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Voltage sag from FET gate driver.

Discussion in 'General Electronics Chat' started by Triode, Nov 27, 2017.

  1. Triode

    Triode Member

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    I actually changed the resistor going into the diodes that charge the capacitors to 5.1Ω as well, I wonder if that's part of the problem. Was 100Ω a good value for that?

    The flyback diodes seem like a good thing to add. I've seen it on several diagrams for similar drivers, I'll give it a try. Would that be parallel with the gate resistor? It sounds like you mean it's in parallel with the body diode, from source to drain?

    I also reduced the gate resistors to 5.1Ω, maybe that's too small?

    Thanks again for all the help! I'll post traces later so there is more to go on, I know without diagrams or traces it's kinda vague.
     
  2. dknguyen

    dknguyen Well-Known Member

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    No, 5 ohm should be good for operation for both bootstrap and gate. That's what I would have used if I bothered to use a resistor at all. OTOH, it does mean if something fails short there's not enough current limiting along those paths to keep traces from burning.

    If you have a scope, scope the gate voltage and see if there's crazy significant ringing happening whenever you switch.

    You said you can't reproduce it right? If you can reproduce it, maybe double and quadruple the gate resistor see if the failure stops at some point. If it does while the resistor is still too small to limit shorting currents from burning something out, then that means that the rise times really are fast enough to excite ringing that is popping stuff. 22AWG wire can handle 8A during normal operation so to burn one out there must have been serious current going through it.

    The flyback diode would be anti-parallel with the drain-source of each MOSFET (the direction that doesn't make a dead short across the motor supply).
     
    Last edited: Dec 29, 2017
  3. Triode

    Triode Member

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    To scope the gate do I connect the ground on the probe to source? Or do I need to do something more complicated? I had been connecting it to Vss on the chip but I just realized this isn't measuring Vgs.
     
  4. dave miyares

    Dave New Member

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  5. dknguyen

    dknguyen Well-Known Member

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    Look at it in both. Source is better but some scopes or probes don't deal so well with the high speed common mode junk going on.
     
  6. Triode

    Triode Member

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    So maybe put a probe to gate and one to source, ground both to Vss and display the difference?
     
  7. dknguyen

    dknguyen Well-Known Member

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    I'd personally only connect one probe at a time due to that common-mode issue that some scopes have a hard time dealing with. But if your scope can handle both, go go for it.
     
  8. dave miyares

    Dave New Member

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  9. Triode

    Triode Member

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    EDIT: before you put too much time into this, I just found that the supply capacitor (1000uF) was failed open. This means that there was not much to smooth out the demand on the power supply. Will update once I've fixed that.

    Well, I traced it. And it looks pretty nasty. This is with two FETs installed in series.

    C1 through C3 on top frame are the high side FET gates, C4 through C6 are the low side gates. The bottom also displays F1 which is not relevant here (it was diffing the G and S, but C1-C2 are now on two different high side legs)

    I can see there is a lot of waving going on during the off period of the waveform that should not be occurring. This could be how shoot through would occur. I could be wrong but it looks like that low side "buzzing" is enough to trip the low side fets to turn on unintentionally. I'm thinking the first thing I'll try is to add more resistance to the gates.

    Switching.jpg

    Heres another if it adds any clues. In this one I put a probe on the gates of each of the two parallel high side FETs. You can see that during the high period they track eachother almost perfectly. But at the end of the fall they both do their own noise. Is that ringing?

    Here since C1 C2 are the two gates F1 now shows their difference. And its pretty noisy.

    Looks like I have a lot of work to do.

    Thanks for all the help!

    upload_2017-12-29_16-51-3.png
     
    Last edited: Dec 29, 2017
  10. Triode

    Triode Member

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    EDIT (same as edit above): before you put too much time into this, I just found that the supply capacitor (1000uF) was failed open. This means that there was not much to smooth out the demand on the power supply. Will update once I've fixed that.

    -The board I was showing below did not have a failed supply capacitor.

