00884a Microchip Op Amps and Capacitive Loads

10
2003 Microchip Technology Inc. DS00884A-page 1 M AN884 INTRODUCTION Operational amplifiers (op amps) that drive large capacitive loads tend to have peaking and oscillation problems when they are not properly compensated. Other problems include: reduced bandwidth, lower output slew rate, and higher power consumption. This application note explains why these problems occur, how to modify the op amp circuit for better performance, and how to quickly compute circuit values. SIMPLIFIED OP AMP MODEL In order to understand how capacitive loads affect op amps, we must look at the op amp output impedance and bandwidth. Figure 1 shows a simplified AC model of an op amp configured for a non-inverting gain of G N . The open-loop gain is represented by the dependent source with gain A OL (s), where s = jω = j2πf. The output stage is represented by the resistor R O (open-loop out- put resistance). FIGURE 1: Op amp model. We will include gain bandwidth product (f GBP ), the open-loop gains second pole (f 2P ) and the non- inverting gain (G N ) in our open-loop gain (A OL (s)) model. Low frequency effects are left out for simplicity. f 2P models the open-loop gains reduced phase margin (PM < 90) at high frequencies due to internal parasitics. Both f 2P and the capacitive load (C L ) reduce the feedback loops phase margin. The op amp feedback loop (R F and R G ) causes its closed-loop behavior to be different from its open-loop behavior. Gain bandwidth product (f GBP ) and open- loop output impedance (R O ) are modified to give closed-loop bandwidth (f 3dBA ) and output impedance (Z OUT ). We can analyze the circuit in Figure 1 to give: Figure 2 shows Z OUT s behavior. At low frequencies, it is constant because the open-loop gain is constant. As the open-loop gain decreases with frequency, Z OUT increases. Past f 3dB , the feedback loop has no more effect and Z OUT stays at R O . The peaking at G N = +1 is caused by the reduced phase margin due to f 2P . FIGURE 2: MCP6271s Closed-Loop Output Impedance vs. Frequency. Figure 3 shows a simple AC model that approximates this behavior. The amplifier models the no-load gain and bandwidth, while the inductor and resistor model the output impedance vs. frequency. Author: Kumen Blake Microchip Technology Inc. V E V IN V OUT R O V E A OL(s) R F R G Z OUT A OL s () 1 s G N ω GBP ------------- 1 s ω 2P --------- + G N 1 R F R G ------ + = f 3dBA f GBP G N ----------- Z OUT R O 1 A OL s () G N ----------------- + = , R F R + G R O » 0.001 0.01 0.1 1 10 100 1000 1.E-01 1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 Frequency (Hz) Output Impedance ( ) 0.1 10k 100k 1M 10M G N = +1 G N = +10 G N = +100 1k 100 10 1 Driving Capacitive Loads With Op Amps

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op amps and driving capacitive loads

Transcript of 00884a Microchip Op Amps and Capacitive Loads

  • M AN884

    INTRODUCTIONOperational amplifiers (op amps) that drive largecapacitive loads tend to have peaking and oscillationproblems when they are not properly compensated.Other problems include: reduced bandwidth, loweroutput slew rate, and higher power consumption.This application note explains why these problemsoccur, how to modify the op amp circuit for betterperformance, and how to quickly compute circuitvalues.

    SIMPLIFIED OP AMP MODELIn order to understand how capacitive loads affect opamps, we must look at the op amp output impedanceand bandwidth. Figure 1 shows a simplified AC modelof an op amp configured for a non-inverting gain of GN.The open-loop gain is represented by the dependentsource with gain AOL(s), where s = j = j2f. The outputstage is represented by the resistor RO (open-loop out-put resistance).

    FIGURE 1: Op amp model.We will include gain bandwidth product (fGBP), the

    f2P models the open-loop gains reduced phase margin(PM < 90) at high frequencies due to internalparasitics. Both f2P and the capacitive load (CL) reducethe feedback loops phase margin.

