3 phase inverter bridge

Discussion in 'The Projects Forum' started by ShaunManners, May 9, 2008.

  1. ShaunManners

    ShaunManners Thread Starter Active Member

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    Hi All,

    I am still trying to get my 3 phase inverter working, I have successfuly managed to get my PIC MCU working great, and have used it to generate 3 signals which I then fed into 3 L6384's (half bridge drivers). This has all been tested and works a treat...

    I have put together a bridge using 6 P55NF06 MOSFETs... When I connected them to 24v (nothing else connected) they quite literally blew up and caught fire... see attached photo.

    The high side drains are all connected to +24v via the heatsink, and the source is bolted to the drain of the low side. The low side source is soldered to a wide track and bolted to ground.

    All gate pins go to the pins on the left of the board.

    I do not have any protection diodes in - partly because I wasn't planning on putting a load on it yet.. and also, I'm not sure what diodes to use, any suggestions? bearing in mind I'm planning on using an alternator as a 3phase induction motor.

    I also have a circuit breaker on the positive side... which I reset when I want to turn it on. This is one of my theories... could resetting a breaker produce a large enough spike to damage the MOSFETs?

    Any suggestions, observations, ideas or help would be welcome!

    Cheers
    Shaun

    Attached Files:

  2. SgtWookie

    SgtWookie Expert

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    My first knee-jerk reaction is that your upper side and lower side were ON at the same time, resulting in a "shoot through" condition.

    This can happen when the turn-off time is less than the turn-on time, or vice-versa, which is usually the case.

    "Shoot-through" means that the H-bridge drivers are trying to short the supply to ground. This is a very unhappy situation, as smoke and fire may be emitted from the weakest side of the driver, and causing the circuit to not function as desired. However, it can be quite entertaining here in the U.S. around Independence Day, the 4th of July.

    I suggest that you re-visit your code, and ensure that an appropriately long delay is inserted between the "off" commands and the "on" commands so that you will no longer encounter the "shoot-through" condition.

    Unless you like smoke and flames. ;)
  3. S_lannan

    S_lannan Active Member

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    wow.. nice job!

    Some what reminscent of my own prototyping :D

    Anyhow you can get the old scope out to check this too.

    Perhaps you might of made an error when connecting the floating source of the high side. I did it once, luckily enough i had current limiting but both the fets were red hot, but lived.
  4. ShaunManners

    ShaunManners Thread Starter Active Member

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    I don't particularly like flames ;) it costs me £6 each time they blow! and that photo was of my 3rd attempt!
    However I do agree its quite good fireworks ;)

    Your intial reaction is the same as mine, I went back to the code to make sure it was leaving a large enough deadtime... and it was giving me about 1uS... and considering the MOSFETs switch in the order of 8-16nS! it should be fine... Now I am using the drivers, they have a deadtime thingy that you control using an external resistor, which I have chosen to give me roughly 1uS...

    The problem with this theory is that I didn't have any of the MCU/driver circuit connected when I turned the 24v on for the bridge... so all of the MOSFET gates were floating... which to me means they should all have been off and no short circuit or shoot through conditions.. yet they still go bang.

    This is why I was wondering if it was a spike created by resetting the breaker that damaged the gate (I read something like that on a site) and then shortly after it makes it blow up... see here: http://www.4qd.co.uk/serv/MOSFETfail.html#dvdt

    Otherwise I'm stumped... I appreciate your input :)

    Cheers
    Shaun
  5. ShaunManners

    ShaunManners Thread Starter Active Member

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    Nice job blowing it up, or just in general? ;)

    What could I check with the scope?

    The floating source goes to the drain of the low side, that is corerct right? what other error could I have made?

