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MOSFET 12VDC Relay?

Discussion in 'Power Electronics' started by adamq, May 4, 2021.

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  1. adamq

    adamq

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    Dec 17, 2013
    I'm trying to traditional relays with something more modern and efficient.

    The signal voltage is 12VDC. The design load is 12VDC/10A.

    Efficiency is a big priority. I see there are a bunch of inexpensive pre-made "arduino" type MOSFET driver modules which would work for this application. Unfortunately most seem to use IFR520 MOSFETs which introduce voltage drop and wasted power.

    Any thoughts?
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Trying to drive, trying to replace?
    Why do you think any other MOSFET wouldn't? To varying degrees any MOSFET has losses. Even a seemingly "perfect" relay contact has some resistance (if only a few mΩ) and will therefore have losses.

    The IRF520 has a max. RDSon of 270 mΩ. But MOSFETs with much lower RDSon are available, e.g. for automotive applications.
     
  3. adamq

    adamq

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    Dec 17, 2013
    Woops. Trying to "replace" traditional relays.

    Right, the IFR520's 270milliohm causes a comparatively higher voltage drop than other MOSFETS.

    I've been looking over datasheets. Some are in the area of 12 milliohm.

    I agree that every relay has a power cost. I'm looking for recommendations on how to make it most efficient :)
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    The irf520 isn't good for 10 A anyway.
    These modules for example use a D4184 MOSFET with less than 10 mΩ RDSon worst case.
     
    adamq likes this.
  5. adamq

    adamq

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    Dec 17, 2013
    Well, my anticipated load is actually 5A but I am hoping to build multiple circuits (and spares) that will meet all my power needs.

    This is for a fully electric sailboat.

    The D4184 looks like a great choice. Thank you!
     
  6. adamq

    adamq

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    Dec 17, 2013
    In my mind it seems impossible to use the same VGS as VDS when switching a MOSFET. In this case, 12VDC for VGS, and also for VDS, but it could be any voltage. Am I missing something?

    This is my thought:

    MOSFET is off. VDS is 12VDC and VGS is 0VDC. 12VDC is applied to the gate, and now VGS becomes 12VDC. MOSFET turns on. VDS becomes basically zero. At this point, VGS is also becomes zero.

    Won't the the MOSFET end up in a continuous off-on cycle?

    In that case, is there an elegant workaround for continuous "on" duty?
     
  7. adamq

    adamq

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    Dec 17, 2013
    For other readers:

    The answer is yes, it's a problem. Google n-channel MOSFET high side switching.

    For on-off cycling, the solution can be "bootstrapping" the gate with a capacitor, but it appears that it won't work for continuous duty as the capacitor will eventually drain.

    For continuous duty, it looks like IC to drive the MOSFET gate is the elegant solution.
     
    Last edited: May 4, 2021
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Why should it? That would happen if you connected the Gate to the drain, but that is not how you drive a MOSFET. The principle is like this:
    upload_2021-5-4_19-10-21.png
    The driver is able to maintain VGS[SUB] > 0 V even with V[SUB]DS ~ 0 V.
    No its not, see above. Although high side switching is an issue, but it is not the same as you described.
    An IC is only an integrated realization of a boost technique that can be realized by a discrete circuit. Of course it looks elegant, because you need only a chip and a few peripherals instead of a whole bunch of components. But if you happen to have another, even higher voltage available. you wouldn't need a specializes high side driver IC:
    upload_2021-5-4_19-20-17.png
    Such a voltage could be present in the system for whatever other purpose. Or it might be generated once to support a few high side drivers without having to resort to specialized high side driver ICs.
    You see, being as specific in your question and supplying details can prevent misunderstandings and misconceptions on our side (and consequently on yours).

    Remember: always look ak the voltage differetials (VGS, VDS), not at the absolute voltages versus ground.
     
    adamq likes this.
  9. adamq

    adamq

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    Dec 17, 2013
    I am obviously misunderstanding something then.

    More information: it's a 12VDC circuit only. No higher voltage is currently available. The gate is controlled with a single pole toggle switch. The MOSFET must be on the high side of the load.

    If I am switching a 12VDC load, and providing 12VDC to the gate, once the MOSFET is on, won't the voltage differential be less than the threshold (nearly 0V)?

    I understood the voltage differential between gate and load should remain at around 10V for the MOSFET to be at its lowest resistance.

    I did consider a boost converter to send 20V to the gate. That would leave 8V VGS differential, which is pretty good but not the lowest RDS(on) according to the datasheet.
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    finally ...
    So you'll need a high side driver, right.
    It is actually between gate and source. In your case as the source is connected to the load it is the same. Nevertheless stick to correct terminology for a clear understanding on all sides.
    The actual value of the voltage depends on the specific MOSFET used. You'll have to examine the datasheet. There are also MOSFETS for application in digital circuits that require only around 2 V - 3 V gate drive voltage (Logic level MOSFETs).
    This is essentially what a specialized high side driver IC does.
     
    adamq likes this.
  11. adamq

    adamq

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    Dec 17, 2013
    OK... I came up with this schematic.

    To turn on the MOSFET, the mechanical switch is closed. This sends 6V to the gate via a Zener diode and resistor.

    The MOSFET is now on. Since the system voltage is 12V, and the gate is at 6V, there still remains 6V potential between gate and source. From my reading today, it doesn't matter that the gate is lower than the source. The MOSFET will remain "on" until the gate drops below the threshold voltage.

    To turn the MOSFET off, the mechanical switch is opened. This causes the gate voltage to bleed to 0 via the resistor.

    Right? Thoughts?