    The op amp feedback loop (RF and RG) causes itsclosed-loop behavior to be different from its open-loopbehavior. Gain bandwidth product (fGBP) and open-loop output impedance (RO) are modified to giveclosed-loop bandwidth (f3dBA) and output impedance(ZOUT). We can analyze the circuit in Figure 1 togive:

    Figure 2 shows ZOUTs behavior. At low frequencies, itis constant because the open-loop gain is constant. Asthe open-loop gain decreases with frequency, ZOUTincreases. Past f3dB, the feedback loop has no moreeffect and ZOUT stays at RO. The peaking at GN = +1 iscaused by the reduced phase margin due to f2P.

    FIGURE 2: MCP6271s Closed-Loop Output Impedance vs. Frequency.

    Author: Kumen BlakeMicrochip Technology Inc.

    VE

    VIN

    VOUTROVEAOL(s)

    RFRG

    ZOUT

    GN 1RFRG-------+=

    f3dBAfGBPGN-----------

    ZOUT RO 1AOL s( )

    GN-----------------+ = , RF R+ G RO

    0.001

    0.01

    0.1

    1

    10

    100

    1000

    1.E-01 1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

    Frequency (Hz)

    Out

    put I

    mpe

    danc

    e ( :

    )

    0.1 10k 100k 1M 10M

    GN = +1GN = +10GN = +100

    1k100101

    Driving Capacitive Loads With Op Amps 2003 Microchip Technology Inc. DS00884A-page 1

    open-loop gains second pole (f2P) and the non-inverting gain (GN) in our open-loop gain (AOL(s))model. Low frequency effects are left out for simplicity.

    Figure 3 shows a simple AC model that approximatesthis behavior. The amplifier models the no-load gainand bandwidth, while the inductor and resistor modelthe output impedance vs. frequency.AOL s( ) 1

    s GNGBP-------------- 1 s

    2P---------+

  • AN884FIGURE 3: Simplified AC model.The equations for LOUT and ROUT are:

    Note that ROUT is larger than RO in order to includef2Ps phase shift effects, especially at low gain (GN).

    CAPACITIVE LOAD COMPENSATIONOur discussion on compensating capacitive loads isdivided into three topics. First, we will show the effectof capacitive loads when there is no compensation.Second, we will show a simple compensation method,and how it improves circuit behavior. Last, we will showhow to deal with inverting circuits.

    Capacitive LoadFigure 4 shows a non-inverting gain circuit with anuncompensated capacitive load. For small capacitiveloads and high gains (typically CL/GN < 100 pF), thiscircuit works quite well.

    FIGURE 4: Uncompensated capacitive load.Replacing the op amp in Figure 4 with the simplified opamp AC model gives a LC resonant circuit (LOUT andCL). When CL becomes large enough, ROUT does apoor job of dampening the LC resonance, whichcauses peaking and step response overshoot. Also,the overall closed-loop bandwidth (f3dB) is reduced.

    We recommend setting RF + RG >> ROUT for bettercircuit performance. A simplified transfer function is:

    Now that we have estimates of fP and QP, we can usethe equations in Appendix A to estimate bandwidth(f3dB), frequency response peaking (HPK/GN) and stepresponse overshoot (%ovrsht). Note that f3dB is not thesame as the op amps no load -3dB bandwidth (f3dBA).

    MCP6271 Example (Uncompensated)The equations above were used to generate the curvesin Figure 5 and Figure 6 for Microchips MCP6271 opamp. The parameters used are (see Appendix B):fGBP = 2.0 MHz, f2P = 4.5 MHz and RO = 360. As canbe seen at GN = =+1 V/V and CL = 100 pF, theresponse peaks enough to be a concern.

    FIGURE 5: Estimate of MCP6271s AC response with GN = +1.

    FIGURE 6: Estimate of MCP6271s AC response with GN = +10.

