I'm working through the excellent text on SMPS design "Switching Power Supplies A to Z", by Sanjaya Maniktala. Right now I'm reading the section on the feedback loop and stabilization. As a first-order rule of thumb to achieve stability, Mr. Maniktala suggests that the loop gain transfer function of the switcher intersect the 0 dB point at approximately 1/6th the switching frequency, with a slope of -1. To achieve this, one needs a compensation network transfer function that has a pole at the origin, and at least two zeros to cancel the double pole of the switcher's equivalent LC output network. A type 2 compensation network consists of a pole at the origin, a pole, and a zero. Supposedly, the zero of the compensation network plus the zero created in the loop transfer function by the output capacitor ESR is used to cancel the double pole created by the equivalent output network. However, in one of the diagrams the author suggests the Type 2 compensation be set up this way: set the compensation network zero at the LC pole, and set the compensation network pole at the approximate position of the ESR zero. I'm not sure I understand how this will work. The pole at zero gives a slope of -1. Plus the LC double pole, gives a slope of -3. The compensation network zero at the LC pole location then gives a -2...but if the ESR zero is then canceled with a pole in the compensation network, how does one obtain the required -1 slope?

# Type 2 compensation network design

Started by ●September 30, 2011

Reply by ●September 30, 20112011-09-30

On Thu, 29 Sep 2011 23:02:37 -0400, Bitrex <bitrex@de.lete.earthlink.net> wrote:>I'm working through the excellent text on SMPS design "Switching Power >Supplies A to Z", by Sanjaya Maniktala. Right now I'm reading the >section on the feedback loop and stabilization. As a first-order rule of >thumb to achieve stability, Mr. Maniktala suggests that the loop gain >transfer function of the switcher intersect the 0 dB point at >approximately 1/6th the switching frequency, with a slope of -1. To >achieve this, one needs a compensation network transfer function that >has a pole at the origin, and at least two zeros to cancel the double >pole of the switcher's equivalent LC output network. > >A type 2 compensation network consists of a pole at the origin, a pole, >and a zero. Supposedly, the zero of the compensation network plus the >zero created in the loop transfer function by the output capacitor ESR >is used to cancel the double pole created by the equivalent output >network. However, in one of the diagrams the author suggests the Type 2 >compensation be set up this way: set the compensation network zero at >the LC pole, and set the compensation network pole at the approximate >position of the ESR zero. > >I'm not sure I understand how this will work. The pole at zero gives a >slope of -1. Plus the LC double pole, gives a slope of -3. The >compensation network zero at the LC pole location then gives a -2...You ultimately want one pole, and that's from the inductor. The compensation network should be flat up there (its pole+zero has maxed out) and the cap has been dominated by its ESR. Nothing left but the L and resistors.>but >if the ESR zero is then canceled with a pole in the compensation >network, how does one obtain the required -1 slope?Don't do that last part. John

Reply by ●September 30, 20112011-09-30

On Sep 30, 4:02=A0pm, Bitrex <bit...@de.lete.earthlink.net> wrote:> I'm working through the excellent text on SMPS design "Switching Power > Supplies A to Z", by Sanjaya Maniktala. Right now I'm reading the > section on the feedback loop and stabilization. As a first-order rule of > thumb to achieve stability, Mr. Maniktala suggests that the loop gain > transfer function of the switcher intersect the 0 dB point at > approximately 1/6th the switching frequency, with a slope of -1. To > achieve this, one needs a compensation network transfer function that > has a pole at the origin, and at least two zeros to cancel the double > pole of the switcher's equivalent LC output network. > > A type 2 compensation network consists of a pole at the origin, a pole, > and a zero. Supposedly, the zero of the compensation network plus the > zero created in the loop transfer function by the output capacitor ESR > is used to cancel the double pole created by the equivalent output > network. However, in one of the diagrams the author suggests the Type 2 > compensation be set up this way: set the compensation network zero at > the LC pole, and set the compensation network pole at the approximate > position of the ESR zero. > > I'm not sure I understand how this will work. The pole at zero gives a > slope of -1. Plus the LC double pole, gives a slope of -3. The > compensation network zero at the LC pole location then gives a -2...but > if the ESR zero is then canceled with a pole in the compensation > network, how does one obtain the required -1 slope?Sounds like bollocks. You need a bode plot and measure the phase margin. It will then become clearer. You can't tell by looking at poles and zeros. Use the frequency domain. Hardy

Reply by ●October 3, 20112011-10-03

On 9/29/2011 11:29 PM, John Larkin wrote:> On Thu, 29 Sep 2011 23:02:37 -0400, Bitrex > <bitrex@de.lete.earthlink.net> wrote: > >> I'm working through the excellent text on SMPS design "Switching Power >> Supplies A to Z", by Sanjaya Maniktala. Right now I'm reading the >> section on the feedback loop and stabilization. As a first-order rule of >> thumb to achieve stability, Mr. Maniktala suggests that the loop gain >> transfer function of the switcher intersect the 0 dB point at >> approximately 1/6th the switching frequency, with a slope of -1. To >> achieve this, one needs a compensation network transfer function that >> has a pole at the origin, and at least two zeros to cancel the double >> pole of the switcher's equivalent LC output network. >> >> A type 2 compensation network consists of a pole at the origin, a pole, >> and a zero. Supposedly, the zero of the compensation network plus the >> zero created in the loop transfer function by the output capacitor ESR >> is used to cancel the double pole created by the equivalent output >> network. However, in one of the diagrams the author suggests the Type 2 >> compensation be set up this way: set the compensation network zero at >> the LC pole, and set the compensation network pole at the approximate >> position of the ESR zero. >> >> I'm not sure I understand how this will work. The pole at zero gives a >> slope of -1. Plus the LC double pole, gives a slope of -3. The >> compensation network zero at the LC pole location then gives a -2... > > You ultimately want one pole, and that's from the inductor. The > compensation network should be flat up there (its pole+zero has maxed > out) and the cap has been dominated by its ESR. Nothing left but the L > and resistors.Hi John, sorry for the delay in my reply. I did some more reading and discovered that the guidelines for the type 2 compensation network in question were for current mode control, not voltage mode control. For a type 2 network under voltage mode control, it looks like one is free to set the second compensation network pole somewhere between the crossover frequency and ten times the crossover frequency, to improve load transient response.>> but >> if the ESR zero is then canceled with a pole in the compensation >> network, how does one obtain the required -1 slope? > > Don't do that last part. > > John >