    By contrast, here is what the version with just one FET looks like. A lot cleaner and less noisy. Though I am still concerned with that dip at the end of switching. Note that this board does not have the flyback diodes I added to the other one.
    1 fet board.jpg
     
    Last edited: Dec 29, 2017
  11. dknguyen

    dknguyen Well-Known Member

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    The second graph in post #27 where the edge undershoots on the falling edge and oscillates (or conversely, overshoots on the rising edge and oscillates) is what ringing looks like. There is ringing in the graph in #28 as well, with that huge dip followed by a bunch of smaller oscillations.

    That huge initial dip doesn't look like it's being caused by ringing per se, since it's totally out of proportion with the much smaller and much faster oscillation that follow afterwards. It almost looks like something is receiving insufficient power. But notice that it only happens on the switch off, so that just might be remedied by adding in the flyback diodes. You should take a look at the voltage across the source-drain too and see if it looks clean.

    Ideally, you want the waveform to be critically damped for a balance between minimizing ringing while maximizing transition speed (which minimizes switching losses).

    Ringing can cause damage and other unpredictable behaviour like latching the transistors and stuff like that.

    Failing of the main bypass capacitor will cause an increase in the inductive spikes experienced by the system, especially if the wires between power supply and switches are longer. This can damage the transistors.
     
    Last edited: Dec 30, 2017
  12. Triode

    Triode Member

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    To revisit this concept, if you are still seeing this thread, I have two questions:

    Is it ok that these only supply 83 mA? If they were supplying the transistors with no capacitors I figure they would need to supply the gate charge current in the desired switching time. So I get Amperage = Charge/Seconds, or 108 nC/300 ns = 0.36 amps. But I realize that they don't need to supply this much current all the time, presumably once the fets are on keeping them on requires very little current, so if you have capacitors to handle that switch on boost it can charge them up the rest of the time. After all 300 ns per switch at 30 kHz is only 1% of the time. Is that why it works, or am I way off?

    The other question is about the DC-DC converters themselves. See I tried to pick out a 15 volt one to do the job, but I screwed it up. I got a single output one that I thought was the same thing but at 15V.
    https://www.digikey.com/product-detail/en/texas-instruments/DCP021515P/DCP021515P-ND/307875
    DCP021515P (rather than ending in D)

    Now to my mind, this would mean it can output 15 volts above whatever you connect the low side of the output to.

    But it looks like this

    upload_2018-1-22_17-37-32.png

    And since those outputs are -15 and +15 apparently, they seem to actually be 30 volts apart, and they would probably kill my transistors which have a Vgs max of 20.

    Is there anything I can do to make these usable. I got 10 and they were $10 a piece, so it kinda sucks to shelve them, but I suppose I can use them later on another project.

    Thanks again.
     
  13. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You might consider a bridge driver with a built-in charge-pump to generate the high-side gate voltages, such as this.
    That way you don't have to worry about duty-cycle or the sequence of the bridge operation.
     
  14. Triode

    Triode Member

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    That does sound a lot simpler, do they have one with 1 or 3 half bridges? I did some searching but didn't find any. With this one I could use 2 chips and just not use the 4th half bridge driver I suppose. My mosfets have a total of 324 nC of gate charge. Would the 1A drive current of this be enough to switch at 30 kHz?.

    Thanks!
     
  15. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    I could not find out if the charge pump will go to 100% and 0% duty cycle.
    There should be a steady state current draw for the top side driver. 100uA maybe 1mA. (?)
    Then there is the 1A gate current for 1% of the time. Looks like you know how to find that.
    With out doing the numbers I don't know but I think 80mA will work.
    If you got the "DCP021515P" then it will output 15V across the (+) and (-) pins. It will work.
    In your case you will put the (-) pin on the Source of the top MOSFET. and Put the (+) on the top Gate driver supply.
    ....Put the output pins across the bypass pins for the top Gate driver.

