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MOSFET switching problems

Discussion in 'General Electronics Discussion' started by zappyton, Aug 8, 2013.

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  1. zappyton

    zappyton

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    May 20, 2013
    I'm building a push pull converter, and my MOSFETs aren't switching properly when given the signal to. I'm assuming that it's some kind of snubber problem, but I don't know. Maybe a TVS diode would help?

    The circuit is attached; the MOSFETS are IXFX30N100Q2 and they are being given a signal from a MC34152P driver chip. I've also attached a photo of the waveforms without the RC snubber across the transformer coils to Vin; I appear to have not taken with the snubber (i.e. the 6,8nF cap, and the 80R resistor), but the sharp peak was exactly the same, but the oscillation before the circuit goes high again was reduced in frequency.

    In the scope picture, channel one is the input to the MOSFET gate, channel 2 is the drain voltage. I was expecting them to be almost opposing square waves, but this is clearly not the case. Can anyone offer any suggestions?

    Oh, and although the diagram shows a 50V input, as does the scope, it will need be running off of mains, i.e. 320V; at the moment I have it connected to a variac.

    It so nearly works, but not quite! Any help would really be appreciated.

    Thank you!
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    What's the use of C2, R5 and R6? I don't think they are appropriately placed here. At turn-on, the rising source current will lead to a voltage drop across these components, effectively reducing the gate-source voltage. Also it is not possible to see how the transformer primaries are coupled to which secondaries.

    Also, R1 and R2 are recommended to reduce EMI in sensitive environments. To get the circuit working you can short-circuit them.

    Do not run the circuit without the snubber. At turn-off of the MOSFET the energy stored in the transformer will create the high peak in the drain voltage you see in the scope trace. This voltage will couple back to the gate via the D-G capacity of the MOSFET.
    Also try experiomenting with different snubber networks. Apart from your configuration (in parallel to the coil, there is also an often founf configuration parallel to the transistor. Here's an article on snubber networks worth reading.

    For ease of debugging I suggest you reduce the circuit to the minimum of only one transistor plus respective load and get that circuit up and running. Only then complement the rest of the circuit. This increases your chances to locate and fix the problem and minimizes the risk of destroying components.
     
  3. zappyton

    zappyton

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    May 20, 2013
    Hi, thanks for helping.

    C2, R5 and R6 are the current sense resistors for the control of the push pull. I omitted that bit for clarity as it seems to be working. There should never be more than 200mV across there, or the chip starts reducing the duty cycle .
     
  4. zappyton

    zappyton

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    May 20, 2013
    Sorry I accidentally clicked the reply button!

    But with the snubbed the peak voltage didn't change, or does that mean the snubber I've used is just really wrong?

    The reason I did it this way as opposed to ground is that I was concerned about the power loss in the resistor going from mains to ground. Do you think that this would be OK?

    And the primaries are all coupled with all the secondaries, which are all equal, sorry that I couldn't make that clear on the diagram.
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    power loss in the resistor will be almost the same regardless of the connection to GND or Vsupply, since only the AC part is effective due to the decoupling capacitor.

    I can't tell you whether your snubber is correct or not. Try tweaking the component values and observe the behaviour of the voltage.

    How are the primaries and secondaries coupled (start of winding?) Is this a flyback converter?



    Try adding a load to the output. What happens to the signals? (Since a load would drain energy from the secondary, more energy will be taken from the transformer to power the load circuit, which means less energy is available to create the spike at the primary.)

    I think you may also get rid of the many rectifiers and inductor on the secondary. Connect all secondary coils in series (observing phase!) and use just one rectifier and inductor.
     
  6. zappyton

    zappyton

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    May 20, 2013
    It's a push pull converter, but I don't understand what you mean by how are they coupled. They're all on a big iron core, but I suspect that's not what you mean!

    The problem I have with the last two suggestions is that the circuit is designed to chatge a capacitor (the fifty micro farad one) to 3500V. The transformer expert I talked to recommended doing it this way to lower the inter winding capacitance of the secondary, and also 3500V diodes are expensive!

    The high voltage gives me a problem with adding a load as well, because the power losses quickly become ridiculous.

    Am I wrong in thinking that the d-s volatge should go high again when the MOSFET is off? Or will it not because of transients? Is there any way to tell if a the FETs are actually switching?
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    I could have seen that had I looked a bit more into the details. What I meant was that energy transfer in a transformer can depend on where the windings start (those black dots often seen on coils) and how the starting points relate to each other between primary and secondary. Also it is not obvious from your schematic that this is a single core transformer (you didn't show the symbol for the core). It could as well have been two transformers operating back to back.

    O.K., that wasn't obvious from the schematic. I agree with your friends then.

    I wonder why then you use this rather complex setup in the first place. If you don't draw power from the circuit, you don't need a push-pull drive. A simple flyback converter could do the job as well.
    The power loss doesn't have to become "ridiculously high",Use a few 100kOhm to generate 1W of load. At 50V primary voltage the output will be considerably less than 3500V.
    Be that as it may, since you have this circuit, the flyback pulses are inherent to the circuit (as far as I can see). You can only minimize them by applying a proper snubber circuit (experiment!), You can not eliminate them completely.
     
