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no image Thursday, December 07, 2017 by A B

A simple "power resistor" MOSFET module (DC load bank)

Hi everyone,


In order to do a 30kW load bank out of 1kW power resistors, I wanted to fabricate small "control modules" that would commutate each resistor (around 36V/30A on each) through a power MOSFET.

The command would come from a µC with a MUX. Each MUX output would be connected to the entry of an optocoupler that would himself saturate the gate of a N-Channel MOSFET (through a PNP transistor, the optocoupler output being NPN).

I'm attaching a first "draft" that I wanted to submit to you guys in order to have some feedback and maybe know whether I'm completely wrong and out of line with my design (If I can call putting 4 discretes together that).


HERE IT IS

Thanks for the feedback!


A

Comments

  • by  Aubrey Kagan (edited)

    A B

    I am sorry to tell you that I don't think your circuit will work. But before I attempt to get into that let me pose a question to you: are you aware that at 30A each transistor will be dissipating just under 40W. That will get REALLY HOT! Have you considered the heatsinking necessary for 30x40= 1200W. 

    In principle the emitter of your PNP transistor should go to the +12V isolated supply and the collector through a resistor to the gate of the MOSFET. But that is very basic and will probably cause real problems. Firstly the rise (and  especially) the fall times on the gate are going to be very slow meaning that the transistor is going to be in the linear region for many, many milliseconds and as a result it will have a much higher resistance and the heat dissipation will go up astronomically and likely blow the MOSFET.

    I am not an expert on high power MOSFETs, but I think you would need ones with much greater heat dissipation and in packages that can be mechanically bolted to heatsinks. I would also recommend that you use a gate driver IC to speed up the gate transition times.

    We have some self built electronic loads in house that use dirty great MOSFETs, and I must tell you that when they blow it is a real song and dance to find spares and then reinstall them with silicon grease etc.  If you look at my blog on the "Burn In Chamber" you will see the kind of heatsinking you will need.

    Or make it easy for yourself and use relays or contactors! If you socket mount them, it is much easier to replace if one burns out, especially if this is for use in small volumes.

    I just came across this  You-tube videoon driving MOSFETS
  • by  David Ashton
    Not much I can add to Aubrey's comments - he is far more knowledgeable than I in this area.  But I can make one suggestion.  Have 8 loads increasing in steps of 2, ie in your case

    156.25W,   312.5W,   625W,   1250W,    2500W,   5000W, 10000W,   20000W

    By switching them in with an 8-bit port you will be able to select any load from 150W to nearly 40 KW.

    Of course you may not need this level of control....

    I'd also go along with Aubrey's comments on using relays.  You can pick up automotive relays with 12V coils pretty cheaply and they will take 30A.  Unless your supply is very sensitive to short glitches in the load as it changes you will be OK with relays.  You could use MOSFETS for the smaller loads (say up to 1KW) and relays for the larger ones.

  • by  A B
    Hi Aubrey,

    Thanks for the reply,

    I'm afraid I haven't myself understood regarding the MOSFET : it will not be used to dissipate the power but just as a switching element for the 1kW power resistor.


    And in that case, even a very common IRFB3077 with a 3mOhms RdsOn will have an on-state loss of : 0,003*30*30=2,7W. Almost low enough for case dissipation in a TO220!

    The linear region operation will probably not be a problem since they will only rarely switch, only during ramp-up/down, no high frequency switching.

    Regarding the PNP, I might have made a mistake with the schematic and put it the other way around.

    I pondered with relay operation for simplicity's sake but 30A DC even at only 36V are already quite pricey. The cheaper I found were some Gigavac and retail for around 50USD.... a bit on the pricey side!
    Of course, I could use regular 30A AC rated but that's just a recipe for welded contacts, even if they come cheap-ish...


    A

    • by  Aubrey Kagan

      A B

      It seems you are going to go with your (corrected) schematic and I should point out that you need a pull down resistor from the collector (in the revised design) to ground. Without it when the PNP turns off, the gate to the MOSFET will be floating and almost inevitably will put the MOSFET into the linear zone rather than cutoff.  It also means that the turn off will be slow as the gate capacitance has to drain through the series drain resistor and the pull down resistor.

      Remember to purchase several spare MOSFETs!
  • by  Aubrey Kagan

    A B

    It seems you are going to go with your (corrected) schematic and I should point out that you need a pull down resistor from the collector (in the revised design) to ground. Without it when the PNP turns off, the gate to the MOSFET will be floating and almost inevitably will put the MOSFET into the linear zone rather than cutoff.  It also means that the turn off will be slow as the gate capacitance has to drain through the series drain resistor and the pull down resistor.

    Remember to purchase several spare MOSFETs!
  • by  A B
    Hi David,


    Thanks for the answer,

    Indeed automotive relays are quite cheap but they are 12V rated... As soon as you up the voltage above 24V, things start to get expensive...