    VINGN

    RF+RG

    ROUT

    LOUT

    1 + s/3dBA

    LOUT GNRO 2fGBP( )=

    ROUT RO 1fGBP

    GN f2P----------------- ,

    fGBPGN

    ----------- f2P

    2---------

    VIN

    RG

    CL

    VOUT

    RF

    MCP6xxx

    VOUTVIN

    ------------- GN 1s

    PQP-------------- s

    2

    P2

    -------+ +

    where:P 2fP 1 LOUTCL= =

    QP ROUT CL LOUT= , RF R+ G ROUT

    -20

    -15

    -10

    -5

    0

    5

    10

    15

    1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

    Frequency (Hz)

    Gai

    n (d

    B)

    10k 1M 100M

    GN = +1

    CL = 10 pFCL = 100 pF

    CL = 1 nF

    10M100k

    0

    5

    10

    15

    20

    25

    30

    35

    1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

    Frequency (Hz)

    Gai

    n (d

    B)

    10k 1M 100M

    GN = +10

    CL = 100 pFCL = 1 nF

    CL = 10 nF

    10M100kDS00884A-page 2 2003 Microchip Technology Inc.

  • AN884

    The peaking (HPK/GN) should be near 0 dB for the bestoverall performance. Keeping the peaking below 3 dBusually gives enough design margin for changes in opamp, resistor and capacitor parameters overtemperature and process. However, the performance isdegraded.

    For this example, our formulas give the estimatedresults shown in Table 1. As CL increases and gaindecreases, there is more peaking.

    TABLE 1: RESPONSE ESTIMATES

    Compensated Capacitive LoadThe simplest compensation method for capacitiveloads is shown in Figure 7. The resistor RISO is used toload down the LC resonant circuit, which reducesfrequency response peaking. As can be seen, RISOdoes not change the DC response or power and onlycosts one additional resistor.

    FIGURE 7: Compensated capacitive load.The simplified op amp AC model produces thefollowing transfer function, where, again, we requireRF + RG >> ROUT.

    With these equations, we can now find a reasonableRISO value. When QP = 1/2, the response has thehighest possible bandwidth without peaking, and theequations are in their simplest form.

    MCP6271 Example (Compensated)These equations were used to compensate theMCP6271 circuits represented in Figure 5 andFigure 6. The same parameters were assumed, withthe only change being the addition of RISO. Theimproved results are shown in Figure 8 and Figure 9.Table 2 shows much better results than Table 1.

    FIGURE 8: Estimate of MCP6271s compensated AC response with G = +1.

    FIGURE 9: Estimate of MCP6271s compensated AC response with G = +10.

    Circuit Response

    GN(V/V)

    CL(F)

    fP(Hz)

    QP( )

    f3dB(Hz)

    HPK/GN(dB)

    %vrsht(%)

    1.0 10p 9.4M 0.31 3.2M 0 0100p 3.0M 0.98 3.7M 1.1 15

    1n 940k 3.1 1.4M 9.9 6010.0 100p 940k 0.22 220k 0 0

    1n 300k 0.71 295k 0 410n 94k 2.2 140k 7.2 49

    RISOVIN

    RG

    CL

    VOUT

    RF

    MCP6XXX

    VOUTVIN

    ------------- GN 1s

    PQP-------------- s

    2

    P2

    -------+ +

    where:

    P 2fP 1 LOUTCL 1RISOROUT-------------+ = =

    QP 1 PLOUTROUT------------- RISOCL+ = , RF R+ G ROUT

    QP 1 2 0.707=

    RISO 0= , CL LOUT 2ROUT2( )

    RISOLOUT

    ROUTCL---------------------

    2ROUT2 CL

    LOUT------------------------ 1= , CL

    LOUT2ROUT

    2----------------->

    -20

    -15

    -10

    -5

    0

    5

    10

    15

    20

    1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

    Frequency (Hz)

    Gai

    n (d

    B)

    10k 1M 100M

    GN = +1

    CL = 10 pFRISO = 0

    CL = 100 pFRISO = 523

    CL = 1 nFRISO = 232

    100k 10M

    0

    5

    10

    15

    20

    25

    30

    35

    1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

    Frequency (Hz)

    Gai

    n (d

    B)

    10k 1M 100M

    GN = +10

    CL = 100 pFRISO = 0

    CL = 1 nFRISO = 0

    CL = 10 nFRISO = 226

    100k 10M 2003 Microchip Technology Inc. DS00884A-page 3

  • AN884

    TABLE 2: RESPONSE ESTIMATES.

    Figure 10 shows the RISO values for the MCP6271estimated by the above equations. It is shown versusnormalized load capacitance (CL/GN) for ease ofinterpretation. Measured data for one representativepart is shown in Figure 11.

