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High Current MOSEFT Circuit Resistors

Discussion in 'General Electronics Chat' started by Silntknight, Nov 5, 2010.

  1. Jaguarjoe

    Jaguarjoe Member

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    You got the right parts.
    When the uP output is low, the LED in the opto is off. This puts the photoxistor in the dark which means it has a very high resistance. That means the collector (output) of the PT is high. A high input to the gate driver will be inverted, making its output and the gate of the FET low.
     
  2. Silntknight

    Silntknight New Member

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    You are wiring it differently from me. Go back to my original diagram. I wired it on the low-side, I think (if I am using that term correctly). I connected the gate driver to the PT emitter. When I connected it on the breadboard, it didn't invert. If I switch it, I should get inversion? That would mean I have to shift around a lot of stuff. Can you upload a schematic for that? I'll test it on the breadboard to see if it works. If it does, I'll switch it to be like you say to have it (so I don't need new parts). Otherwise, I'll just get a non-inverting gate driver. I didn't know inversion depended on where the driver input was placed.
     
  3. Silntknight

    Silntknight New Member

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    Another thing, do I have to connect all the components in a circuit to the same ground pin (same ground potential of course)? For example, in this schematic: http://www.electro-tech-online.com/custompdfs/2010/11/Hall-US1881EUA.pdf (the 3-wire automotive circuit). All the components' grounds are wired together before going to the same ground pin. Can I just have all of them connected to a ground plane which is connected to the same ground pin? It just seems like a larger trace version of what they depict in the schematic.
     
  4. dave

    Dave New Member

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  5. Jaguarjoe

    Jaguarjoe Member

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    The gate driver performs its own inversion independent from the opto.
    You can wire the opto 4 ways:
    The LED anode is connected to +5v thru a current limiting resistor and the uP output grounds it. It will light when the uP output goes low.
    The LED cathode is connected to GND thru a current limiting resistor and the uP takes it to 5v. It will light when the uP output goes high.
    The PT's collector is connected to 12v and its emitter goes to gnd thru a resistor. The output at the emitter goes high when its lit up.
    The PT's emitter is connected to GND and the collector goes to +12v through a current limiting resistor. The output at the collector goes low when its lit up.

    Grounding is tricky. I worked at one place where they ran all of the component grounds to one big ground stud on the chassis. This cured their ground loop problems.
    You are not generating much current on the uP board so I wouldn't fret over it. When you have high currents on the board you have separate grounds for each high current device.
     
  6. Silntknight

    Silntknight New Member

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    I like the first setup. That shouldn't add complication to my board. I also do not need to get non-inverting drivers! I didn't know there were so many ways to wire an opto. The same 150 ohm resistor I have now is the current limiting resistor you mentioned, right?
     
  7. Jaguarjoe

    Jaguarjoe Member

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    I don't recall 150 ohms, but after 13 pages of posts things get blurry. To determine the current limiting resistor for the LED you need to look at the opto's data sheet. Find Vf and If. Vf is probably around 2 volts, If should be 10ma. R = E/I = (Vsupply - Vf)/If. For example, R = (12-2)/0.01 = 10/0.01 = 1k ohms.
    The load resistor (Rl) for the PT half of the opto is a little bit trickier. We need the PT to saturate in order to be a good switch. This occurs when Vsupply < (Rl)(Ic). For Vsupply = 12v and Ic = 1ma(a typical value), Rl > 12/.001 = 12k or more. 15k would be OK.
    The opto runs a tad faster if you ground the emitter and put Rl up to Vsupply.
     
  8. Silntknight

    Silntknight New Member

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    Well, I guess I should have gone with a ~65 ohm resistor, 65=(3.3-2)/.02. In any case, I know that the setup I have not works because I breadboarded it and it works. I switched the anode to +3.3V and the cathode to the pin, like you said, and that solved the inversion problem.