    • by  Rick Curl
      Automotive relays with 12 volt coils would be fine.  The higher voltage will be on the contacts.  Most of them are good up beyond 100 volts on the contacts. 


      Have you considered solid state relays?  Look Here


      -Rick

      • by  David Ashton (edited)
        Thanks Rick, you beat me to it.  I'd be happy with 30-50V on the contacts of an automotive relay, probably even more.  But those solid state relays you linked to are magic!  40A @ <1.5V drop - that would still be potentially 60W - 45W at AB's desired 30A load - so you'd still have a lot of heat to dissipate.  But for $11.95 it's hardly worth re-inventing the wheel....

        For reliability I'd be tempted to use more loads at a lower power - ie limit your loads to 5KW max.  You can still switch more than one at a time with one processor port pin to make up higher loads. For example you could have 8 x 5KW loads which would let you do 0-40KW in 5KW steps...


  • by  Rick Curl
    Automotive relays with 12 volt coils would be fine.  The higher voltage will be on the contacts.  Most of them are good up beyond 100 volts on the contacts. 


    Have you considered solid state relays?  Look Here


    -Rick

    • by  David Ashton (edited)
      Thanks Rick, you beat me to it.  I'd be happy with 30-50V on the contacts of an automotive relay, probably even more.  But those solid state relays you linked to are magic!  40A @ <1.5V drop - that would still be potentially 60W - 45W at AB's desired 30A load - so you'd still have a lot of heat to dissipate.  But for $11.95 it's hardly worth re-inventing the wheel....

      For reliability I'd be tempted to use more loads at a lower power - ie limit your loads to 5KW max.  You can still switch more than one at a time with one processor port pin to make up higher loads. For example you could have 8 x 5KW loads which would let you do 0-40KW in 5KW steps...


  • by  David Ashton (edited)
    Thanks Rick, you beat me to it.  I'd be happy with 30-50V on the contacts of an automotive relay, probably even more.  But those solid state relays you linked to are magic!  40A @ <1.5V drop - that would still be potentially 60W - 45W at AB's desired 30A load - so you'd still have a lot of heat to dissipate.  But for $11.95 it's hardly worth re-inventing the wheel....

    For reliability I'd be tempted to use more loads at a lower power - ie limit your loads to 5KW max.  You can still switch more than one at a time with one processor port pin to make up higher loads. For example you could have 8 x 5KW loads which would let you do 0-40KW in 5KW steps...


  • by  A B
    Hi Rick,


    Indeed the coil is 12V rated and the the higher voltage on the contacts but a 30A @12VDC will barely cut 1,5A @ 50VDC... tough life for DC contats!

    See there

    • by  Rick Curl

      There are tons of higher current automotive relays out there.  for example: Hasco

      Note that it's good all the way up to 277 volts AC!


      -Rick

    • by  David Ashton
      I wouldn't take maximum contact voltage ratings too seriously below about 50V.  DC is more difficult to switch than AC because it is not self-extinguishing - so higher DC voltages can spark more than AC.  However below about 50V this is not much of a problem.  Take this relay - very similar to all the others but rated at 70A at 75 VDC,  I would reduce the maximum current at higher voltages - on this one at 35V I'd probably use a max of 40 or 50A - but that's just to extend the life.    I think most manufacturers just give 12V because they're intended for use at that voltage. 

      At DC or AC,  Snubbers can be used to reduce arcing across the contacts.  They consist of a low value resistor and a medium value capacitor.  For what you're doing I'd use 10 ohms or thereabouts and about 1 uF or more.  When the contacts open there is a current path through the resistor until the capacitor is charged. They can keep the voltage down until the contacts have moved slightly apart, which can reduce arcing on the contacts.

      • by  A B (edited)

        Hi David and Rick,

        - Indeed there are a lot of higher currents automotive relays but even the Hasco reference you mentioned (nice one BTW) is only rated at 28VDC max (even though 277V in AC!) and for 30A... a bit short.

        - David, regarding contact ratings, I would -at the contrary- tend to be more conservative after having welded one too many contacts in high-current DC switches (>500A) even though the ratings said they were up to the job.
        The relay you linked is very similar to the one I linked up there and while it can indeed switch north of 100VDC (according to the datasheet), the current allowed even at 35VDC will only be around 1,5A thanks to the square law of current...
        The snubber seem like an interesting solution indeed but not very safe since the current will not be "galvanically" interrupted because of that...


        • by  David Ashton
          AB - points taken but the truth lies somewhere in between I think.  Your linked relay you will note has two current ratings (NO/NC - ie normally open (closing contacts) and normally closed (opening contacts).  The contacts can break a lot less than they can make.  Temperature also plays a factor.  But I'd question whether the ratings go down as much as you say.   I tried to find some documentation on this but it is very hard.  None of the relays you, I or Rick cited have ratings at different voltages.  My understanding has always been that up to about 50VDC you don't have to downrate much (but I'm open to correction...) And 500A is a LOT of current at any voltage.