    Cheers
    Shaun
  6. SgtWookie

    SgtWookie Expert

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    Floating gates? Oops...
    I suggest that you add pull-down resistors on the gates, so that if you happen to power the thing up again without the drivers connected that the gates will all be pulled to 0v. Floating is floating - you don't really know what voltage might be there. But from the looks of things, I'd say it wasn't zero!
  7. S_lannan

    S_lannan Active Member

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    nice job in general. I like blowing up stuff too!

    the high / low side driver circuit won't work unless you connect it up properly.
    I'm assuming you might have left out a connection or two.

    on the driving ic
    pin 4 should connect to the ground of the bridge (eg low side source).
    pin 5 should connect to the low side gate.
    pin 6 should connect to the high side source, (which is the same point as the low side drain.)
    pin 7 should connect to the high side gate.
    pin 8 should connect to the 'bootstrap cap' which connects to pin6.

    and guess what?
    with this ic it seems you can't connect VCC to a low impedance supply!!!
    so you have to take a resistor from your high voltage supply (assuming you have one ) and feed it to the VCC pin. It's basically a zener regulator without the resistor. you could feed it of the 24v supply (with a resistor of cause).

    I would say...

    Use current limiting when testing the bridge = no more blown fets :D
    Possibly try a different gate driver eg.ir21362.
  8. S_lannan

    S_lannan Active Member

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    oh you left the gates floating.....
  9. ShaunManners

    ShaunManners Thread Starter Active Member

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    ok then... I'm still not sure why it would blow the first time when I did have it connected though... hmm.. mind you that was before I implemented the drivers... it could be that I have made two errors that ended the same way!

    That aside, the pull-down resistors... for the hgh side, do I connect it between the gate and the high side source, or to ground? and what value would you reccomend?

    Cheers
    Shaun
  10. ShaunManners

    ShaunManners Thread Starter Active Member

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    Thanks for saying its a good job... when it blows up all the time I wonder!!

    I have checked all those pins, they are exactly as you say... and I used a 100ohm resistor on Vcc... as I am using 3 of them and they take 25mA each... on a 24v supply... please tell me if this is wrong!

    When you say current limiting... presumably you mean put a resistor in line with the supply to the high side drain? what sort of value? and for the moment, with no load on the bridge, is 1/4 watt ok?

    Thanks both of you for your help... and sorry if I am making stupid mistakes!

    Cheers
    Shaun
  11. ShaunManners

    ShaunManners Thread Starter Active Member

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    Ok, I had another go with just a half bridge (in case it blew again I didn't want to waste another 6 MOSFETS)...

    No fireworks this time which is good! I put a 1k 1W resistor in series with the high side drain, connected both gates to the drivers and made sure the drivers were running ok before switchng the power on to the bridge.

    I have attached a schematic of basically what I have - apart from the fact the alternator isn't connected, and I dont yet have protection diodes across the FETs.

    I put my scope between ground and the centre of the bridge and got the result you can see in the second attachment. I slowed the frequency down to somewhere around 1hz. Does it look right? the two blips seem to be where the other two phases should start... not sure why its doing that...

    Cheers
    Shaun

    Attached Files:

  12. SgtWookie

    SgtWookie Expert

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    Well, if you used a 2k resistor from gate to source, that'll keep the gate at 0v when you don't have the driver powered up.

    Before you DO power it up with your drive circuit, I suggest that you go to International Rectifier's site, and download AN-944.
    http://www.irf.com/technical-info/appnotes/an-944.pdf
    Compare those notes with how you've constructed your bridge driver, the goal being to avoid smoke!

    Were I you, I'd give your drivers extra time to ensure that the high/low side MOSFETS were absolutely turned off before turning on the low/high side - and check it with an O-scope to verify turn on/turn off times, before decreasing that margin.

    In looking at the ST Microelectronics datasheet for those MOSFETS, it only shows the nC's at 48V, max of 60. Thing is, this can vary quite a bit with voltage - you'll read about it in App Note 944.
  13. ShaunManners

    ShaunManners Thread Starter Active Member

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    Ok I'll put the 2k resistors in...

    And thanks for the advice on the driver circuit... I'll print that App note off now and go have a read :)

    Cheers
    Shaun
  14. SgtWookie

    SgtWookie Expert

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    Oops, we cross-posted.

    See where the voltage drops to 0, then goes up to 24? That's good.

    Right after that, see it going down to about 12? That's not good. It looks like you have both the high and low side on simultaneously.

    Then there are two blips - looks like the low side kicked off briefly, twice. Finally, the high-side driver turned off.