    Anyway, I still have a problem with heat dissipation. For this project, I know I need to dissipate about 5W of heat, at most. I've decided to use a heatsink/fan combo that I already have. It is actually used for a CPU so I know it can dissipate enough heat. I will use aluminum or copper heatpipes from the MOSFETS to the case. The heatsink/fan will be mounted on top of the case. I might even throw in a Peltier junction.
    For another project, I am using this (http://www.infineon.com/dgdl/IPD09N...90004&fileId=db3a304412b407950112b429dca44214) MOSFET. I need to dissipate about 30W of heat continuously. The package has a rated dissipation of 63W, so does it need a heatsink? What does the power dissipation rating (Pd or Ptot) mean?
     
  9. Jaguarjoe

    Jaguarjoe Member

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    Last edited: Nov 27, 2010
  10. Silntknight

    Silntknight New Member

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    For the second case (30W dissipation), can I simply parallel 3 MOSFETS to get about 3W dissipation for each MOSFET? I believe the current is shared roughly equally, so W=(42/3)^2*.015=2.94W. That kind of dissipation can be easily dealt with using a small bolt-on heatsink with a fan for all 3 MOSETS. Does that sound right?
     
  11. Jaguarjoe

    Jaguarjoe Member

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    Yep, parallel them together. Only caveat is driving the gates. All of the gate capacitances add up making the trio harder to drive than just one mosfet.
     
  12. Silntknight

    Silntknight New Member

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    The second project (using parallel MOSFETS) is controlling high-torque motors, so capacitance isn't a problem. I don't need high switching times, largely because the motor will be almost constantly on at 42A (stall). As long as voltage doesn't add (and it doesn't in parallel), then I am fine. The MOSFETS are logic level, so the only worry is whether or not the Arduino can drive the MOSFET. It should be able to easily if they are in parallel.
     
  13. Jaguarjoe

    Jaguarjoe Member

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    The mosfets need to be totally on or totally off. The region in between on and off (triode region) is where the heat is generated so we switch as fast as possible through the triode region. The AVR in the Arduino supplies/sinks 25 ma (I think). It will drive the mosfets on or off sooner or later (because it takes time to charge/discharge the gate capacitance and get through the triode region). If the mosfets get hot, add gate drivers, or just put them in now, they're cheap.
    Motors are an inductive load, don't forget a big back emf diode. Its current rating must be equal to the motor current-42A.
     
  14. Silntknight

    Silntknight New Member

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    AFAIK, you are right about the Arduino. I might add a gate driver because 1.5A should be plenty of current (compared to 25mA). I can't seem to find a 42A Zener on DigiKey, though. I hadn't forgotten about back EMF actually. I was planning on using this one 1N5352BGOS-ND. I thought back EMF was high voltage, low current and only occurred from sudden collapse of a magnetic field. There shouldn't be any sudden collapse in this application. Could I put several Zeners in parallel? That *should* work.
     
  15. Jaguarjoe

    Jaguarjoe Member

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    If you are running the motor in one direction only, you do not need a zener. A silicon rectifier will work like a 1N5406. Its good for 200 amps for 8.3ms (Ifsm). They're about 10 cents each. Another device to look at is the 1N6272 series of zener diodes made to absorb spikes. I don't know if you already bought the zeners for the other project but it would be good to look into these:

    http://www.electro-tech-online.com/custompdfs/2010/11/1N6272.pdf

    An inductor will do all it can trying to maintain the flow of current that existed prior to the transistor turning off. Voltage across it will keep rising trying to get to that current. Its only for a short while though.
     
    Last edited: Nov 29, 2010
  16. Silntknight

    Silntknight New Member

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    I looked at the Zeners I bought and I realized that they aren't rated for 7A of current. In the datasheet (http://www.electro-tech-online.com/custompdfs/2010/11/1N5333B-DPDF.pdf) it says that they are rated for .9A of current, maximum non-recurrent surge current (they are the 1N5388BG at the very bottom of the list). Even the part you sent me isn't rated for that much current at 200V. In fact, according to the datasheet I would need a 30V Zener to absorb 7.5A for 1ms. I don't know how long the pulse is though. I can't seem to find a way to get 400V worth of Zeners in series and 7.5A of maximum current. With my Zeners, I get 2A maximum, and yes, I already ordered them. Should I just try using these Zeners with an ignition coil? They are adequately rated for the fuel injectors.
    Otherwise, I found an ignition coil driver (http://www.st.com/stonline/books/pdf/docs/9671.pdf), but it won't run properly for my application. I need it to run on 12V. It is limited to 7V, which is strange. Still, I'd rather keep to the current setup if I can find a way to manage the higher current. I *think* the diodes I have will be fine, but I'm not sure.
    For the other project, why would I need a rectifier? To prevent backflow or something?
     
