LM25118 switching frequency problem

Bob Lee62837

Mastermind 7230 points

Other Parts Discussed in Thread: LM25118, LM5118

Dear support member,

My customer use LM25118.

Specifications Condition

Vin=19V(min)、23V(typ)、31V(max)
Vout=23V
Iout=1.6A(typ)、8.57A(max)

fsw=65kHz
L=61.4μH
Cout=470μF×2

When Increasing the outputcurrent from 1.6A, switching frequency become half frequency. (65kHz →32kHz)

Customer tried switching frequency 300kHz.

When switching frequency is 300kHz, switching frequency don't become half frequency.

Why switching frequency half become?
CCM, DCM, transition, relationship?

Customer must Specifications condition switching frequency 65kHz.

Best regard.

over 10 years ago

0 David Baba over 10 years ago

TI__Mastermind 40040 points

Have you used the quick start calculator? Please check your Cramp/Rramp values.

0 Bob Lee62837 over 10 years ago in reply to David Baba

Mastermind 7230 points

Dear David Baba,

Thank you very much for your answer.

I was based on LM25118EVM.
I calculated in Excel.

Cramp：4700pF
Rramp：OPEN(NAND)

Cpole：2200pF
Czero：0.1μF
Rcomp：2.7kΩ

Best Regard.

Bob Lee.

0 David Baba over 10 years ago in reply to Bob Lee62837

TI__Mastermind 40040 points

Hi Bob,
I put in your design parameters and it suggested an Rramp of ~100k and Cramp of ~6.8nF. can you try at least install 100k ohm and recheck?
Thanks.David.

0 Bob Lee62837 over 10 years ago in reply to David Baba

Mastermind 7230 points

Dear David Baba,
Thank you very much.

Although I try.
Could you teach reason Rramp install 100kΩ?

Best regard.

Bob Lee

0 David Baba over 10 years ago in reply to Bob Lee62837

TI__Mastermind 40040 points

Hi Bob,

A 50uA current source into the Cramp capacitor is used as a corrective ramp to the current sense to avoid sub harmonic oscillations which occurs when the duty cycles is greater than 50%. However we typically need additional Slope compensation when you have a higher vin as in your case. Rramp adds additional slope compensation.

I scrutinized this part of the design because you mentioned you were switch at Fsw/2. Sometimes this can be due to Sub harmonic oscillation where the subsequent pulse is missing entirely?

0 Bob Lee62837 over 10 years ago in reply to David Baba

Mastermind 7230 points

Dear David Baba,

Thank you very much for answer.
I will try it.

I will contact the result.

Best regard.
Bob Lee

0 Bob Lee62837 over 10 years ago in reply to Bob Lee62837

Mastermind 7230 points

Dear David Baba,

I tried R7=100kΩ.
Frequency became 65kHz constant. Thank you very much.

May I ask additional question?
Although you taught R7=100kΩ reason a sub-harmonic in the previous e-mail,Could you tell the operation contents with Data sheet 8.2 Function Block Diagram?

Best Regard.
Bob Lee.

0 David Baba over 10 years ago in reply to Bob Lee62837

TI__Mastermind 40040 points

Hi Bob,

The LM5118 provide a slope compensation ramp internal by charging a capcitor (Cramp) with a current proportional to Vin (in the Buck Boost mode). This is shown in the "Ramp Generator" Block. With a higher Vin (<20V) it is advised to increase the current sourced into the Cramp by adding a resistor from Cramp to VCC thereby increasing the Ramp current which increases Slope Voltage (VSL). This voltage (VSL) is added to the reconstructed signal representing the Inductor current.

Hope this helps?

David Baba.

0 Bob Lee62837 over 10 years ago in reply to David Baba

Mastermind 7230 points

Dear David Baba,

Thank you very much for answer.

I understand R7 is role by Ramp current which increases Slope Voltage (VSL).

May I ask additional question?

When frequency 65kHz,
When Increasing the outputcurrent from 1.6A, switching frequency become half frequency. (65kHz →32kHz)

When switching frequency is 300kHz, switching frequency don't become half frequency.

Is R7 add 65kHz only?
300kHz also need add?

What is there a threshold that need to add the R7?

Best regard.
Bob Lee.

0 David Baba over 10 years ago in reply to Bob Lee62837

TI__Mastermind 40040 points

Hello Bob,

When you increase the switching frequency you in effect decrease the Down slope of the inductor current because Vout/L = DiL/DT; DT = 1/Fsw (Switch Frequency). Your slope compensation volage requirement has now been reduced as a result. i.e. you will no longer need a R from Vcc to Ramp pin when Fsw is increased.

The amount of Slope compensation required is proportional to the down slope of the inductor current. In a Buck Boost this is Vout/L. We provide guidence in the datasheet for output voltages greater than 12V you may need additional slope compensation.

If you Fsw is increase, then you may not need additional Slope compensation. The Quick Start calulcator that can be downloaed provide a great starting point for the R Ramp value and when it needs to be used.

0 Bob Lee62837 over 10 years ago in reply to David Baba

Mastermind 7230 points

Dear David Baba,

Thank you very much for answer.
Customer was very pleased.

Best regard.
Bob Lee.

0 Bob Lee62837 over 10 years ago in reply to Bob Lee62837

Mastermind 7230 points

Dear David Baba,
Thank you very much for answer.
May I ask a add question?

Because customer want to check the Excel(snvu065.xlsm) formula,
Could you tell protection code?

Best regard.
Bob Lee.

0 Garrett Roecker over 10 years ago in reply to Bob Lee62837

TI__Mastermind 21720 points

Bob,

What equation are you looking for?

