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LM46002 regulator output voltage drop when connect to TPA3116 circuit

Other Parts Discussed in Thread: LM46002

Hi,

I just used the WEBENCH for a regulator design(from 36-12V, chose suggest chip LM46002) to support my amplifier(TPA3116). And got some problems when combine these two circuits together.

Here are the lists of things I checked which works fine:

1. With no load, the output voltage of regulator is around 12V

2. With load of resistance, output voltage also around 12V

3. Not use regulator, just use 12V directly from constant volatage source. Apply on amplifier, the amplifier part works fine

However,

when I use the output from regulator to support my amplifier circuit, the voltage between regulator output (amplifier input) drop to around 5.5V. 

I am so wondering why this thing happen. Just change the load from resistance to amplifier typical application circuit, why this happen?

Here is the regulator circuit:

Thank you in advance.

  • Hi Molly,

    Do you know how much current that the amplifier consumes when it is operating? You can check this when you connect it to a regular 12V constant voltage supply.
    The maximum current that the LM46002 can support at the load is 2A.

    Thanks
    -Arief
  • Hi Arief,

    When I turn the highest volume, it's around 0.8A. So basically, it's not higher than 2A when the amplifier works.

    Also, I use the constant voltage supply to support my current circuit(regulator+amplifier), the current is basically lower than 10mA, so from my opinion, the amp is not working.

    Thanks,
    Molly
  • Hi,

    Were you able to use an oscilloscope on the SW pin of the LM46002 to see the waveform when it is connected to the amplifier?

    Thanks
    -Arief
  • I got the waveform now, but haven't find a way to attach the image><
  • Here it is, find the way now: 

  • When you reply, on the bottom right corner there is an option to choose rich formatting. When you click on that button, you will have an option to attach the image
  • Can you change the timebase to be 5us per division and see the difference between connected to amplifier and connected to a resistor?
    Also what resistor value do you use

    Thanks
    -Arief
  • Here are the images with amp circuit, even got wield voltage like 1A

  • Here is the image when connecting the resistance, the value I use now is 1k. But you mentioned pin SW right? It's not the final output value

  • Molly,

    The waveform looks correct for 1k resistor. Can you lower the resistance to 24 ohm and recapture the waveform? This will simulate a 0.5A load at the output.

    Also what input voltage do you use?
    I am suspicious of the inductor that you use, do you have the part number of the inductor? It could be saturating

    Thanks
    -Arief
  • I don't have a 24 ohm resistance so I use two 47 ohm resistances in parallel(which should be 23.5 ohm), hope it's fine. Here is the image I got.

    The input voltage I am using is 36V now. The inductor is SRR1240-220M from Bourns Inc.

  • Hi Molly,

    The last waveform does not look correct for normal operation at 0.5A. On the other hand, your inductor seems to be correct.
    Do you have a layout file or maybe a picture of the circuit? There are some pins that are quite sensitive to noise.
    Also can you make sure that the voltage on VCC pin is about 3.3V

    Thanks
    -Arief
  • Hi,

    Do you got the thing working? Let me know if you still need assistance on the LM46002. I can send you an EVM if you like to connect with your amplifier. If you have already make your board, make sure the sensitive trace is short and the FB node is away from the SW node. I can try to visually inspect it for you if you have a picture of it. A good indicator of noise on FB pin is when you start loading more and more current, the voltage will droop slowly as you increase the load. See page 44 on this document
    www.ti.com/.../lm46002.pdf

    Thanks
    -Arief
  • Hi,

    I am still working on this. Get short circuit last night, so rework on it at this moment. I will get you the information once got all the circuit done. Thank you for the waiting.

    Best Regarts,

    Molly

  •  Hi,

    I just got the new one done. I check the voltage of pin VCC, it's 3.26V(around 3.3V). Also, I am not using a layout now, just using the breakout to test. Attached is the figure of breakout layout

  • Hi,

    are you still here~ Here is my BOM list also.

  • Ah yes, i did not see anything wrong with the BOM. On the layout, yes i can see potential issue.

    Move the feedback resistor RFBT, RFBB close to the pin of IC. You can have the cable from VOUT to RFBT long but put the resistor very close to the IC (red breakout board). You might do this one first and see the behavior. I think this will improve regulation.

    Second you need to put the CIN capacitor close to the pin of IC. At least one of them the 1uF. Put it as close as you can between VIN pin and GND pin. I would put it on the red breakout board. Between pin 14 and 15

    Third is the VCC capacitor also need to be put close to the IC, I would move this also on the red breakout board.
    Keep the length of wire from SW pin short to the inductor.

    Thanks
    -Arief
  • I see, will try, give feedback to you in half to an hour
  • One more question, as you mentioned, feedback resistor RFBT, RFBB should be close to the pin of IC. But as I am using the red breakout together with the green breakout. If thinking from the distance from the pin on green board to my resistance, they are not far away. This will also have effect on?
  • Hi,

    Just tried the first one you mentioned, all the situations remains the same, nothing changed
  • Hi,

    The reason i ask for them to be closer to the pin is because the sensitive node is the Feedback pin and if you have long traces it might pick up noise and mess with the load regulation. The FB pin is high impedance pin so its quite sensitive. Normally we make the loop from FBpin to RFBB to GND is tight as mentioned on datasheet to keep the noise coupling small.