  17. Silntknight

    Silntknight New Member

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    This part actually seems pretty good. The MegaSquirt uses them, so I could too if I wire it in the same way: VB921ZVFI and http://www.megamanual.com/ms2/megasquirt_ShemV3.00.pdf. I wouldn't need the gate driver between the circuits, but I'm not sure how I'd make the circuit work. Somehow, I'd need to limit that voltage to 8V. I know there is a maximum input current of 10mA, so can I just put a 800 ohm resistor in series with the gate on the ignition driver? This is assuming that I can even get the part. It appears to be rather rare.
     
    Last edited: Nov 29, 2010
  18. Jaguarjoe

    Jaguarjoe Member

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    I think I mentioned that VB921 part a few pages ago, but its kinda blurry.

    Vin refers to the control signal voltage coming from your uP. If you look further down the specs you'll see that it is 4.x to 5.x volts. You'll need a level shifter to go from 3.3 to 5 volts, not to exceed 8 volts. It requires 10ma of signal current to work.

    The output stage of this part automatically maintains 7.5 amps of current through the coil. The output stage has a 2.5 volt drop across it. The coil will be ~0.5 ohms. At 7.5 amps thats E = IR = 7.5*0.5 = ~4 volts across the coil. 2.5 + 4 = 6 volts required to operate. This will work with the car battery.

    This part that you mentioned in your other post is just about the same as the VB921 part. Either one will work:

    http://www.st.com/stonline/books/pdf/docs/9671.pdf

    Vcc refers to the power for the logic built into the chip. It wants to be 5 volts. There is no spec for main coil voltage, but I'm sure its made for a 12 volt system.

    A 400 volt zener could be troublesome. The ones I mentioned could flow 6 amps in pairs but that's only for 4 cycles/minute. Your engine won't run that slow :)

    Here is a IGBT coil driver transistor. It has a gate instead of a base so it is a direct replacement for a mosfet. As a bonus, it has its own zener built in.

    ISL9V5036S3S

    This is an obsolete part but Cross Components in Florida has many of them.
     
    Last edited: Nov 29, 2010
  19. Silntknight

    Silntknight New Member

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    Well, if Cross Components carries it, I am willing to get it. "Without the need for external components," that's a phrase I like to hear. The MegaSquirt uses an IGBT system (the one previously mentioned), but it has its own logic circuits and stuff. I am doing all the programming on my end, so this IGBT should be great. Internal clamping means I don't need more Zeners, correct? I am going to order two tonight. I like to have spares.

    Also, when you mentioned the rectifier diodes, 1N5406, you meant that as a normal diode would operate- to prevent reverse polarization, right?
     
  20. Silntknight

    Silntknight New Member

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    It has a gate-emitter voltage of 10V, but I am driving it with 12V. It says continuous voltage of 10V, so is 12V OK or should I use a small voltage divider? Also, I should use a Mica washer to insulate the tab, I know, but where do you get those?

    Oh, and a MOSFET gate driver can also drive an IGBT gate without damage? I don't see why not.
     
    Last edited: Nov 29, 2010
  21. Silntknight

    Silntknight New Member

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    Also, I'm thinking about this part ISL9V3040P3 because I know it clamps at 400V and the tab is not electrically active. What you you think about coil-on-spark?

    Going back to ignition coils, I read that ballast resistors are bypassed during cranking but are introduced into the circuit during normal operation. That seems like a rather complex circuit to design. Is it necessary?
     
    Last edited: Nov 29, 2010

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