High-Power switching using MOSFETs

Discussion in 'The Projects Forum' started by Nuggetchris, Aug 7, 2018.

  1. Nuggetchris

    Thread Starter New Member

    Aug 7, 2018
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    For a project, I need a means of switching a high-current circuit between two inductors. The purpose is, each inductor serves as an electromagnet, and they are switched every 15s or so in order for one of the two to cool before starting again. The system I've designed uses arduino as a controller and a couple of MOSFETs wired inline in order to electronically switch the circuit. The current delivered to each inductor should be around 75A at 12VDC. In this specific example, I used three 35A rated MOSFETS in each 'bank' which would function similar to a single relay. Here is the schematic:

    [​IMG]

    Bank B of MOSFETs is a splice from green line to Bank C. The idea is that Bank A will act as a supply switch (Open = NO Power), Bank B (Closed = IND1 ON, IND2 OFF), Bank C (Closed = IND1 OFF, IND2 ON). **The supply switch acts to isolate IND2 from IND1** . To the contrary, with not having the supply switch and splice, the only way I could supply power to the second inductor is if I were also supplying it to the first. That is why I have included the leftmost *third* bank instead of a simple 2, since I need to bypass power past the first inductor.

    I am asking for a few aspects:

    1) I feel I need more diodes in the circuit to protect against kickback. Have the diodes been added in the right places?

    2) Are all the connections correct for the switches and the circuit to function? (Specifically the MOSFET banks and Source/Drain)

    Here is an image i've found of an example NCH circuit:

    [​IMG]


    It would appear from this image, that I *do* in fact have the connections of source and drain for each MOSFET wired correctly, with source on the positive end and drain on negative end. Please clarify if it is wrong.

    3) Is there a solution to the issue of MOSFETs not turning on at the same VGS, leading to overload/explosion?
     
  2. AlbertHall

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    Jun 4, 2014
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    I don't see why you need the leftmost mosfets.
    In your circuit the mosfets will not get the correct gate/source voltage and will not work correctly.
    Connect them as in your second diagram. Source connected to battery negative with the coil(s) connected between the drain and battery positive.
    Then you will also need a connection between arduino negative supply and battery negative.
    The mosfets will need to be able switch on fully with just 5V between gate and source. Check the gate voltage specified for the Rds on.
     
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  3. Nuggetchris

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    Aug 7, 2018
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    Hi Albert, I’m a little confused at why the mosfets would not get the correct gate/source voltage. What I gather from your claim of positioning would lead me to believe I should move the mosfets onto the negative side of the circuit past each inductor. Can you clarify?

    And the point of the leftmost mosfets is described, to allow me to bypass past IND1 to IND2 without operating IND1. If I only had two switches, the only way I could deliver power to IND2 is along the lines that power IND1.
     
  4. AnalogKid

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    Nope. Correct in terms of positive and negative, but not in terms of how you want the circuit to function.

    In the 2nd schematic, the FET is acting as saturated switch pulling the load down to GND. In the first schematic, the FETs are acting as source-followers, a non-satured condition that will cause massive heating in the transistors. Depeiding on the FET type and current sharing, the "on" voltage across the Fets could be 4 V or more. Either change to p-channel FETs acting as saturated switches that pull the load up to V+, or (better) reconfigure the circuit with the n-channel FETs switching to GND.

    Also, what is the output voltage from the Arduino? For the n-channel FETs to turn on correctly, it should be 10 V above the source, or approx, 22 V. This would cause saturation, which is what you want, and then you wouldn't have to change the circuit configuration.

    ak
     
  5. AlbertHall

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    The leftmost mosfets switch the battery onto the green wire which feeds the drains of both the other groups of mosfets. If you ditch those left hand mosfets and connect the battery positive directly to that green wire you have exactly the same functionality.

    When the mosfets are switched fully on the source voltage will be the same as the drain voltage - that is at +12V. The gates need to be positive of that to turn the fets on, say +17V. How is the arduino going to do that?

    In your second diagram with the sources connected to battery negative the source voltage doesn't change whether the fets are on or off. So the arduino can easily supply 0V or +5V to switch them off or on respectively.
     
