High Voltage H-Bridge

Discussion in 'The Projects Forum' started by pyrokid73, Jul 28, 2011.

  1. pyrokid73

    Thread Starter New Member

    Jul 28, 2011
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    I am using an H-bridge for output to a UV lamp. Attached is the circuit for the H-bridge, 555 timer to produce signal for switching, optocouplers to isolate the gate drivers and keep all four MOSFETs turning on and off at the same time, and the low side gate drivers with their relative ground tied to the respective MOSFET's source.

    I am using two power supplies, one for the timer and drivers at +15V and the other is at the terminal BUCK-OUT. To ensure I had the timer and other chips working I set the BUCK-OUT terminal to 0V, the gates of each MOSFET were turning on and off at the right times and quickly. I first started applying about +30V to BUCK-OUT, this made the signal to the gate worse, not as clean, and brought the duty cycle further away from 50%, but by adjusting the potentiometers I was able to get the signals back to a near 50% duty.

    The problem is that this is going to be a constant current output, the voltage will be changing constantly. Open circuit voltage will be a max of around 600V-800V and the on state will be 80V-140V. I have tested the circuit with and without a load, both show the same outcome, so the thing messing up the signal is the voltage across GND and BUCK-OUT.

    Please let me know what you think could be the problem, or if you need any other information about this circuit.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The MOSFETs' gate charge will change depending on Vds and load.

    You say the gate signal gets worse, not as clean - but not what it looks like. I imagine you're seeing some ringing on it that gets worse when you increase BUCK-OUT.

    A single 555 timer is a hard way to try to get an exact 50% duty cycle. It'll change over temperature. If you really wanted a true 50% duty cycle that'll stay that way, you would need to use a /2 circuit, like a D-type F/F with the Q\ wired to the D input and clocked with the 555 output at twice the desired frequency out, or a JK F/F with both set/reset tied high, and clock it at 2x the desired frequency out.

    You're trying to use FAN3121's and FAN3122's to do your inverting for you; but the problem is that you don't seem to have taken "dead time" into account, unless I'm mistaken - or is that supposed to be performed by R22/R23?

    Speaking of which, you're feeding all of the optocoupler IR emitters with R22/R23, so at any one time there will be two IR emitters in parallel with just 1 resistor limiting their current. What'll happen there if their Vf is not identical, it'll throw your timing off - as one will be getting more current, which will heat it up more, which will give it yet more current.

    If you're seeing "ringing" on the gate, and your dead time is insufficient (which I think it will be insufficient) you can help the situation by using a fairly low-value resistor from the gate driver to the gate, and a fast diode like a 1N5817 Schottky with the cathode towards the driver. That way, the MOSFETs will turn on more slowly, turn off more quickly, and the resistor will tend to "snub" the ringing.

    The datasheet for your IXFH6N100 MOSFETs is here:
    http://ixdev.ixys.com/DataSheet/91529.pdf
    Take a look at the Td(on), Tr, Td(off) and Tf specs. These MOSFETs take a lot longer to turn off than they do to turn on.

    Looking at your high-side driver, I see you're charging their caps via a BA159, which is fine. However, I don't know how much current you're putting through the low-side MOSFETs. Those MOSFETs have a rather high Rds(on) of 2 Ohms - so if you're putting a couple Amperes through them, you'd lose 4v in Vds from the low side MOSFET, and another volt across the BA-159 - so that would pull your charge on the high-side caps down by at least 5v. I have no idea how much current your UV lamp requires.

    I see to the left of your BA159's used to charge the driver caps, just a connection to +15v. Are their rather large and small caps in parallel next to the diodes? If not, and the trace/wiring from the 15v supply to the BA159 is more than an inch or so, you'll have a problem getting the driver caps charged due to the inductance of the wiring/traces, and the traces will be adding noise to the 15v supply.
     
    Last edited: Jul 31, 2011
  3. pyrokid73

    Thread Starter New Member

    Jul 28, 2011
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    The R22/R23 pots that I am using is to limit the current in the diodes, at any one time there are actually four of the optocoupler diodes in parallel on at once. Would you recommend using 8 different but same value resistors for each of the opto's diodes?
    What size resistor would you recommend I use between the MOSFET gates and the gate drivers?
    The current once the lamp is on should be about 800mA, peaking at about 1.5A depending on the light output of the lamp. This would mean about a 3V drop max across the low-side MOSFETs, assuming another 1V drop across the diodes, the Vgs of the high-side FETs will still be around 11V, which would be enough to fully turn the on.
    The +15V supply is from a 15V linear regulator that just happened to not be on this part of the schematic. There is probably about 2 inches between a 22uF electrolytic cap and the diodes, but there are 0.1uF ceramics less than a quarter in from the diodes.

    Also, at this point I would be fine with getting somewhat close, but not exactly 50% duty cycle. I would much rather get the rest of the circuit working before doing the smaller/less important tuning.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, and put the resistors next to the optocouplers.

    Well, you may have to experiment. The datasheet mentions a 4.7 Ohm resistor in the turn-on/turn-off specifications, but you may need more than that. I don't have SPICE models for your MOSFET nor the FAN312x's, so it would be tough to get an idea from here. Somewhere between 4.7 and 22 would be likely numbers, though.

    I don't know what your load "looks like" to the bridge. Is it purely resistive? Reactive? Capacitive or inductive?

    By the way, are you really operating the 555 on 3V? What voltage is IC2 actually putting out? You have it as a 78L03. Are you actually using a CMOS TS555?

