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LM5164: LM5164: Is there a waveform issue with the output of LM5164 after turning off the input power?

ling li
Prodigy 40 points
Part Number: LM5164
Other Parts Discussed in Thread: LM5161

Tool/software:

As shown in Fig

It is planned to design a power supply, the input voltage is: 9V~40V support (maximum AC28V input), and the output is: 8V@0.4A.

The 220uH inductor is designed to allow the LM5164 to operate in CCM mode at the time of the output 8V@40mA.

9V-40V is ideally operated in CCM mode, and the output ripple is also within the design range.

However, when the input input is turned off, the output waveform is detected as shown in the following figure, the yellow waveform is the output, and when the input voltage is turned off (green), the 8V output will be glitched when the output supply voltage drops (the load is 20ohm false load)

Figure 1 shows the input DC40V (AC28V) waveform with the input voltage test turned off, and the output voltage waveform is yellow..

（The green waveform is the input voltage, please ignore the fluctuating DC voltage after rectification filtering）

Figure 3 As shown below, when the input voltage is turned off at DC24V, there is a spike in the output voltage waveform. (Output waveform is yellow: DC8V)

Figure 4 As shown below, when the input voltage is turned off at 12 VDC, there are multiple spikes in the output voltage waveform. (Output waveform is yellow: DC8V)

Figure 5 As shown below, when the input voltage is turned off at DC9V, there are multiple spikes in the output voltage waveform. (Output waveform is yellow: DC8V)

I want to know:

1. How to eliminate this pulse during the output drop process;
2. What is the reason for this situation?

Thank you so much..

  • 0
    TI__Mastermind 39790 points

    Hello Ling,

    It looks like the device is not being disabled correctly and chattering at the UVLO of the internal VCC regulator.  Suggest design Enable divider resistor to turn off just below the 9V threshold when VIN is falling.  Please use Equations 13 and 14 to set resistor divider, thanks.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    Thank you very much for your reply，As you suggested, I changed the R2 to 200k and set it to shut down below 9V。

    You look at the picture and it doesn't seem to solve the problem.

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    It looks like the Enable pin is chattering, it is turning on and off at it is passing the threshold.  you can see the green waveform bounce as it passes the threshold.  the specification of the EN pin is spec'd in the datasheet.  if the Enable voltage exceeds the turn on limit as its turning off, the device will turn back on again.  you only have a band limit voltage of 0.6V at VIN, if this is true, you will need an active pull down to control the turn off to prevent it turning back on again.

    Hope this helps,

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    As shown in the figure

    lm5164.pdf

    page 22   F 7-9

    The output voltage waveform doesn't seem to be the same as the situation I encountered.

    Do not know why?

    1、Is this a problem with the chip itself?

    2、Can I ignore it?

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    The reason I think you are having an issues is due to the supply.  The issues is not related to the IC.  You can confirm this by measuring the EN pin voltage turning the on/off transitions to confirm that the device is doing what it is supposed to do when the chatter at the EN pin exceed the hysteresis thresholds, hope this helps and makes sense.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    As shown in the figure

    Purple is the voltage of the EN_pin.

    Green is the input voltage.

    yellow is the output voltage.

    Just as you have judged, this phenomenon may be caused by the jitter of the EN_pin. But how to eliminate this jitter? Especially how to eliminate the jitter during the process of the input voltage decreasing?  

    Thank you so much..

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    I rechecked your schematic, looks like the input LC filter is not correctly Damped.  please try adding damping cap in parallel with the CIN.  Rd = ~1ohm and Cd = 22uF

    If this does not help, suggest a diode from EN pin to VIN, this may help.

    Thanks.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    hi，

    1、“ looks like the input LC filter is not correctly Damped.  please try adding damping cap in parallel with the CIN.  Rd = ~1ohm and Cd = 22uF”，I don't quite understand what that means? How should I do that?
    2. I removed the L2 inductor, and the test did not solve.
    3. According to your suggestion, as shown in the figure, there is no solution to the .

    Thank you so much..

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    The only thing I can suggest that will force this to work is placing an active pull down on the EN pin.  the issue is with your supply as the power is removed the voltage rises turning on the device again.  this means you have a "soft" supply.  

    It looks like you have high impedance between the input power source and the VIN of the Device, when current stops flowing the voltage at VIN of the device and EN of the device increase and turn the device back on.  if you can't reduce the inline impedance, you will have to actively pull the EN pin low once you detect the input voltage is at a point you wish to disable the device.

