{"id":397678,"date":"2024-06-29T13:36:48","date_gmt":"2024-06-29T13:36:48","guid":{"rendered":"http:\/\/savepearlharbor.com\/?p=397678"},"modified":"-0001-11-30T00:00:00","modified_gmt":"-0001-11-29T21:00:00","slug":"","status":"publish","type":"post","link":"https:\/\/savepearlharbor.com\/?p=397678","title":{"rendered":"Compensation for Error Caused by Limited Gain-Bandwidth of Operational Amplifiers in Low-pass Filters<\/span>"},"content":{"rendered":"
<\/div>\n
\n
\n
\n
\n
\"Amateur<\/div>\n

An operational amplifier has the internal compensation circuit for stability which limits its working bandwidth. Frequency response of the compensated Op Amp has slope of \u22126\u00a0dB\/octave or \u221220\u00a0dB\/decade. Unity gain frequency defines the bandwidth where the Op Amp is able to amplify a signal. If we multiply the gain and frequency at any point, the result is the same, allowing us to use this parameter to select the appropriate Op Amp. It is called Gain-Bandwidth Product, GBW or GBP. The limited open-loop gain introduces a closed-loop gain and phase error.<\/p>\n

But we want to optimize our circuits, right?
<\/a> <\/p>\n

\n\n\n\n
\"'Closed-loop<\/td>\n<\/tr>\n
Closed-loop gain of the Non-Inverting and Inverting Amplifier<\/td>\n<\/tr>\n<\/table>\n<\/div>\n

The closed-loop gain of the non-inverting amplifier is:<\/p>\n

        A G = \u2014\u2014\u2014\u2014\u2014\u2014\u2014     1 + A \u03b2 <\/pre>\n
  and for the inverting amplifier:<\/p>\n
       A (1 \u2212 \u03b2) G = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014        1 + A \u03b2 <\/pre>\n
  where:<\/p>\n
  A is the open-loop gain;
  \u03b2 is the feedback fraction.<\/p>\n
  Let\u2019s calculate, for example, an inverting amplifier with desired gain of 1 when A = 100<\/nobr>, \u03b2 = R1 \/ (R1+R2) = 0.5<\/nobr>:<\/p>\n
      A (1 \u2212 \u03b2)     100 \u00d7 (1 \u2212 0.5)     50 G= \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2212 \u2014\u2014 \u2248 \u22120.98       1 + A \u03b2       1 + 100 \u00d7 0.5      51 <\/pre>\n
  and A=1:<\/p>\n
      A (1 \u2212 \u03b2)     1 \u00d7 (1 \u2212 0.5) G= \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 \u22120.333       1 + A \u03b2       1 + 1 \u00d7 0.5 <\/pre>\n
  But to make a real Op Amp stable, A is frequency dependent and there is a phase shift, how you can see in the picture below.  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
Frequency response of the compensated Op Amp without a feedback network<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Gain-Bandwidth Product (GBW) = A \u00d7 F<\/nobr> is a constant, and the greater the GBW, the faster and expensive the Op Amp. Of course, A cannot be infinity, so we see a shelf at low frequencies due to a finite gain.  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
The inverting amplifier with ideal and compensated Op Amp.<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The figure shows the difference between the ideal and compensated Op Amp with GBW = 1\u00a0MHz. You can see that the cyan line of the compensated Op Amp is always below the yellow line showing its open-loop gain and there must be some margin.<\/p>\n
  However, the simulator shows that gain at 1\u00a0\u041c\u0413\u0446 is not -9.55\u00a0dB, but about -7\u00a0dB due to a phase shift at the output.<\/p>\n
  The closed-loop gain error versus gain margin for the non-inverting amplifier looks likes that:  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Closed-loop gain error vs Gain margin of the non-inverting amplifier.<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  <\/p>\n
Error compensation for 2-nd order low-pass filter<\/h2>\n
  This equation is usually used to compute the GBW of an Op Amp:<\/p>\n
GBW(Hz) = 100 \u00d7 Q \u00d7 G \u00d7 F3<\/pre>\n
  where:<\/p>\n
  Q is the quality factor of the filter;
  G is the specified gain;
  F3 is the cutoff frequency at -3\u00a0dB;
  100 is the gain margin.<\/p>\n
  Why? A low-pass filter with Q > 0.707<\/nobr> has a peak and we need to multiply the gain with this peak value to account it. The peak value is:<\/p>\n
               Q \u03b2 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 \u2248 Q     sqrt(1 \u2212 1 \/ (4 Q^2)) <\/pre>\n
  The most of used filters are low-pass filters for ADCs and DACs. So even for an LPF with 150\u00a0kHz an Op Amp with GBW of 15\u00a0MHz is desirable.<\/p>\n
  Texas Instruments\u2019 engineers shared a method to reduce the requirement in [1][2].