Thanks,

Garrett

0 Bob Lee62837 over 10 years ago in reply to Garrett Roecker

Mastermind 7230 points

Dear Garrett Roecker,

Thank you very much for answer.

It becomes "Bode Plots" sheet of Excel (snvu065.xlsm).

Could you tell equation?

Best regard.
Bob Lee.

0 Garrett Roecker over 10 years ago in reply to Bob Lee62837

TI__Mastermind 21720 points

Bob,

Attached are the equations that we use for the quickstart. If you want more details send me a message and I can explain futher.

Thanks,

Garrett

LM5118_Quickstart_Control_Equations.docx

0 Bob Lee62837 over 10 years ago in reply to Garrett Roecker

Mastermind 7230 points

Dear Garrett Roecker,

Thank you very much for answer.

Also Could you tell "Error Amplifier Configuration" equation?

Best regard.
Bob Lee.

0 Garrett Roecker over 10 years ago in reply to Bob Lee62837

TI__Mastermind 21720 points

Bob,

I've added in the error amp equation to the attached document.

2364.LM5118_Quickstart_Control_Equations.docx

Thanks

Garrett

0 Bob Lee62837 over 10 years ago in reply to Garrett Roecker

Mastermind 7230 points

Dear Garrett Roecker,

Thank you very much for answer.

I will try calculation.

Best regard.
Bob Lee.

0 John Kurian over 10 years ago in reply to David Baba

Prodigy 115 points

Hi David,

I am facing the same problem Bob faced. I used the TI Webench for designing a converter with inputs ranging from 13.5V to 18V and the output 14.5V(current 4A) and the operating frequency 100KHz(set the efficiency as high). These are the values :
Rramp = 309K
Cramp = 1.5 nF
Ccomp2 = 3.9nF
Rcomp = 41.2K
Ccomp = 470pF
Rt = 61.9K
Rfbb = 1K
Rfbt = 10.7K

The voltage is stable when it there is no load connected, but it drops when the load starts drawing current(more than 0.24 A). The frequency at the Ramp terminal also changes(to around 30Khz). Even if we reduce the current, the voltage won't go back to the stable value. We need to remove the load to make the voltage to go back to the stable value. In an effort to find the problem, I removed the current sensing connections and connected them to the ground. It did not make any difference.

Best regards,
Rohan

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

What inductance value are you using? Output Capacitance? Does this happen at all input voltages?

Here is a link to the quick start calculator. When I plugged in your values the circuit looks pretty unstable.

www.ti.com/.../snvu065

-Garrett

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Hi Garrett,

Thanks for the reply.

Inductance is 40uH (10A). Output capacitance is 550uF. This happens only when the input voltage is lowered. At higher input voltages it is capable of giving a bit higher current, around 1A.

Best Regards,

Rohan John

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

I think stability is your issue which is why you see regulation at light loads. Please try these values:
Rramp = DNP
Cramp = 1500nF
Comp2 = 2.2nF
Rcomp = 13.2kOhm
Ccomp = 68nF
Current sense resistor: 15mOhm (value calculated since not provided)

I got these values using the quick start.

-Garrett

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Garrett,

I apolofize for the delay in rsponding. The problem was with the layout of the board. Got hold of a TI's evaluation board. Now it is working fine. I have one more doubt though, Will it be possible to control the output voltage through a digital potentiometer? Will it cause any instability(due to pin capacitance of the digital pot-comes to around 100pF and frequency is 340kHz). I gave it a try but I am getting low regulation and its picking up all the enviornmental noise.

Best Regards,

Rohan

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

The digital pot on the low side feedback resistor can cause noise due to the capacitance. One way to check this is take a 100pF and tie it to GND when you are still using a fixed capacitor. Hopefully this will give you an idea of the operation.

Another way you could possibly do this is to sink/source current to the low side resistor. This would make it possible to change the output voltage while using fixed feedback resistors.

-Garrett

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Garret,

By saying current/sink source, did you mean connecting the digital pot parallel to the low-side resistor? I tried it like that and its working. Up till now, I had used a regulated power supply for powering the device. But when I connect it to a solar panel(12V 40W) and load it, it becomes unstable and output voltage drops.

Best Regards,

Rohan

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

Yes you can do it that way. Are you saying that the digital pot is being supplied by the solar panel?

-Garrett

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Garrett,

No, the entire converter. The digital pot and the associated microcontroller is being powered by the converter's output. I plan to use the converter as an mppt battery charger. It works when there is good sunlight, but during low power levels, the output voltage does not reach a stable value even when there is no load(other than the small control circuit which takes about 0.03A). I checked the EN pin and its being pulled low periodically during this time. A small noise also comes from the inductor.

Rohan

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Garrett,

What are the conditions which will cause the device to pull down the EN pin periodically(pulled down at a frequency of 70 Hz, normal operating frequency 340kHz).

Rohan

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

How is the enable pin connected? Is it connected to VIN or are you controlling the EN signal?

-Garrett

0 John Kurian over 10 years ago in reply to Garrett Roecker

Prodigy 115 points

Garrett,

The EN pin is connected to the input voltage VIN through 1 Mohm resistor.

Rohan

0 Garrett Roecker over 10 years ago in reply to John Kurian

TI__Mastermind 21720 points

Rohan,

I don’t suspect that the EN pin is your issue. From everything that you have described it sounds like the part is entering shutdown due to low input voltage.
Another place to check is to make sure you are not current limiting. If you current limit the part will pull UVLO low.

-Garrett

Power management

Power management forum

LM25118 switching frequency problem