    Have you try to put the CIN closer together to the pins and VCC as well? The other loop that is important in buck regulator is the VIN to CIN and to GND pin as mentioned on the datasheet.
    Or can i see your components location so it will give me a better idea because i only see one capacitor from your previous picture

    Thanks
    -Arief
  • Hi,

    I think I solve the problem after change the position of CIN, thank you so much for your time and patience
  • No problem. Just in case you decide to make a PCB, you can follow the sample layout on the datasheet to keep important loop and pins are taken care of

    Thanks
    -Arief
  • Thank you, PCB is the next step. I will try to follow the rules on datasheet and will contact you again if got any other questions
  • Hi,

    Got another problem now. When I connect my amp with the speaker and use the regulator to power them. Looks like the regulator gone immediately. But the current I checked is just around 36mA, should lower than the regulator. The voltage I check now is all 0V from the regulator pin, why this happen?

  • I think the layout still could be the issue. Can i see your latest layout/adjustment that you made to the board?
    Is VCC capacitor is close to the pin?

    Thanks
    -Arief
  • HI,

    Seems like situation get worse when I get all the components on the same board. Got no output (0V) now and trying to solder a new layout to see if it works or not. Here is the image of layout now. Checked all the pins, they are not shorted. Sorry for the interrupt again.

  • Hi Molly,

    I would say you can check the EN pin and what is the voltage on that pin to GND. It needs a minimum 2.2V to operate. Or you can also connected directly to VIN.

    Can you indicate which is the VIN capacitor? Also the boot capacitor to SW node should be located closely. Can you indicate which one is the boot capacitor? This is the capacitor located from SW to BOOT pin

    THanks
    -Arief
  • Hi Arief,

    Thank you so much for your help several days ago. You are so correct that the layout plays an important role in this design. Successfully get the output each time(whether with load or without load). But, another strange thing happen which bothers me. It's like the voltage and current looks good each time when it powers on. But after some time, the current drops to half and then recover then half then recover, the time it remains stable of good drop as the load becomes smaller.

    For the load part, in order to get certificate load to check the regulator working current. I am using 20oum, 0.5w, 4 in serial, and get more pairs in parallel. So each resistance can get around 0.45w. 

    I am trying to explain, please let me know if there is anything not explained specifically

  • Hi Molly, 

    It seems like a current limit issue or stability. Try to double the output capacitor and see if that helps. Also make sure that you are not current limited on the power supply. By the way how much resistance you have in total? Also how did you change the layout

    Thanks

    -Arief

  • Sure. First, for double the capacitor, if I put two 47uF(output capacitor from the schematic) in parallel, is that ok for the double the output capacitor as you mentioned?

    Second, just checked the limitation of current limitation of that channel, it's 2.36A, the maximum of this channel is actually 2.2A.

    Third, as I mentioned, in order to meet the power of each resistance, so I got 4 * 20 ohm in serial, and in order to check current tolerance of the regulator, so I got this kind of set in parallel increasingly. So the number of resistance I am testing now is from 4 to 36 now. When the number came to 36, the time it became unstable become short then I find this problem, after around 10-15 seconds

    Last, I changed the layout as you suggested. Like the last image I send you. Just I little change, moved the Cout on another breakout, others are all on the small red breakout.

    Thanks,

    Molly

  • Yes that should be ok to double the capacitor. Make sure the capacitor is rated for 25V or so.
    So if i understood you correctly your equivalent resistance value is from 4 ohm to 36 ohm? I dont think you can run the 4 ohm on this part because the maximum current is 2A. You should be ok running with 36 ohm

    Thanks
    -Arief
  • 4*20 in serial is 80 in each line, then parallel the 80's. I think the resistance value now is range from 80ohm to 8ohm. Is my calculation wrong?
  • By the way, the capacitor needs to rate for 25V? actually what I have keep using now is 16v. is it possible the volatge tolerence make effect on the results now?
  • Nope your calculation is correct. Before i do not know how many you put in parallel thats all. So your current should be in range.
    The capacitor 16V is ok for 12V output but sometimes the capacitor value is derated based on voltage. I would just add 1 or 2 more 47uF and see if there is any difference.

    Is there any way you can show the behavior? Oscilloscope shot maybe?

    Thanks
    -Arief
  • I see, will try adding capacitor in parallel in first. Check first and if still not work, will try to upload image or short video. Thank you in advance
  • One short small video shows the phenomenon. Once the current changed, cannot read the voltage from multimeter. After add one more capacitor, looks like the time it's stable becomes a little longer, from 20s to 30s?

  • Molly,

    Could i see the SW node on the oscilloscope when it happens from stable to unstable?

    Thanks
    -Arief
  • I remember the output capacitor what the schematic from simple wrench first recommend was 8 * 4.7uF in parallel, will that make huge difference?
  • I dont think that makes a difference. It looks like you have some kind of oscillation.
    I would like to see the oscilloscope probe on
    1) Input voltage
    2) SW node
    3) output voltage
    4) Inductor current if you can

    Thanks
    -Arief
  • This is SW pin. there will be a gap on the waveform once the current drops.

  • Wow i never seen that happen before. Do you have input voltage waveform, and output voltage?
    I am also curious about inductor current if you can have that too
  • Happens on input pin too!!!!! The hight of waveform becomes quite flat when current drops

  • inductor current:

  • output waveform:

  • Hmm the input dissapears for some reason. I am not sure why, how about dialing it down to 24V on the input. I am worry something happen to the input supply
  • I believe that when I do the design, the input voltage I choose is from 28 to 42v, will that be fine for me to drop that one to 24v?
  • Yes that would be ok, as long its higher than 12V

    Thanks
    -Arief
  • Just tested, same thing happen
  • The time for this kind of situation need to happen is much shorter when the load is 9ohm rather than 80ohm. Haven't did long time test yet. But for 80ohm severeal minutes is definitely good. But for 9 ohm, 1 minute is too long