    Hope this helps,

    david.

  • 0 in reply to David Baba
    Prodigy 40 points

    Thank you very much for your patient reply.

    According to your suggestion, I added a switch S1 at the input terminal of the following circuit to turn off the input power supply. The test waveforms are as follows:

    Use S1 to turn off the power input.

    Just as you have said, it may be a compatibility issue between my experimental power supply and the LM5164 test board.

    This circuit may need to be adapted to multiple types of power adapters.

    1、“”“The only thing I can suggest that will force this to work is placing an active pull down on the EN pin” 

    Are there any recommended reference circuits?

    2、“It looks like you have high impedance between the input power source and the VIN of the Device, when current stops flowing the voltage at VIN of the device and EN of the device increase and turn the device back on.  if you can't reduce the inline impedance”

    How should I reduce the inline impedance?

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    it could be the supply itself, best to check with the system supply that would be actually used in application, this would be my first recommendation.

    Hope this helps.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    Thank you very much for your patient reply.

    I'm not considering adding a circuit to control the EN_PIN for the moment.

    As shown in the figure, after removing the rectifier bridge today and testing the output waveform, the following figure will not appear. Actually, I want to know why there is a waveform at the place marked as A by me (the voltage goes up). Does it have anything to do with the rectifier bridge?

     I replaced it with a laboratory power supply, but the situation remained the same.I changed the rectifier bridge to a half-wave rectifier, and the situation was the same.My application is AC28/50Hz, and both DC9 - DC40 need to be supported. So the rectifier circuit should not be removed.

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    When you have a current through a diode there is a voltage drop, when the current reduces, the voltage drop reduces.  follow the IV curve of the diode. 

    Same with a resistance, if you have a resistor and current flows through resistor there will be a voltage drop when the current stops the voltage drop will reduce.  this is basic ohms law.  you need to actively pull down EN pin for this device in your given circuit unfortunately.

    Sorry.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    hi David，

    Thank you very much for your patient reply.

    Sorry, I overlooked the IV curve of the diode.However, I am thinking that if I do not actively control EN and use a diode to achieve the anti-reverse connection function, the problem I encountered will definitely occur?I wonder if the EN_PIN of LM5164 should use a hysteresis comparator? I'm not sure.

    Are there any other chip solutions recommended? 

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    Yes, the EN has a comparator with hysteresis, but the set hysteresis is not enough to overcome your on/off variation.  The LM5161 has programable hysteresis to suit your needs, please consider this for your application.

    Thanks,

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    hi David，

    Thank you very much for your patient reply.

    My application can easily trigger EN. I tried to set the band limit voltage to be greater than VF (according to the I-V curve). The situation I encountered did not occur again, but the input range of my design is greatly reduced.

    I found a problem that the  band limit voltage of the hysteresis comparator inside the EN_PIN of LM5164 is variable and not fixed, as shown in the following figure.

  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    Sorry for the misunderstanding, you are restricted by the limits of the equations below.

    you can adjust the values of the Ton and Toff off values, but you cannot adjust the hysteresis voltage, this is what is restricting your ability to overcome the issue, I believe.

    Hope this makes sense.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    Hello David.

    Thank you very much. I agree with your opinion.

    I calculated it according to this document.
    6330.LM5163, LM5164 Quickstart Calculator rev3.xlsm
    1、
    Ruv1=1Mohm,Ruv2=200kohm.
    Vin(on)=1.5*(1+1M/200k)=9V
    Vin(off)=1.4*(1+1M/200k)=8.4V
    hysteresis voltage: Vh =  9V-8.4V = 0.6V 
    2、
    Ruv1=1Mohm,Ruv2=80.6kohm.
    Vin(on)=1.5*(1+1M/80.6k)=20.11V
    Vin(off)=1.4*(1+1M/80.6k)=18.77V
     hysteresis voltage:  Vh =  20.11V-18.77V= 1.34V 
    Is there any problem with the above calculation?
  • 0 in reply to ling li
    TI__Mastermind 39790 points

    Hello Ling,

    No, no issues with your calculations, if you can live with the 1.34V hysteresis this may help you situation.  Hope this helps.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    Hello David.

    Thank you very much. I agree with your opinion. If the input is less than 20 volts, it cannot meet the requirements. However, I am still very grateful for your tireless help. I have decided to change to another chip solution. Thank you very much.