    FIGURE 10: Estimated RISO for the MCP6271.

    FIGURE 11: Recommended RISO for the MCP6271.The main difference between Figure 10 and Figure 11is at GN = +1. The reduced phase margin at low GN(caused by f2P) requires additional compensation atlow CL. The simplified equations in this application notegive reasonable estimates in this condition, but are notexact.

    When large capacitive loads give lower op amp band-width than desired, refer to Microchips line of PowerMOSFET Drivers at www.microchip.com.

    Inverting GainsInverting gain circuits (see Figure 12) arecompensated in the same way as non-inverting gaincircuits. Since the inverting input of the op amp is atvirtual ground, the load presented to the output by thefeedback network is now RF instead of RF + RG. Thus,we need to set RF >> ROUT. Use the noise gain:

    in the previous equations, even though the invertinggain is -RF/RG. For example, an inverting gain of-1 V/V gives GN = +2. The reasons for this behaviorcome from op amp feedback theory [1, 3].

    FIGURE 12: Compensated inverting gain circuit.

    SLEW RATEIn Figure 7 and Figure 12, the op amp will produce anoutput current (IOUT) that goes into CL. This currentcannot exceed the op amps output short circuit current(ISC). This current limit causes the output slew rate tobe limited (SRCL). Note that SRCL is independent of theop amps internally-set slew rate (SR). We can deriveSRCL as follows.

    Slew Rate and Sine WavesSine waves with edge rates faster than SRCL or SR willcause signal distortion problems. The sine wave

    Circuit Response

    GN(V/V)

    CL(F)

    RISO()

    fP(Hz)

    QP( )

    f3dB(Hz)

    HPK/GN(dB)

    %vrsht(%)

    1.0 10p 0 9.4M 0.31 3.2M 0 0100p 523 2.1M 0.71 2.1M 0 41n 232 780k 0.71 780k 0 4

    10.0 100p 0 940k 0.22 220k 0 01n 0 300k 0.71 300k 0 410n 226 74k 0.71 74k 0 4

    10

    100

    1,000

    1.E-11 1.E-10 1.E-09 1.E-08

    Normalized Load Capacitance; CL/GN (F)

    Estim

    ated

    RIS

    O ( :

    )

    10p 1n 10n

    1k

    100p10

    GN = +1GN = +2GN t +5

    100

    10

    100

    1,000

    10 100 1,000 10,000

    Normalized Load Capacitance; CL / GN (F)

    Rec

    omm

    ende

    d R I

    SO (

    )

    GN = +1GN = +2GN +5

    10p 100p 1n 10n

    One representative part

    GN 1RFRG-------+=

    VIN

    RISO

    CL

    VOUT

    RF

    MCP6XXX

    RG

    dVOUT t( )dt

    ------------------------IOUT t( )

    CL-------------------=

    SRCL maxdVOUT t( )

    dt------------------------

    ISCCL--------= =

    where:SRCL is in units of V/s

    VOUT t( ) VM 2ft( )sin=DS00884A-page 4 2003 Microchip Technology Inc.

  • AN884

    has a maximum edge rate of

    Thus, we need to keep

    One solution is to low-pass filter the signal before itreaches CL. The filter bandwidth needs to satisfy

    Another solution is to add RISO, as shown in Figure 7and Figure 12. The maximum current occurs whenVOUT(t) = 0; at this point the voltage across RISO is VM.Thus, we need

    This choice will also reduce the signal bandwidth tomeet the limit given above. The equations in theCompensated Capacitive Load section andAppendix A can be used to find the resultingperformance as long as the signals slew rate does notexceed SR or SRCL.When large capacitive loads cause a lower slew ratethan desired, refer to Microchips line of PowerMOSFET Drivers at www.microchip.com.

    Sine Wave ExampleLets look at the MCP6271 with G = +1 V/V andCL = 1.0 F. In Appendix B, we find SR = 0.9 V/s andISC = 25 mA. This gives:

    which is much lower than SR. With a maximum peakvoltage of 2.5VPK, we need an input signal with abandwidth less than 1.8 kHz.If we use RISO to limit the output current, we need it tobe > 100. Setting RISO = 130 gives:

    Note that if we used the RISO value for responsepeaking elimination (7.6), we would achieve a widerbandwidth (29 kHz), but would need to keepVM < 0.15 VPK to avoid output current limiting andsevere signal distortion.

