LM5041: Feedback loop Compensation

Kubendran G1

Intellectual 330 points

Part Number: LM5041
Other Parts Discussed in Thread: LM6132

Hi Every one:

I need to clarify LM5041 application circuit compensation design.

Application circuit error opamp pole and zero placed below freq.

Error op amp zero 1 placed by R35,C36 @ 1.326KHz

Error op amp zero 2 placed by R32,C40 @ 22.9KHz

Error op amp pole 1 placed by R42,C40 @96.7KHz

Error op amp pole 2 placed by R35,C35 @279KHz

My question 1) Op amp first zero placed in plant double pole L=60uH, Cout: 932uH F=673Hz?

2)Op amp second zero placed freq where should get?

3) I dint find Cout ESR value. Pole 1 placed in plant zero? I think Pole 2 placed almost in buck switching freq.LM5041..docx

3) How to compensate opto pole by error opamp in this application?

4)Kindly share any document available for compensation design for this application. (Opto coupler feedback with external opamp)

5) R11 and C11 form the opto pole?

Kindly reply as soon as possible.

Regards,

Kubendran

over 5 years ago

0 Kubendran G1 over 5 years ago

Intellectual 330 points

Hello TI Team,

I am expecting your valuable reply.

Regards,

Kubendran

0 Kubendran G1 over 5 years ago in reply to Kubendran G1

Intellectual 330 points

Hello TI Team,

We are facing loop compensation and buck switching node voltage oscillation issue with respect to LM 5041 application circuit design. Always i seen LM5041 related query response very delayed.

Kindly reply previous query.

Regards,

Kubendran

0 Peter Meaney over 5 years ago in reply to Kubendran G1

TI__Mastermind 31580 points

Hi Kubendran,

Sorry for the delay, the office that supports this controller was on public holiday, I have assigned an AE to this post.

Regards

Peter

0 Kubendran G1 over 5 years ago in reply to Peter Meaney

Intellectual 330 points

Hi Peter,

When will i expect the reply.

Regards,

Kubendran

0 Kubendran G1 over 5 years ago in reply to Kubendran G1

Intellectual 330 points

Hello TI team,

No reply?

Regards,

Kubendran

0 Peter Meaney over 5 years ago in reply to Kubendran G1

TI__Mastermind 31580 points

Hi Kubendran,

My mistake, there was a mistake in the email address for the AE in the system and this post did not appear on his list. I am checking if the AE assigned can support if not I will assign a new AE to help.

I should have an update tomorrow.

Regards

Peter

0 Teng Feng over 5 years ago in reply to Peter Meaney

TI__Genius 13660 points

Hi Kubendran,

Unfortunate, I didn't find any document for the compensation design of this application.

In this application, the designer used much complex two stage network with two LM6132 op amps and a LM4041 reference. I guess this circuit demonstrates the flexibility that can be used in electronic circuit design. I think it can be designed simpler with only one amplifier LM6132 and one reference LM4041.

Since in this application, there is no internal error amplifier used, so it relies on the designer to add the error amplifier. It is targeted at isolated designs the error amp is usually on the secondary side driving an opto-coupler through the isolation barrier to the control pin.

The schematic in page 14 of the LM5041 datasheet shows such a scheme. U4A is the error amplifier. C35, R35, C36, R42, C40 and R32 make up the type III compensation network.

Regards,

Teng

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

Hi Sir,

Please read carefully my previous query. What you are said error op amp R and C value corresponding pole and zero frequency already calculated. My question how you are placed compensator pole,zero and opto coupler pole to the plant response.

If i consider plant is double pole that value not matched with actually what you are used in compensator pole and zero. Actually current mode plant having one pole and one zero. In this way also checked compensator value did not matched.

Kindly look this one serious and give best solution.I am facing converter output oscillating issue.

Regards,

Kubendran

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubendaran,

Did you made a measurement of the gain, phase or if you have done a load transient test and measured the response of the output voltage waveform to see if the response with well damped or under damped.

I don't think we can get a final value through calculation, actually it is complected and difficult to establish the signal model.

My suggestion is: please adjust the compensation according the measured transient performance.

Regards,

Teng

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

Hi Sir,

I will share the waveform to your mail. Please reply.

Regards,

Kubendran.

0 Kubendran G1 over 5 years ago in reply to Kubendran G1

Intellectual 330 points

Hi Sir,

Measured loop response waveform shared to you.Please reply the mail.

Kindly share how to make external slope compensation to LM 5041 controller IC. May be sub harmonic oscillation causes instability in my circuit.

Regards,

Kubendran

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubendran,

The LM5041 has integrated a slope compensation by buffering the internal oscillator ramp and summing a current ramp generated by the oscillator internally with the current sense signal. If you want to add additional slope compensation, you can increase the source impedance of the current sense signal, i.e. increase the resister connecting to CS pin.

Regards,

Teng

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

HI Sir,

How to calculate required resistance value against Selected buck inductor ?.Kindly provide clear details.

Already failure mode waveform sent to you. Please reply it.

Regards,

Kubendran

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubendran,

Attached two document related the slope compensation of a buck converter for your reference. Once the desired slope compensation ramp is calculated, then we can calculate the needed external resister.

2425.An Intuitive Approach to Nonlinear Slope Compensation.pdf

5857.slua101 Slope compensation.pdf

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

Hi Sir,

Thank you for sharing the useful document. Please find the attached file of slope compensation value.

Kindly share how to calculate required resistor value.

Regards,

KubendranSlope compensation.docx

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubendran,

The internal compensation ramp current is 45 uA and internal impedance (Rint) is 2 kΩ, the slope compensation value can be calculated by Vcom= 45 *(2 + Rext) mV, Rext is the external resister between CS pin and current sensing voltage signal in kΩ.

Regards,

Teng

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

Hi Sir,

Thank you for sharing calculation.

Vcom=45*(2+Rext) Vcom=139mV , from that Rxt=1K Ohm

Please ensure calculated slope amplitude 139mV is correct ? which is posted in previous post.

Regards,

Kubendran

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubenddran, I am calculating slope compensation value for you. will reply you later.

0 Kubendran G1 over 5 years ago in reply to Teng Feng

Intellectual 330 points

Hi Sir,

Thank you for giving good support. Same kind of support i am looking for loop compensation also.Because i checked loop compensation details for cascaded push pull converter in TI seminar note also. Not yet get proper document for this topology. Kindly check, how did you done for evaluation board otherwise share EVAL board loop analysis report.

Regards,

Kubendran

0 Kubendran G1 over 5 years ago in reply to Kubendran G1

Intellectual 330 points

Hi Sir,

Kindly share the calculated value.

Regards,

Kubendran

0 Teng Feng over 5 years ago in reply to Kubendran G1

TI__Genius 13660 points

Hi Kubendran,

To guarantee current loop stability when the duty is above 50%, the slope of the compensation ramp must be greater than one-half of the down slope m2 of the current waveform. For the buck regulator, m2 is equal to Vout * Rs / L, therefore, the amplitude A of the compensating waveform should be chosen such that

A > T * Vout * Rs / L. Your result is right.

Some publications suggest adding half as much ramp as the down-slope of the inductor current. some others recommend much. My answer is — to make sure your current loop won’t oscillate.

Besides, a paper about designing Type III Compensators for DC/DC switching converter for your reference.

http://www.ti.com/lit/an/slva662/slva662.pdf?&ts=1589272510511

Regards,

Teng

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LM5041: Feedback loop Compensation