  Let\u2019s try to figure out how to use their equations in practice.<\/p>\n
Multiple FeedBack LPF<\/h3>\n
  Let\u2019s compute a 2-nd order LPF with bandwidth of 150\u00a0kHz, gain of 1 and quality factor of 0.707.  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
2-nd order MFB LPF<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The required GBW is:<\/p>\n
GBW(Hz) = 100 \u00d7 Q \u00d7 G \u00d7 F3 = 100 \u00d7 0.707 \u00d7 1 \u00d7 150 kHz \u2248 10.5 MHz<\/pre>\n
  Let\u2019s try to use an Op Amp with GBW of 1\u00a0MHz.<\/p>\n
  Add R4 in series with C2:<\/p>\n
             1                   1 R4 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 2122 \u2248 2.1k (E96)      2 \u00d7 \u03c0 \u00d7 GBW \u00d7 C2   2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 75pF <\/pre>\n
  Change R3:<\/p>\n
R3' = R3 \u2212 R4 = 4990 \u2212 2100 = 2868 \u2248 2.87k (E96)<\/pre>\n
  And finally enter the values into a simulator:  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
Error compensation for 2-nd order MFB LPF<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The top circuit and green color on the charts: the ideal amplifier.
  The middle circuit and cyan color on the charts: the Op Amp with GBW of 1\u00a0MHz.
  The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1\u00a0MHz and with the error compensation.<\/p>\n
  When R3 is too low, the R3\u2019 value may be negative. In such cases you should recalculate the filter with a greater R3 value.<\/p>\n
Sallen-Key LPF<\/h3>\n
  Let\u2019s compute a 2-nd order LPF with bandwidth of 150\u00a0kHz, gain of 1 and quality factor of 0.707.  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
2-nd order Sallen-Key LPF<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The required GBW is:<\/p>\n
GBW(Hz) = 100 \u00d7 Q \u00d7 G \u00d7 F3 = 100 \u00d7 0.707 \u00d7 1 \u00d7 150kHz \u2248 10.5 MHz<\/pre>\n
  Let\u2019s try to use an Op Amp with GBW of 1\u00a0MHz.<\/p>\n
  Add R5 in series with C1:<\/p>\n
             R3 + R4                  \u221e + 0                      1 R5 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 1061 \u2248 1.07k (E96)      2 \u00d7 \u03c0 \u00d7 GBW \u00d7 C1 \u00d7 R3   2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 150pF \u00d7 \u221e   2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 150pF <\/pre>\n
  Change R2:<\/p>\n
R2' = R2 \u2212 R5 = 4990 \u2212 1070 = 3920 = 3.92k (E96)<\/pre>\n
  And finally enter the values into a simulator:  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
Error compensation for 2-nd order Sallen-Key LPF<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The top circuit and green color on the charts: the ideal amplifier.
  The middle circuit and cyan color on the charts: the Op Amp with GBW of 1\u00a0MHz.