    Slew Rate and Square WavesSquare waves with fast edges can also cause prob-lems with capacitive loads. The maximum edge rate ofa square wave with a (10% to 90%) rise time of tr anda peak-to-peak voltage of VPP, can be approximatedas:

    Thus, we need to keep

    One solution to this problem is to use square waveswith lower edge rates (higher tr). Filtering the squarewaves (lowpass filter bandwidth < 0.35/tr) is anotherapproach. Using slower logic gates may be a solutionin some cases. It is also possible to add RISO, as shownin Figure 7 and Figure 12. The maximum currentoccurs when the ideal output just reaches the new leveland VOUT(t) is still slew rate limited. To keep IOUT < ISC,we need:

    Using RISO will both slow the edges down and changethe shape of the transitions.

    When large capacitive loads cause a lower slew ratethan desired, refer to Microchips line of PowerMOSFET Drivers at www.microchip.com.

    Square Wave ExampleLets use the MCP6271 with G = +1 V/V andCL = 100 nF. In Appendix B, we find SR = 0.9 V/s andISC = 25 mA. We can then calculate:

    which is significantly slower than SR. With a maximumvoltage swing of 5.0VPP, we need an input signal with arise time > 16 s.Filtering the input square wave at the input of the opamp would require a bandwidth less than 22 kHz.If we use RISO to limit the output current, (with amaximum voltage swing of 5.0VPP and an input risetime of 10 s), we need RISO > 75. SettingRISO = 100 gives:

    QP = 0.046f3dB = 1.2 kHz

    maxdVOUT t( )

    dt------------------------ 2fVM=

    where:VM is the peak output voltage

    2fVM min SRCL SR,( )

    SRCL 0.028V/s=

    QP = 0.18f3dB = 16 kHz

    maxdVOUT t( )

    dt------------------------

    0.8VPPtr

    -----------------

    0.8VPP tr min SRCL SR,( )

    SRCL 0.25V/s= 2003 Microchip Technology Inc. DS00884A-page 5

  • AN884

    Note that if we used the RISO value for responsepeaking elimination (24.0), we would achieve a widersmall signal bandwidth (92 kHz), but would need tokeep VPP < 3.7VPP to avoid output current limiting andreduced rise and fall times.

    POWER DISSIPATIONIt is well known that reactive elements (ideal capacitorsand inductors) do not dissipate power. However, an opamp driving a reactive load will dissipate power. Thishappens because load current in the output stagealways flows in a direction that dissipates power. Theoutput transistors rectify the load current.Figure 7 and Figure 12 show the circuits underdiscussion. There will be no DC load current becauseCL blocks DC. At low frequencies, IQ (op ampsquiescent current) and CL will dominate the outputcurrent behavior. At high frequencies, RISO willdominate.Given an output voltage of

    it can be shown that the average power dissipated bythe op amp at low frequencies is:

    The power dissipation increases with frequencybecause CL dominates the load.At high frequencies, the average power dissipated bythe op amp becomes constant because RISOdominates:

    In the frequency range where neither CL or RISOdominates the load (f 1/(2RISOCL)), a somewhatconservative estimate of POA is the minimum valuefrom the two formulas above.

    DESIGN VERIFICATIONWe recommend that you always verify the performanceof your circuit design with SPICE simulations and bybreadboarding it on the bench. SPICE macro models ofMicrochips op amps are available on the Microchipweb site at www.microchip.com for your convenience.

    SUMMARYWe have seen that op amps that drive large capacitiveloads tend to show peaking and oscillation, reducedbandwidth, lower output slew rate, and higher powerconsumption. Adding one resistor to the circuit cangreatly improve the performance. The resulting band-width is a little under the no load bandwidth.

    Simple formulas were given that allow a circuitdesigner to quickly evaluate the impact of capacitiveloads. The fix is easy to implement and understand.