    Well, if your load is inductive, you can get in trouble pretty quickly if you don't have exactly a 50% duty cycle. What can happen when driving transformers or other types of inductors is a phenomenon known as "flux walking"; it starts off with mostly AC current in the inductor, but as the imbalance builds, a large DC component comes into play. You'll wind up zapping MOSFETs really quickly if it's really imbalanced, but if it's "mostly" 50%, you can wind up with the FETs just occasionally blowing up for no apparent reason.
     
  5. pyrokid73

    Thread Starter New Member

    Jul 28, 2011
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    Alright, resistors are next to the optocouplers. I got my hands on another scope probe, so I am now able to test two things at once.

    I now know why the voltage applied was messing up the signals. You were correct in your assumptions, i didn't have enough dead time, actually none at all. The gate signals are crossing each other by about 500ns.

    Correct me if I'm wrong, but isn't the 4.7 Ohm resistor in the MOSFET datasheet just to show what they were using? The gate driver specifies that you CAN use a resistor if you need a slower rise or fall time. Although this would solve some issues of the overlapping, it would bring up a whole new issue of efficiency as the MOSFETs would be turning on and off slowly and have a lot of voltage across them while doing so. I need the rise and fall time to be as fast as possible to reduce wasted energy in the form of heat. Any suggestions of somewhere else there could be something added for dead time?

    I am using a TS555IN, which is powered by 3V. This is a low power timer chip and can be used with a Vcc between 2V and 18V.

    For the load, I am not sure exactly what it "looks like," but I may be able to figure that out for you soon if that will help. As of right now, I am just using dummy loads, aka large resistors, this is the main reason I wasn't too worried about the exact 50%, but thanks a lot for the info; it will be good to know when other situations arise.
     
  6. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    The function of the gate resistors and how to dimension them is quite well explained in these two documents.
    http://www.irf.com/technical-info/appnotes/an-978.pdf
    http://focus.ti.com/lit/ml/slup169/slup169.pdf

    I would not try to solve an overlapping problem by changing the values of the gate resistors. It might work, but I'd prefer determine the deadtime BEFORE I pass the signals to the drivers. This way I'm safe and free to play around with gate resistor values, turn-on/off times, snubbers etc.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Good deal!

    Not good. You'll also find that the MOSFETs take a good bit longer to turn off than they do to turn on.

    There's another reason for slowing the rise/fall time on the gates; and that is the parasitic inductance of the MOSFET leads combined with the effective capacitance of the gate itself creates a series LC resonant circuit, that will ring like a struck bell when a rapid transition of the gate voltage level occurs. The resistor in series with the gate drive serves as a snubber to dampen the oscillations.

    That's also why I was asking if your load was inductive. I should have also asked if it is an inductive load, was current flow continuous or discontinuous.

    If no current is flowing through an inductor, and you turn on a switch to supply current to the inductor, there is no power loss through the switch initially because there is no current flowing through it. However, as current starts flowing through the inductor, if the switch has resistance, power will be dissipated in the switch as the current increases.

    Same thing with a MOSFET; if current isn't flowing when you turn on your MOSFET switch, you can pretty much take your time about it, as the inductive load will be dropping all the voltage initially. It's the turn-off that needs more attention in that case.

    I don't know where you are with the design stage - are you still on a breadboard? (lousy way to try to test something like this, btw - far too many parasitics to deal with) If you're on a PCB already, it would help to see how it's laid out.

    I'm curious why you went with 3v? I saw that in your schematic earlier, and I asked if you were actually operating it on 3v. I didn't know that anyone made a 7803; the lowest I've seen is a 7833 (3.3v).

    OK, but if they are wirewound resistors, they will have some inductance unless they specifically state "non-inductive"; the non-inductive wirewound resistors are made differently from the standard ones.
     
  8. pyrokid73

    Thread Starter New Member

    Jul 28, 2011
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    I read up a little more about the gate resistor and added a 4.7Ohm resistor, which cleaned up the gate signal while voltage was applied to BUCK-OUT.

    I was not trying to solve the problem with a gate resistor, although the small one I am using cleaned up the ringing in the signal a larger one may slow the rise and fall time of the MOSFET, adding power losses due to the MOSFETs being a "resistor" more of the time. Is there something simple that I could add in before the gate drivers, or even before the optocouplers that would help separate the gate signals that the gate drivers are producing?

    As I said just now, added the 4.7Ohm resistor which took out the ringing.

    I was using a PCB, which I am still using, but adding stuff to it, just sky wiring everything until I can get this working correctly.

    With the 555 timer chip, as I decreased voltage current also decreased, and I was trying to use the least power with the circuit. The regulator I am using is a 3.3V. I had another one that was 3V that I thought I had been using, but once I actually looked it was a surface mount that I didn't want to use (FYI: they also have a regulator that outputs 2.6V).
     
  9. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    The simplest thing I can think of right now is having two schmitt trigger inverters in series, and in between them a resistor, a diode and a capacitor.
    The capacitor to + rail and input of the second inverter, the resistor between the two inverters, the diode in parallel with the resistor pointing towards the second inverter.
    The RC gives you a delay for your rising signal, the diode is there to maintain the fast turn-off.
     
  10. pyrokid73

    Thread Starter New Member

    Jul 28, 2011
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    Can you put up a schematic of this because I can't figure out how this would be hooked up? This seems like it could work, but I can't figure how it would be used. Any other suggestions are welcome also.
     
  11. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    The FAN already has schmitt-trigger inputs but it also has pull-up/down resistors, so I can't use them directly :-/

    I used a 4050 which is just a hex-buffer, I think there won't be a problem since your driver signal is already "digital". The left side of the circuit goes to the 1R resistors.

    (If you want to use a schmitt trigger you could use a 40106 which is an inverter, but then you'd have to swap the FANs (fan3121 for fan3122 and fan3122 for fan3121).)

    I didn't use two 40106s because you'd need 8 gates which is 2 ICs.
     
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