Step-down converter output noise, switch point leakage voltage and transient ring issue

Discussion in 'Power Electronics' started by Benengineer, Jun 17, 2016.

  1. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    I have a Step-dwon converter circuit. My output T3 is a big noise, which is out of spec. Meanwhile, I see it rises up to a certain level, around 3.4V, when the diode is turned on, not totally turned off. The test point is at Pin 2. Please see the following ciruit and output waveforms.

    upload_2016-6-17_8-32-14.png

    upload_2016-6-17_8-35-9.png
    T3 output for noise

    upload_2016-6-17_8-37-11.png
    Pin 2 output

    Please let me know why it rises up after the diode is turned on. How can I reduce the noise if I want to improve the ciruit.

    Thanks,
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Use a Low ESR Cap (C6) at the output and check
     
  3. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    Why
    why isn't the diode turned off totally?
     
  4. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    My input is 24V and output is 3.3V. Why isn't diode turned off totally?
     
  5. benta

    Member

    Dec 7, 2015
    101
    24
    First, the MC34063 is a 1st generation switching regulator and will in certain circumstances not run well. There are newer and MUCH better ICs available.
    Second, your trace on pin2 looks like what I remember from the MC34063, it's quite typical when operating without a load. Try to load the regulator, and you'll see a different picture.
    Third, the output noise: running without load brings the MC34063 to skip cycles and do other funny stuff. But I suspect you have bad ground/neutral routing. You need to keep the high current grounds (D2, C6) and the signal grounds (U1, R3) separate and bring them to a star point at (C5).

    Benta.
     
  6. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    Benta,
    I added the load to the converter. And my ground is at R3. The noise is better, but the switch node output at Pin2 is not improved. The C6 esr is about R = 0.3. Please see the captures I got.
    upload_2016-6-17_16-23-12.png
    The converter output at T3
    upload_2016-6-17_16-24-28.png
    The switch node at Pin 2

    Can I increase the C6 value or C2 ?

    Thanks
     
  7. crutschow

    Expert

    Mar 14, 2008
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    I suspect you have a grounding problem and need to do as benta suggested. If you are sloppy with the ground routing you can get a lot of noise with a switching regulator.
    How is the board laid out? Show us a picture.
     
    Lestraveled likes this.
  8. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    The ground should be fine. I focus on why the switch node, which is diode is turned on in shourt time, it rises up to 4.7V. Where is this voltage coming from? Idealy speaking, this should be ground during the diode on. I will show the pcb layout to you on Monday.
    Thanks
     
  9. crutschow

    Expert

    Mar 14, 2008
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    You are operating in the inductor non-continuous mode, as indicated by the short time the diode is on and the inductor is supplying current to the output. This may mean the diode current is very high.
    What is the output load?
     
    Last edited: Jun 18, 2016
  10. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    I don't know what output load is. I even don't know how to measure it because they are not very professional in this small company. Eevery time when I measure any parameters, I have to solder a lot of wires to test it on PCB board. They don't have specifical test boards to test. Can you give me suggestion to measure the output load?

    So, this is not continuous mode, right?

    Thanks
     
  11. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    By the way, I forgot to tell you that I changed R2 value from 5.1 to 2.21. On PCB board, R2 was 5.1 when I measured it. After I changed, noise becames smaller. Why don't this happend?

    Thanks
     
  12. crutschow

    Expert

    Mar 14, 2008
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    So what is the converter powering?
    You can measure the load by disconnecting the converter from the load, powering the load from an external power supply, and measuring the current.

    R2 controls the peak or short-circuit current, so not sure how it changes the noise unless the inductor is saturating.
    Do you know what the inductor current rating is?
     
  13. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Why doesn't OP show the layout?
     
  14. benta

    Member

    Dec 7, 2015
    101
    24
    Quite honestly, all this is absolutely normal behaviour for the MC34063 and its predecessor uA78S40.
    These ICs will only operate if a certain ripple level is present at the output, that's the way they are designed.
    Your trace shows the "on" pulse (high voltage pulse charging the coil at ~Vin).
    Then the "off" period where the diode turns on ( ~-0.7 V) and coil current decreases.
    The third part of the trace is where coil DC current has dropped to zero and its inductance and parasitic capacitance (winding-to-winding) makes it resonant.
    As the coil at this point carries no current, it will voltage-wise reflect the DC output voltage as maintained on C6.
    If you want to reduce the ripple (what you call noise), you could try a second filter after the VCC point.
    Or better: use a modern controller/regulator. The SimpleSwitchers from TI work very well.

    Benta.


    Benta.
     
  15. crutschow

    Expert

    Mar 14, 2008
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    Why do you think it's not turned off?
    Look at the voltage.
    When the controller switch turns off, the diode turns on due to the inductor current flow and there's a negative voltage of a few tenths of a voltage (the forward voltage drop of the Schottky diode) .
    When the inductor current stops flowing, the diodes turns off and the diode voltage rises to the positive output voltage (which is feed back through the inductor).
     
  16. ronv

    AAC Fanatic!

    Nov 12, 2008
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    If your concern is the ringing as the diode (D2) turns off you can probably make it go away with a snubber across the diode or a ferrite bead on it's lead.
     
  17. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    I have post the PCB board pictures.
    Ideally, when the diode is turned on, the darlington transistor should not have voltage leakage. If that is the case, then transient ringing are longer. I suspect 3.7V comes from the darlington transistor when the diode is turned off. As you said, pin 2 is negative voltage the diode is turned on; while it is position voltage, then the diode is turned off in short time. If I add a snubber, i will add a parallel capacitor with the diode. Is that ok?

    upload_2016-6-20_10-36-38.png
     
  18. crutschow

    Expert

    Mar 14, 2008
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    As I said, the DC voltage is your output voltage feeding back through the inductor when it stops conducting in the forward direction.
    Once current stops flowing in the forward direction through the inductor there's nothing to stop it from flowing in the reverse direction to charge the stray capacitance across the diode.
    No need to add a transistor leakage current to account for that.

    If you add a snubber capacitor, it's also usual to add a small resistor in series with the diode to absorb the ringing energy from the LC tank (L from the inductor and C from stray capacitance), otherwise you will just change the ringing frequency but not get rid of the ringing.
    Values of a few tens of ohms up to about a 100 ohms are typical. You can experimentally determine the best value.
     
  19. Benengineer

    Thread Starter Member

    Feb 6, 2016
    101
    2
    Are those two circuits equivalent?
    upload_2016-6-20_12-8-22.png

    or
    upload_2016-6-20_12-9-1.png
     
  20. crutschow

    Expert

    Mar 14, 2008
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    Not really.
    The first adds resistance to the diode, which significantly reduces efficiency, and only damps the ringing from the diode capacitance, not from any other stray capacitance.
    You need the second circuit, with the resistance in series with the capacitor, to absorb the tank energy.
    The added capacitance swamps the stray capacitance so that most of the energy is absorbed in the resistor.
     
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