    How about showing your drive signals for a high/low side pair using dual traces?
    [eta]
    Ok, now I finally see your circuit diagram :rolleyes: ;)

    I completely missed that you were using L6384 drivers. You don't even have to worry about the turn on/turn off times or AN-944; the L6384 should take care of all of that for you.
    It appears you only have a single decoupling cap for all pin 2's - you should have a separate cap for each L6384.
    What size and type caps are you using for your bootstrap?
    What value resistor for SD/DT? Are you keeping that line low on your uC, or are you monitoring it as an input?
    Last edited: May 10, 2008
  15. ShaunManners

    ShaunManners Thread Starter Active Member

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    Ok, I'm not sure I understand why it would go to about 12v with both sides on? surely as long as the low side is on, it should drop to pretty much zero? or is the Rds higher than I think?

    I'm not sure what the two blips are... they are quite evenly spaced though...

    That would be ideal, unfortunately my scope only has a single channel :(
  16. ShaunManners

    ShaunManners Thread Starter Active Member

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    heh we did it again ;)

    I only used the one decoupling cap because that was how they did it in the original circuit diagram.. I did think it was a bit odd I must admit...

    The bootstrap caps are 0.47uF Tantalum

    and the resistor I am using for the SD/DT pin is 100k... it is supplied with 5v from the MCU... this was just a pure guess... I think if the voltage drops below 0.5v it shuts down... although I must admit to being a little unsure about this part.
  17. SgtWookie

    SgtWookie Expert

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    Ok, now that I'm actually reading the datasheet for the L6384... :rolleyes:
    SD/DT, when below 0.5v, shuts down the IC.
    When ABOVE 0.5v, it sets the "dead time"; ie; the high/low side both off time.
    This is contrary to what I stated before. I've been reading too many datasheets recently. There was one that did ensure that there couldn't be shoot-through, and now I don't recall that IC's part number.

    The description for the rest of the functionality of this pin I find very confusing. No wonder that you're confused.
    Partial quote:
    "...The connection of the components between pin 3 and ground has to be as short as possible. (that's pretty clear) This pin can not be left floating for the same reason.
    The pin has not be pulled through a low impedance to VCC, because of the drop on the
    current source that feeds Rdt. The operative range is: Vdt....270K × Idt, that allows a dt range of 0.4 - 3.1ms."

    OK, so you're supposed to take Vdt and do something with it.... 270k times the current required during dead time for a range of 0.4 to 3.1ms.

    Later:
    "Iref (typ) 20uA"

    Perhaps it's just because I'm really tired, but this isn't helping me understand it a whole lot more.

    Ok, now this helps:
    "dt, pin 3 vs pins 5,7
    Dead Time Setting Range (**)
    Rdt = 47k, 0.4-0.5mS
    Rdt = 146k, 1.5mS
    Rdt = 270k, 2.7-3.1mS
    (**) Pin 3 is a high impedance pin. Therefore dt can be set also forcing a certain voltage V3 on this pin. The dead time is the same obtained with a Rdt if it is: Rdt × Iref = V3."

    So, basically - the higher the value of Rdt, the longer the dead time. It's not quite that simple, but you get the picture, right?

    I suggest you cut the trace from your uC to the L6384's pin 3, and use a 270k resistor on each of the L6384's pin 3 to ground, keeping the leads short as possible. Noise on that pin will cause unexpected shutdown operation of the L6384. That might be why you were seeing the two spikes.

    The L6384 bypass caps are more than bypass caps. You really need a couple different sizes there.

    Each IC draws 25mA, and it has a Zener clamp on Vcc. It's difficult to tell exactly what the clamp IS, because the datasheet contradicts itself. It might be 14.6v (spec'd as max), or 15.6v (spec'd as typical)

    You could feed all three with one resistor, and use a large and a small bypass cap. But if you want to play it a bit more safe, you should feed them with individual resistors, and have separate pairs of decoupling caps on each IC.

    For one IC Vcc current limit resistor:
    Let's go with 14v for the moment.
    24v-14v=10v
    Let's say you're OK with a bit of overkill on the supply current. Let's go for 35mA.
    Rlimit = E/I
    Rlimit = 10/0.035
    R = 285 Ohms

    Let's go with 270 Ohms, as that's close. What's our new current?
    I = E/R
    I = 37mA - close enough. Wattage needed?
    P = EI
    P = 0.37 Watts. Better use at least a 1/2 Watt resistor. 1 Watt would be more reliable.