  6. Nuggetchris

    Thread Starter New Member

    Aug 7, 2018
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    This is great feedback. Upon reanalyzing my circuit I noticed that Albert was right, I will eliminate the leftmost mosfets. Secondly, it's worth noting that arduino will supply 0-5V to switch the mosfets, so Albert is also right in directing me to mimic the second diagram. My problem however, is figuring out how I will draw the schematic.

    In the second diagram, the 'switch' (which would be arduino in my case) is integrated WITHIN the circuit. If I were to integrate an arduino micro controller into a circuit carrying 75 amps of power I would be essentially creating a homemade bottle rocket. How would I integrate arduino like in the second diagram? And lastly, is this the same case that AK is describing?
     
  7. AlbertHall

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    You will need to either:
    Find mosfets which can be fully turned on by 5V and handle the current - I don't know whether such are available.
    Or use a driver to turn the 5V from the arduino into 12V for the mosfet gates. This could also be isolated to keep 75A away from the arduino circuit.
     
  8. AnalogKid

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    Yes.
    Note - in order for the Arduino to control the FETs, they both must have a common voltage reference (ground). There are specialized isolated FET driver ICs, but I don't see any need for them based on what you've described so far.

    1. What is the power source for the Arduino?

    2. Do that power source and the +12 V magnet power source share a common ground?

    3. If they don't, is it a problem to make them? (system wiring restrictions, connector pin limits, etc?)

    ak
     
  9. Nuggetchris

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    Aug 7, 2018
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    The power source is a single 1000W power supply, with 12V lines leading to magnet and a 5V line leading to Arduino.
    Technically they should share a common ground. However they were on separate circuits as shown.
     
  10. AnalogKid

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    In order for the arduino to control the FETs. both circuits must share a common ground; or, use some kind of isolation technique, such as opto couplers, isolated FET drivers, etc. I vote for a common ground. If the 5 V and 12 V outputs have isolated grounds at the supply, connect them together as close to the load as possible. This will prevent magnet current from modulating the ground potential seen by the arduino and its control signals.

    ak
     
  11. KeithWalker

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    Jul 10, 2017
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    FET SWITCH.jpg
     
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  12. AnalogKid

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    KW - I strongly recommend using reference designators and connection dots in your schematics. It makes it easier to discuss that you have two lines crossing the 12 V - line, but only one of them is a connection. Also, the 2N3904 is an NPN transistor. The part type and connections are correct, but you show a PNP symbol.

    The 10K pullup resistors will turn on the FETs so slowly that there is no need for suppression resistors in series with the gates.

    ak
     
    Last edited: Aug 7, 2018
  13. Nuggetchris

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    Aug 7, 2018
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    Wow... KW good job and thank you for the effort.

    Can someone give me a quick debrief on the use of 74LS05, 2N3904, and the need for those pullup resistors? I also don't see any gate driver or am I mistaken?

    I was working on this schematic in the mean time:

    [​IMG]

    But if KW's does the job I'd be happy to go off that one instead!
     
  14. ebp

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    That should be just 78L05 - it is a common 3-terminal voltage regulator.

    Slow turn-on of the FETs is OK provided that the inductors are discharged at turn-on time, since the current will start at zero. Fast turn-off should be used since the power dissipation in the FETs will total 900 W during the transition, so prolonging the transition isn't a good idea.

    Using NPN transistors driven by microprocessor outputs to turn the FETs off is somewhat perilous. It is necessary to add passive pullup at the outputs of the processor to assure the NPNs are ON during processor reset. The NPNs will be off, allowing the loads to turn on, when the supply voltage to the processor and the pullup resistors is low. This is normally a transient condition but it must be considered. The loads are not blinkenlights where a brief flash is probably of zero consequence. A carefully designed gating scheme using three outputs from the processor, two at one assertion level and the third at the opposite assertion level (e.g. the individual control lines must be HIGH while a common enable line must be LOW) is a better approach for high-power loads or things than mustn't turn on spuriously. The output states during power up and power down and when RESET is asserted must be carefully considered. Bad gate drive levels during power up or down can result in destroyed FETs (probably short-circuiting, meaning the loads get turned on and can't be turned off unless the power is turned off).