  The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1\u00a0MHz and with the error compensation.<\/p>\n
Error compensation for Type II compensation network with Op Amp<\/h2>\n
  Let\u2019s try to apply the same method to the Type II compensation network with Op Amp used in switching-mode power supplies.<\/p>\n
  Consider a Type II circuit with parameters: the zero at 2\u00a0kHz, the pole at\u00a0300 kHz, the middle gain is 0\u00a0dB.  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Type 2 compensation with Op Amp<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  Conservative calculation of the required GBW:<\/p>\n
GBW(Hz) = M \u00d7 Fpole \u00d7 Gfp = 100 \u00d7 300kHz \u00d7 0.707 \u2248 21 MHz<\/pre>\n
  where:<\/p>\n
  M=100 is the gain margin of the Op Amp at a frequency in times;
  Fpole is the pole frequency in Hz;
  Gfp is the gain at the pole frequency in times.<\/p>\n
  Christophe Basso in [4] gives another calculation:<\/p>\n
GBW(Hz) = M \u00d7 (20 Fcross) \u00d7 Gfc = 20 \u00d7 20 \u00d7 5kHz \u00d7 1 \u2248 2 MHz<\/pre>\n
  where:<\/p>\n
  M=20 is the gain margin of the Op Amp at a frequency in times;
  20 Fcross is the frequency with maximum phase boost in Hz with the factor to account the pole position;
  Gfc is the gain at the Fcross frequency in times.<\/p>\n
  The transfer function of the circuit is:<\/p>\n
                          (C1 R1 s + 1) H(s) = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          (s Rfb1 (C1 + C2))(s C1 C2 R1 \/ (C1 + C2) + 1) <\/pre>\n
  Add R2 in series with C2. The new transfer function is:<\/p>\n
                       (1 + C1 R1 s) (1 + C2 R2 s) H(s) = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014          (s Rfb1 (C1 + C2)) (s C2 C1 (R2 + R1) \/ (C1 + C2) + 1) <\/pre>\n
  Change the C2 value and find the R2 value:<\/p>\n
                   1 C2' = C2 \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            2 \u00d7 \u03c0 \u00d7 GBW \u00d7 R1 <\/pre>\n
  <\/p>\n
             1 R2 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014      2 \u00d7 \u03c0 \u00d7 GBW \u00d7 C2' <\/pre>\n
  Now try to use an Op Amp with GBW = 1 MHz<\/nobr> in this circuit.<\/p>\n
  Recalculate the circuit values using the equations above and see the result.<\/p>\n
  Adjust the C2 value:<\/p>\n
                      1 C2' = 56pF \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 40pF \u2248 39pF (E24)              2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 10k <\/pre>\n
  The R2 value:<\/p>\n
              1 R2 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 4k \u2248 3.9k (E24)      2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 39pF <\/pre>\n
  And finally enter the values into a simulator:  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
Error compensation for Type 2 compensation network with Op Amp<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The top circuit and green color on the charts: the ideal amplifier.
  The middle circuit and cyan color on the charts: the Op Amp with GBW of 1\u00a0MHz.
  The bottom circuit and yellow color on the charts: the Op Amp with GBW of 1\u00a0MHz and with the error compensation.<\/p>\n
Error compensation for Type II compensation network with Optocoupler without Fast Lane<\/h2>\n
  There are 2 variants of the circuit: with an Op Amp and a shunt regulator like TL431.
  We will use the same parameters: the zero at 2\u00a0kHz, the pole at 300\u00a0kHz, the middle gain is 0\u00a0dB, the current transfer ratio of an optocoupler is 1.  <\/p>\n
\n\n\n\n
\"'Circuit<\/td>\n<\/tr>\n
Type 2 compensation with Optocoupler without Fast Lane<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The transfer function of both circuits is (C1=Cz, R1=Rz):<\/p>\n
          CTR Rp       (1 + C1 R1) s H(s) = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            Rd   (C1 Rfb1 s) (1 + Cp Rp s) <\/pre>\n
  where CTR is the current transfer ratio of the optocoupler.<\/p>\n
  Add Rc in series with Cp. The new transfer function is:<\/p>\n
          CTR Rp   (1 + C1 R1 s) (1 + Cp CRc s) H(s) = \u2212 \u2014\u2014\u2014\u2014\u2014\u2014 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            Rd   (C1 Rfb1 s) (1 + Cp (Rp + Rc) s) <\/pre>\n
  Change the C2 value and find the Rc value:<\/p>\n
                   1 Cp' = Cp \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014            2 \u00d7 \u03c0 \u00d7 GBW \u00d7 Rp <\/pre>\n
  <\/p>\n
             1 Rc = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014      2 \u00d7 \u03c0 \u00d7 GBW \u00d7 Cp' <\/pre>\n
  Try to use an Op Amp with GBW = 1 MHz<\/nobr> in this circuit.