    If the look at the out put of the DCP021515P, when the MOSFETs are switching it will look funny. From (+) to (-) it will be 15V just fine. When the bottom MOSFET is on the (+) will be about 15V. BUT When the top MOSFET is on the (-) pin will be at supply and the (+) will be 15v above your supply.
     
  16. Triode

    Triode Member

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    I believe you and that should totally make sense given that it says it is 15V. But when I supply it 15V and measure across the Out+ and Out- pins I'm getting 30V. I figured maybe it needs a load so I put a 100 ohm resistor across it. It still was about 30V.

    I'll set up that test again to make sure I did it right. I'm not saying you're wrong because the result I got doesn't make sense to me either.

    I even found another data sheet that shows it like this:
    upload_2018-1-23_12-48-1.png
    https://www.jameco.com/Jameco/Products/ProdDS/1339233.pdf

    And that datasheet shows the same [BB] marking as the chip I have on hand. I just check the chip again and it definitely says DCP021515P.

    Anyway, I'll try the set up again and see. Maybe I did something weird before.
     
  17. dknguyen

    dknguyen Well-Known Member

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    Out+ and Out-? Or Out+ and COM? Because Out+ and Out- sounds like a bipolar supply in which case 30V would make sense since you're skipping past the COM/GND. Remember that there is a single-output version and a dual output version. I think you got the dual.

    http://www.ti.com/lit/ds/symlink/dcp020503.pdf

    Don't count on it. Too specific a component and not enough general use. I can barely find 3-phase gate drivers. The ones I do find have really weak current drive relative to the 2A+ you can get with individual gate drivers.
     
  18. Triode

    Triode Member

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    Ok so, I tested the DCP021515P again and it puts out 15V. I don't know what I must have done before to get 30 volts. Unloaded it was at 26V and then I put a 220 ohm resistor across it and it dropped to 15V on the dot.

    So that mystery is solved I guess.

    So 3 phase is probably too much to ask, but are there ones that are just one high side a low side? I know those are pretty common with the bootstrap style ones, and I just use 3 of them. I'd be happy to use 3 that have built in boost circuits. Right now I'm also trying the isolated supplies and the 555 circuit, but eventually I'll need to go to whatever is least expensive for parts+build, and anything that integrates more parts into one chip is a good thing, if they work. It removes a lot of variables. Of course I could just use two that have 2 half bridges each and not use one of the 4.
     
  19. Triode

    Triode Member

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    Dynguyen:
    Follow up question about the isolated supply based solution. Should I still have the bootstrap diode coming in? It seems like it could help supply power when the low side is off, but maybe it could be detrimental. I suppose with the diode it could never pull the capacitor down. I'm assuming I still want a capacitor there to even out the supply. Is there any problem with a 10u one? I figure that's enough to ride out any high current draws and once switching has occured the DCP02 should keep the power up.

    Here is what I'm designing at the moment, in an eloquently annotated drawing

    upload_2018-1-23_16-48-36.png

    Edit, just noticed this is from the newer version of the board I'm working on. But I figured I'd post it in this thread since the isolated supply circuit was mentioned here.
     
    Last edited: Jan 23, 2018
  20. dknguyen

    dknguyen Well-Known Member

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    I've still never seen one with a built-in charge pump though because it seems to be rarely needed and anyone that does need one can add one onto a regular bootstrap driver.
     
  21. dknguyen

    dknguyen Well-Known Member

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    Remove the diode. Replace with an open. With the diode there, it may never turn on in which means it's pointless and if it does turn on, then the two power supplies might fight with each since their output voltages will never be identical which may also cause the diode to spuriously switch on and off as they fight each other. No need to risk any of that.

    In an isolated supply, the bootstrap is still there but is really the isolated supply's output decoupling capacitor so keep it there. 10uF is fine.
     
    Last edited: Jan 23, 2018
  22. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    I would keep the diode. OR At least have a place for it on the PCB. The isolated supply really is only needed when the dutycycle approaches 100%.
    The isolated supply, see data sheet, wants an external capacitor. Probably a ceramic cap on the top side driver and a ele cap on the supply.
     

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