    Freewheeling diodes will result in slow discharge of the inductors, which may or may not be acceptable. If faster discharge is required then an alternative scheme for clamping the voltage (at higher level) is required and the FETs must be rated appropriately. If a single freewheeling diode is used, a rating of 25 A average current is probably sufficient since the duty cycle of the diode will be very low.

    FETs for switching should almost always be selected on the basis of ON resistance rather than current rating. Current and power ratings are often not realistic in practical circuits.
     
  15. KeithWalker

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    Jul 10, 2017
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    The "74LS05" is a misprint. It should be "LM7805" 5 volt regulator. Make sure that the input to the voltage regulator and the regulator ans Arduino grounds are connected directly to the 12V supply terminals to minimize interference through lead losses.

    The 2N3904 is an NPN low power transistor. (in my haste, I used the wrong transistor symbol. The arrow is in the wrong direction.

    The pull-up resistors make sure that there is enough drive voltage (12V) to turn the MOS FET gates on fully to minimize their power dissipation.

    This is a simple low speed application so there is no need to use FET gate drivers. If you do use them, make sure that the Vdrv is high enough voltage (e.g. 12V) to turn the FETs on fully.

    NOTE: LOW outputs on the Arduino digital pins will turn the loads ON and HIGH will turn them OFF.

    I corrected the circuit diagram and added a diode and capacitor to protect the Arduino from switching transients.

    FET SWITCH.jpg
     
    Last edited: Aug 7, 2018
  16. Nuggetchris

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    Aug 7, 2018
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    So, in summary:

    LM7805 works to convert the 12V system voltage to 5V in order to power Arduino within the same circuit as the working components.

    The NPN transistor is used to amplify the digital output signal from ARD in order to supply enough power to switch the gates.

    What is the small line in the top left @ 25V?

    And lastly, wouldn’t the schematic need a small diode loop around the inductor to account for kickback?

    We’re about there let’s finish this up.
     
  17. AnalogKid

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    I agree with epb - having inverting driver transistors means that there is a chance the system will power up with all FETs on. One solution is to change to logic-level FETs and drive then directly from the arduino. I still see no need for gate resistors.

    ak
     
  18. KeithWalker

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    Jul 10, 2017
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    If you are using MOS FETs that only require a gate voltage of 5 volts or less to fully saturate them, you can dispense with the transistors and drive the FETs directly from the Arduino but most FETs need 6 volts or more to fully saturate them. See the datasheet, or check what is the minimum gate voltage you need for lowest drain to source resistance by measuring the resistance between the drain (+) and the source (-) with a multimeter when you apply a variable voltage between 4 and 7 volts between the gate (+) and the source (-).

    I do not see the small line @25 that you ask about.

    You will need diodes across the loads to prevent damaging fly-back voltages. Most power MOS FETs have protection diodes built in nowadays, but if the ones you are using do not have them (check the spec sheet) you will need diodes across each set of FETs too.

    ak[/QUOTE]
     
    Last edited: Aug 7, 2018
  19. AlbertHall

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    You mean the capacitor 25uF 25V? The 25V is the voltage rating for the capacitor.
     
  20. ebp

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    It isn't difficult to find FETs that will dissipate on the order of 10 W at 75 A with 4.5 V gate-source (that is, for a single FET). With most logic level FETs you will reduce the ON resistance by 20% or so with Vgs of 6-8 V versus that at 4.5 V. You can actually easily find lower ON resistance in surface mount parts than in through-hole, but SM makes managing current and heat harder.

    As I mentioned previously, a free-wheeling diode across the load will slow turn-off, but I suspect that is irrelevant for the application. Without knowing the inductance it is impossible to know if the body diode of the FET(s) can be safely avalanched. If you use a (say) 30 V FET, it will probably avalanche at something around 35 V. At 75 A, that is about 2.6 kW at the start of avalanche, which is rather a lot. A 100 µH inductor would store 281 mJ at 75 A, which is within the rating of many high-power FETs if the die starts out at a very modest temperature.
     
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