  8. zappyton

    zappyton

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    May 20, 2013
    I'm using this setup because the capacitor is going to be discharged at 3Hz, so the power needed is quite high but the load is the capacitor, and then a specialist light bulb is what the cap is discharging into.

    I guess I could use a few hundred kilo ohms of resistance for the load, but at 1W of power dissipation, will the current drawn be significant enough to have an effect?

    Sorry I wasn't clear in my first post!
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    It is the power drawn, not the current, that makes an effect.

    Another way to limit the voltage spike at the primary is a suppressor diode in paralel to the snubber network. This will, however, increase the loss on the primary side by heating the suppressor diode.

    You could add a sense circuit that turns off the primary MOSFETs when the secondary capacitor is fully charged.
     
  10. duke37

    duke37

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    Jan 9, 2011
    What is the purpose of the output chokes L7 to L10?
    If each bridge fed a capacitor, then transient voltages would be reduced.
     
  11. zappyton

    zappyton

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    May 20, 2013
    Ok, I'll try adding a load resistance to draw some power to help the spikes. The control chip does turn off the MOSFETs when the circuit is charged, I'm not sure what the addition of a parallel resistance would do but I think it will be charging almost continuously anyway.

    duke37, do you mean replacing the chokes with just capacitors? Or adding capacitors as well? I thought you need output chokes to stop the current going discontinuous, would the capacitors stop this? Or does it jot really matter?
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    After thinking a bit about the circuit, i think this is what happens:
    Primary coils 1 and 2 are activated alternatingly. Between these two coils there is a mutual inductance since they are wound onto the same core. Therefore activating e.g. coil 1 will invariably induce a current in coil 2. Since the transistor controllng coil 2 is off at that time, this current will lead to a high voltage (the spike you see on the scope trace).
    This high voltage will couple onto the gate via the drain-gate caacitance.

    I don'tknow of any measure to stop this.

    What you can do is:
    1) Remove the gate drive resistor to minimize the driving impedance. This will provide a low impedance path for the injected current to ground, thus keeping the gate voltage low.
    2) Put a suppressor diode from drain to source. The breakdown voltage of the suppressor diode should be at least 10% less than the max. drain-source voltage of the transistor. This diode will limit the voltage at the cost of some power lost during the diode's conducting phase.

    Alternatively you could use a different circuit topology, e.g. a full bridge using 4 transistors but only one primary winding for the transformer.
     
  13. zappyton

    zappyton

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    May 20, 2013
    I'm not seeing a voltage spike on the gate though, am I?
    I can get a suitable suppression diode for tomorrow no problem, as I was thinking it might help.

    I can remove the gate drive resistor as well if you think it will help, and see what happens.
     
  14. duke37

    duke37

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    Jan 9, 2011
    As I understand it, the two primary coils are linked by a mutual inductance. Thus when the drain voltage of one fet rises on turn-off, the drain voltage of the other fet will go down to zero or below. There is a diode in the fet to return this current to the supply.

    Low voltage invertors have bifilar windings (two windings wound on together) to reduce the leakage inductance but with a high voltage invertor, the wire insulation may not be adequate.

    I do not see what the output chokes do except force the magnetic energy to be returned via the opposite fet.

    The current will be discontinuous without the chokes but you are feeding a big fat capacitor so why should that matter? It only falls to a low level during the switch over time when both fets are off.
     
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    Yes you do.
    This is where you see a spike after the drive signal has turned off. This spike is most likely coupled from drain to gate via the D_G capacitance.
     
  16. zappyton

    zappyton

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    May 20, 2013
    Just to let you know progress wise, removing the output inductors has vastly improved the overall shape of the waveform, and I added TVS diodes for the spikes.

    Unfortunately, I was winding up the voltage, keeping a careful eye on the spikes, and when I got to 200V, one of the transistors blew. The spikes were not that high (that I could see), and I had 750V TVS diodes across the MOSFETs, (which are 1kV rated). Any ideas what could've caused this?
     
  17. Harald Kapp

    Harald Kapp Moderator Moderator

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    Have you monitored the current (waveform across the current sense resistors)?
    Normally one would expect a triangular waveform for the current. If the primary inductance becomes saturated, the current will rise steeply. This is just one possible cause for the failure.
     
  18. duke37

    duke37

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    If you have no load, the inductive energy has to be returned to the supply via the diodes in the fets. You could try better external diodes across the fets Will the supply accept this energy?
     
  19. zappyton

    zappyton

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    May 20, 2013
    Ok, I will check the waveform across the current sense resistors and have a look for saturation.

    The supply is rectified mains, with a 470uF cap across it. Will this accept the returned energy?
     
  20. duke37

    duke37

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    Jan 9, 2011
    It will accept energy but without knowing the transformer inductance and frequency it is not possible to say if it will accept it all without a significant rise in voltage.

    The only way that I can see to go forward is to run it at a safe level and measure both voltages and currents.
     
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