    For bypass caps, you might as well use a 10uF and a 0.1uF/100nF cap in parallel for each IC. The small one needs to be ceramic or tantalum, and have short leads.

    Well, I'm getting tired. Better think on this more tomorrow.
    [eta]
    Let's go with 15v (compromise between 14.6v and 15.6v, but a nice round number)
    24-15=9
    Yes, they spec 25mA for current, but that may vary depending on a number of things - temperature, frequency, etc. Let's go with 30.
    So, you need 30ma across 9v.
    R=E/I
    R=300
    P=0.27 Watts
    You could use a couple of 1/4W 150 Ohm resistors in series per IC.
    As things are now, your existing single resistor would need to be 100 Ohms at 1 Watt.
    R= 9v / (30mA*3)
    R= 9v/0.09A
    R=100
    P=EI
    P=9 x .09
    P=0.81
    Last edited: May 11, 2008
  18. ShaunManners

    ShaunManners Thread Starter Active Member

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    I measured the voltage on the supply earlier... it is 14.9 so 15v is a good value to go by ;)

    So thats one 300 ohm, 1/2 watt resistor for each driver supply... I don't have 150ohm resistors.. I do have 220 and 100 both 1/4W would that be ok? I should be able to work it out myself I know...

    I might just order a selection of resistors actually... I have the E3 series in 1/4W plus a few others...

    For the DT/SD resistor on pin 3... I'll replace my single one with three 100k... and I cut the track (cable actually*) from my MCU.

    What do you make of my choice for the bootstrap caps?

    As for the decoupling caps, I have 0.1uF ceramic (thats what I'm using already) so I'll put another one on for each IC... and I have 10uF electrolytic
    would they be ok for the other?

    * I have the MCU on a seperate piece of veroboard from the drivers.

    Thanks for all your help... it is greatly appreciated!

    Cheers
    Shaun
  19. SgtWookie

    SgtWookie Expert

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    Knowns are good :)

    OK, so what would happen if you used one 220 and one 100 Ohm resistor in series across 9v?
    I=E/R
    I=9/320
    I=28.1mA - that's likely enough, it's better than 12% more than specified.
    Let's check the wattage.
    P=IIR
    P=.0281x.0281x220
    P=0.174 Watts
    Your 100 Ohm resistor would dissipate less than half that much power, so you're good with two 1/4W resistors.

    E3? Do you mean E6? E3's are 50% tolerance, and they haven't been used in a while. E6's are 20% tolerance, and they're seldom used either.
    I bought an E24 series of 1/4W resistors at an online auction awhile back. Wound up with something like 5,400 resistors for $18. It was a very good buy. An E6 series wouldn't give you much of a selection at all.

    Well, do you have some combination of resistors where you could wind up with between 220k and 270k?
    If you have some 470k resistors, you could put them in parallel to get 235k.

    Haven't gotten that far yet ;) But actually, the size of the bootstrap cap has to do not only with the gate charge (60nC) and gate capacitance (which varies with voltages) but also the current used by the circuit itself. See if you can use an O-scope probe on your upper gate to observe how the charge is doing.

    Do you mean for the bootstrap caps?
    Last edited: May 11, 2008
  20. ShaunManners

    ShaunManners Thread Starter Active Member

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    Thanks for going over the math... I'm afraid I find it confusing, the math is simple enough! but its understanding how it all works... guess I haven't had a lot of practice.
    but anyways, thats good I can put those in then.

    That really was a good buy! I'll have to have a search around see what deals I can find somewhere :) and no I did mean E3... at least thats what they called it... it was a set of 480 1/4W 5%... maybe its just the E3 values 10/22/47/100/220/470 etc...

    Yep got 470K... ok will put them in instead.

    EDIT: Actually... I have 220K...

    Ok, I'll do that tomorrow... so put the probe between the upper source and gate? I'll take another photo of it for you if you think it'd be useful.

    no.. I meant the decoupling ones on each driver IC... you said I should have two for each IC, one ceramic or tantalum :) I was just checking that an electrolytic would be ok for the 10Uf...
    Last edited: May 11, 2008
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