascodes as well.
• This will increase gain at low frequencies, but not the GBW.
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Fully Differen3al Amplifiers VDD
VBIAS1
VBIAS2
VIN- VIN+
VOUT+ VOUT-
Q1
Q4
Q5
Q3
Q2
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Fully Differen3al Amplifiers
• We can use this amplifier for differen3al opera3on which is desired in most applica3ons.
• However, very good control of biasing voltages is necessary which is typically not possible.
• Therefore Common Mode Feedback (CMFB) should be used.
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Fully Differen3al Amplifiers
• A CMFB circuit senses the common mode level in the two outputs. Then, it feeds back a signal related to this to the tail current.
• A typical sensing circuitry can be two resistors taking the average of the two outputs.
• However, the resistors will load the circuit. You may use source followers to isolate the CMFB circuit.
• Then, the CMFB signal can be compared against a reference and be fed back to the tail.
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Fully Differen3al Amplifiers
• Fully differen3al amplifiers are almost always used in prac3ce. CMFB is also very commonly used.
• The Miller OPAMP or cascode or folded cascode amplifiers can all be made differen3al.
• CMFB will be discussed later.