    Thanks for your patience :)

    relay_mosfet.png
     
    Last edited: May 6, 2021 at 6:43 PM
  12. PETERDECO

    PETERDECO

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    Dec 19, 2019
    Question: I haven't followed this thread from the beginning but if you're using a switch to turn on a MOSFET to turn on a motor, why can't you just use the switch to turn on the motor?
     
    adamq likes this.
  13. adamq

    adamq

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    Dec 17, 2013
    Good question.

    This is for a sailboat. The only source of power is 200 watts of solar. On long passages, every watt is critical.

    Right now, automotive relays are being used. But they have drawbacks, including drawing a couple watts each while in use (about 2-3 watts each). There are 5 onboard right now. 15W is a significant portion of the available power which could be better used for lighting, safety equipment, communications, navigation, etc.

    The control panels for many devices (pumps, refrigeration, winches, etc) are located sometimes 40ft return trip away from the device.

    That means dealing with voltage loss. Big conductors aren't really an option, because of space and cost. Tinned marine wire is spendy, but also it's not good practice to have "large" currents running back and forth (chaffe, corrosion, space).

    That's why automotive-type relays were traditionally used. However we're now in 2021 and there's surely a better way.

    I believe with the right components it can be made much more efficient, solid state, reliable, affordable, and allow for drop-in replacement for automotive relays (which unfortunately means high-side switching).

    A commercially available DC SSR would be another option, but the efficient ones are very expensive. A plain-Jane DC SSR is cheap, but lacks in efficiency.

    If I can figure out this circuit, it should be able to accomplish the same goal at a fraction of a Watt and very inexpensively. It'll replace the current relays, but it will also be used on every new electrical appliance I install. It'll mean the ability to use small conductors for the signal, which is super useful when routing wires.

    As an aside... SSRs and Automotive relays also decouple/isolate. Every conductor in this implementation is fused, including the signal/control wire. So, it's not a critical feature. Nonetheless, I may consider adding an optocoupler once I get this figured out.

    There are a lot of DIY SSRs out there, but none of the ones I've found would work as a drop-in replacement for this application.
     
    Last edited: May 6, 2021 at 7:16 PM
  14. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    @adamq :
    #1: you have the zener diode backwards. In this orientation it works as a normal diode with ~0.7 V pass voltage.
    #2: with the switch on, the gate voltage will be ~ 12 V. The MOSFET will start conducting until the source voltage reaches 12 V - VGSth with VGSth being the threshold voltage of the MOSFET. According to the datasheet VGSth is between 1.2 V and 2.3 V. Therefore the source voltage will rise to between 9.7 V and 10.8 V only, not 12 V as you expect. To achieve this, you need a gate voltage of 12 V + VGSth i.e. 14.3 V minimum, better 15 V or more for good conductance of the MOSFET.

    No, see above. The source voltage will rise when the MOSFET conducts, thus lowering VGS. As I mentioned in post #8: it is the potential difference (voltage) that counts.

    Where did you read this? This is, pardon mem nonsense. With VGS < VGSth the MOSFET will be off.
     
    Tha fios agaibh likes this.
  15. adamq

    adamq

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    Dec 17, 2013
    Hmm... Well, dang.

    None of the gate driver ICs I found were rated for continuous duty. Bootstrap also can't do it.

    It seems like the next solution is a voltage doubler with a either a 555 timer or 4049.

    The gate current of the MOSFET at 20V is 100nA. I believe either IC is up to the task.

    Any other thoughts?
     
  16. Harald Kapp

    Harald Kapp Moderator Moderator

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    If you need a high side switch, why don't you use a p-channel MOSFET?
    upload_2021-5-7_6-28-24.png
     
  17. adamq

    adamq

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    Dec 17, 2013
    That's an idea. The thought to use a p-channel did occur to me. I also considered low-side switching with an n-channel. The problem is that neither of these options are "drop-in" compatible with the current setup.

    The issue with high-side P-Channel: Presently, the 12V relay is normally-open. When the corresponding SPST toggle switch is closed, so does the relay. The opposite would be true for the p-channel MOSFET. Yes, it's just a matter of rotating the toggle switch 180 degrees (on is off, off is on), but I wonder about current leakage at the mosfet, and the idea of multiple live conductors 24/7/365 is not appealing (even though they are all fused).

    The issue with low-side N-channel: The relay currently switches the positive conductor to the load. This is the standard and preferred way to switch devices in boats. It's best-practice, but not mandatory. This would mean the high-current appliance conductor would be live. Low-side switching should not affect the loads, but it's still not a drop-in replacement for the relay.

    I believe they would both be approximately the same efficiency. Every positive conductor is fused, as I mentioned. If I employed high-side p-channel, the gate conductor would be live when the appliance is "off" but I could fuse the gate conductor very conservative tiny fuse, since there should be no current on it.

    Decisions, decisions.
     
  18. Harald Kapp

    Harald Kapp Moderator Moderator

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    Definitely not. Looks like you're mixing up depletion mode and enhancement mode, that is another story, has nothing to do with N-MOS or P-MOS. Learn how a MOSFET works.

    As shown in my diagram:
    - with the switch open VGS = 0 V -> MOSFET off.
    - with the switch closed VGS = -11.9 V -> MOSFET closed.
     
  19. crutschow

    crutschow

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    To turn the P-MOSFET ON when the control signal is high, you can add an NPN driver (example below).

    The MOSFET has negligible current leakage when off (<1µA).

    upload_2021-5-7_7-48-56.png
     
    adamq likes this.
  20. adamq

    adamq

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    Dec 17, 2013
    Let me see if I understand correctly.

    R1 pulls up the gate to keep M1 normally-off.

    When Q1 is on, it pulls down the voltage on M1 Gate, thereby turning it on.

    I guess R3 limits current, while R4 ensures that Q1 receives proper voltage?

    With a low RDSon p-channel MOS, and those 10k resistors, this should be very efficient.
     
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