    Designs that need to drive large capacitors at highbandwidth or rise time may benefit from usingMicrochips line of Power MOSFET Drivers.

    VOUT t( ) VM 2ft( )sin=

    POA VDD VSS( ) IQ 2VMfCL+( )= , f1

    2RISOCL--------------------------

    POA VDD VSS( ) IQVM

    RISO---------------+

    VM2

    RISO-----------= ,

    f 12RISOCL--------------------------DS00884A-page 6 2003 Microchip Technology Inc.

  • AN884

    APPENDIX A: RESPONSE MODELIn this application note, we have seen transfer func-tions of the form:

    This is a 2nd order, low-pass response, which modelsthe op amp circuits in this application note reasonablywell. We will show some simple formulas for sine waveand step responses which help evaluate the perfor-mance of the circuits in this application note [2,4].Given fP (P = 2fP) and QP, we can calculate the band-width (f3dB), peak response frequency (fPK) and gainpeaking (HPK/GN) for a sine wave as follows:

    The step response overshoot (%ovrsht) and rise time (tr)are calculated, as follows:

    It is relatively simple to extract K, fP and QP from fre-quency response simulations or measurements.

    K is the gain at low frequencies (f

    HPKK---------- 1= , QP 1 2

    HPKK---------- QP 1

    14QP

    2----------= , QP 1 2>

    %ovrsht 0%= , QP 1 2

    %ovrsht 100%( )e 4QP

    2 1( )= , QP 1 2>

    tr 0.35 f3dB

    Part fGBP(Hz)f2P(Hz)

    SR(V/s)

    RO()

    ISC(mA)

    MCP6041 14k 45k 0.0030 37k 21TC1034 (Note 1)

    60k 1.1M 0.035 15k 8

    MCP6141 100k 55k 0.024 28k 21MCP606 155k 620k 0.080 4.2k 17MCP616 190k 1.1M 0.080 5.0k 17MCP6001 1.0M 45M 0.6 780 23MCP6271 2.0M 4.5M 0.9 360 25MCP601 2.8M 20M 2.3 680 20MCP6281 5.0M 14M 2.5 170 25MCP6291 10M 27M 7.0 105 25MCP6021 10M 124M 7.0 110 30

    Note 1: These parameters also apply to theTC1026, TC1029, TC1030 and TC1035.

    2: f2P can be estimated from the Open-LoopGain plots in the data sheet. Estimate thefrequency (f-135) where the Open-LoopPhase is -135 (i.e., the phase margin is45). Adjust for the typical capacitive loadused in the measurements (CLtyp):

    CLtyp 2f 135 ROCLtyp( )atan

    f2P f 135 45 CLtyp( )tan , CLtyp 40

    f2P 12f 135 , CLtyp 40> 2003 Microchip Technology Inc. DS00884A-page 7

  • AN884

    APPENDIX C: REFERENCES[1] Bonnie Baker, Operational Amplifier AC

    Specifications and Applications, Microchip Application Note 723 (DS00723).

    [2] Charles Phillips and H. Troy Nagle, Digital Control System Analysis and Design, 2nd ed., Prentice Hall, 1990, pp 192-3.

    [3] Adel Sedra and Kenneth Smith, Microelectronic Circuits, 3rd ed., Saunders College Publishing, 1991, Chapter 8.

    [4] Benjamin Kuo, Automatic Control Systems, 5th ed., Prentice Hall, 1987.DS00884A-page 8 2003 Microchip Technology Inc.

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    There are dishonest and possibly illegal methods used to breach the code protection feature. All of these methods, to our er ou of in

    rned

    er cae.

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    Microchip received QS-9000 quality system certification for its worldwide headquarters, design and wafer fabrication facilities in Chandler and Tempe, Arizona in July 1999 and Mountain View, California in March 2002. The Companys quality system processes and procedures are QS-9000 compliant for its PICmicro 8-bit MCUs, KEELOQ code hopping devices, Serial EEPROMs, microperipherals, non-volatile memory and analog products. In addition, Microchips quality system for the design and manufacture of development systems is ISO 9001 certified.

  • DS00884A-page 10 2003 Microchip Technology Inc.

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