  Recalculate the circuit values using the equations above and see the result.<\/p>\n
  Adjust the Cp value:<\/p>\n
                      1 Cp' = 51pF \u2212 \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 35.1pF \u2248 36pF (E24)              2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 10k <\/pre>\n
  The Rc value:<\/p>\n
              1 Rc = \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 4.421k \u2248 4.42k (E96)      2 \u00d7 \u03c0 \u00d7 1MHz \u00d7 36pF <\/pre>\n
  The extra DC gain is:<\/p>\n
     CTR Rp   1 \u00d7 10k A = \u2014\u2014\u2014\u2014\u2014\u2014 = \u2014\u2014\u2014\u2014\u2014\u2014\u2014 = 1       Rd       10k <\/pre>\n
  And finally enter the values into a simulator:  <\/p>\n
\n\n\n\n
\"'idealCircuit',<\/td>\n<\/tr>\n
Error compensation for Type 2 compensation network with Optocoupler without Fast Lane<\/td>\n<\/tr>\n<\/table>\n<\/div>\n
  The top circuit and green color on the charts: the ideal amplifier.
  The middle circuit and yellow color on the charts: the Op Amp with GBW of 1\u00a0MHz.
  The bottom circuit and cyan color on the charts: the Op Amp with GBW of 1\u00a0MHz and with the error compensation.
  Do not forget that a real optocoupler has a parasitic capacitance and its value should be taken into account.<\/p>\n
Conclusion<\/h2>\n
  The described methods help to reduce the requirement to the Gain-Bandwidth of a used Op Amp and cost of circuits.<\/p>\n
  For filters, they work well with commonly used quality factors below 1.<\/p>\n
References<\/h2>\n
  <\/p>\n
    \n
  1. TI, Vito Shen, Thomas Kuehl, \u00abCompensation Methodology for Error in Multiple-Feedback Low-Pass Filter, Caused by Limited Gain-Bandwidth of Operational Amplifiers\u00bb<\/a><\/li>\n
  2. TI, Vito Shen, Thomas Kuehl, \u00abCompensation Methodology for Error in Sallen-Key Low-Pass Filter, Caused by Limited Gain-Bandwidth of Operational Amplifiers\u00bb<\/a><\/li>\n
  3. Christophe Basso, \u00abUnderstanding Op Amp Dynamic Response In A Type-2 Compensator (Part 1): The Open-Loop Gain\u00bb<\/a><\/li>\n
  4. Christophe Basso, \u00abUnderstanding Op Amp Dynamic Response In A Type-2 Compensator (Part 2): The Two Poles\u00bb<\/a><\/li>\n
  5. Christophe Basso, \u201cThe TL431 in Switch-Mode Power Supplies loops\u201d<\/a><\/li>\n
  6. \u00abidealCircuit\u00bb, simulator, Windows<\/a><\/li>\n
  7. \u00abCircuit Calculator\u00bb, an electronics design tool for Android<\/a><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
    <\/div>\n

     \u0441\u0441\u044b\u043b\u043a\u0430 \u043d\u0430 \u043e\u0440\u0438\u0433\u0438\u043d\u0430\u043b \u0441\u0442\u0430\u0442\u044c\u0438      https:\/\/habr.com\/ru\/articles\/521478\/<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"
    <\/div>\n
    \n
    \n
    \n
    \n
    \"Amateur<\/div>\n
      An operational amplifier has the internal compensation circuit for stability which limits its working bandwidth. Frequency response of the compensated Op Amp has slope of \u22126\u00a0dB\/octave or \u221220\u00a0dB\/decade. Unity gain frequency defines the bandwidth where the Op Amp is able to amplify a signal. If we multiply the gain and frequency at any point, the result is the same, allowing us to use this parameter to select the appropriate Op Amp. It is called Gain-Bandwidth Product, GBW or GBP. The limited open-loop gain introduces a closed-loop gain and phase error.<\/p>\n
      But we want to optimize our circuits, right?  <\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-397678","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts\/397678","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=397678"}],"version-history":[{"count":0,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=\/wp\/v2\/posts\/397678\/revisions"}],"wp:attachment":[{"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=397678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=397678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/savepearlharbor.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=397678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}