          Seems like you have a choice....use relays and worry about current capacity, or use solid state switches (yours or commercial) and be prepared for heatsinks and cooling.  My strong advice would be to switch smaller powers even if it means more loads and maybe driving several loads from one pin, you're less likely to got too near the limits of your switches, of whatever type.

          Can I be rude and ask what this is for?  It does sound fairly hairy stuff.

          • by  A B

            Hi David,


            Sure :  the load will be used to discharge some battery packs (a few kWh each) and thermally characterize them. (dT vs. I)

            We would have used inverter and inject in the grid ideally but cost is too high for that...

            Regarding derating, in the first page of the datasheet I linked, you can find a DC load breaking capacity curve with load limit 1 & load limit 2. In both cases, @40V, you get a little over 1A at the load limit 1 or around 3,5A at load limit 2.

            While you're pretty much asymptotical (infinite) at 12V...

            About the heatsink and cooling in solid state, I just don't see where they lie since the On-state conduction loss will be that of a few Ws and virtually no switching /linear loss...
            • by  David Ashton (edited)

              AB - sorry, being in Australia means I sleep different times from the rest of the world.

              Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

              Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  Rick Curl

    There are tons of higher current automotive relays out there.  for example: Hasco

    Note that it's good all the way up to 277 volts AC!


    -Rick

  • by  David Ashton
    I wouldn't take maximum contact voltage ratings too seriously below about 50V.  DC is more difficult to switch than AC because it is not self-extinguishing - so higher DC voltages can spark more than AC.  However below about 50V this is not much of a problem.  Take this relay - very similar to all the others but rated at 70A at 75 VDC,  I would reduce the maximum current at higher voltages - on this one at 35V I'd probably use a max of 40 or 50A - but that's just to extend the life.    I think most manufacturers just give 12V because they're intended for use at that voltage. 

    At DC or AC,  Snubbers can be used to reduce arcing across the contacts.  They consist of a low value resistor and a medium value capacitor.  For what you're doing I'd use 10 ohms or thereabouts and about 1 uF or more.  When the contacts open there is a current path through the resistor until the capacitor is charged. They can keep the voltage down until the contacts have moved slightly apart, which can reduce arcing on the contacts.

    • by  A B (edited)

      Hi David and Rick,

      - Indeed there are a lot of higher currents automotive relays but even the Hasco reference you mentioned (nice one BTW) is only rated at 28VDC max (even though 277V in AC!) and for 30A... a bit short.

      - David, regarding contact ratings, I would -at the contrary- tend to be more conservative after having welded one too many contacts in high-current DC switches (>500A) even though the ratings said they were up to the job.
      The relay you linked is very similar to the one I linked up there and while it can indeed switch north of 100VDC (according to the datasheet), the current allowed even at 35VDC will only be around 1,5A thanks to the square law of current...
      The snubber seem like an interesting solution indeed but not very safe since the current will not be "galvanically" interrupted because of that...


      • by  David Ashton
        AB - points taken but the truth lies somewhere in between I think.  Your linked relay you will note has two current ratings (NO/NC - ie normally open (closing contacts) and normally closed (opening contacts).  The contacts can break a lot less than they can make.  Temperature also plays a factor.  But I'd question whether the ratings go down as much as you say.   I tried to find some documentation on this but it is very hard.  None of the relays you, I or Rick cited have ratings at different voltages.  My understanding has always been that up to about 50VDC you don't have to downrate much (but I'm open to correction...) And 500A is a LOT of current at any voltage.

        Seems like you have a choice....use relays and worry about current capacity, or use solid state switches (yours or commercial) and be prepared for heatsinks and cooling.  My strong advice would be to switch smaller powers even if it means more loads and maybe driving several loads from one pin, you're less likely to got too near the limits of your switches, of whatever type.

        Can I be rude and ask what this is for?  It does sound fairly hairy stuff.

        • by  A B

          Hi David,


          Sure :  the load will be used to discharge some battery packs (a few kWh each) and thermally characterize them. (dT vs. I)

          We would have used inverter and inject in the grid ideally but cost is too high for that...

          Regarding derating, in the first page of the datasheet I linked, you can find a DC load breaking capacity curve with load limit 1 & load limit 2. In both cases, @40V, you get a little over 1A at the load limit 1 or around 3,5A at load limit 2.

          While you're pretty much asymptotical (infinite) at 12V...

          About the heatsink and cooling in solid state, I just don't see where they lie since the On-state conduction loss will be that of a few Ws and virtually no switching /linear loss...
          • by  David Ashton (edited)

            AB - sorry, being in Australia means I sleep different times from the rest of the world.

            Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

            Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  A B (edited)

    Hi David and Rick,

    - Indeed there are a lot of higher currents automotive relays but even the Hasco reference you mentioned (nice one BTW) is only rated at 28VDC max (even though 277V in AC!) and for 30A... a bit short.

    - David, regarding contact ratings, I would -at the contrary- tend to be more conservative after having welded one too many contacts in high-current DC switches (>500A) even though the ratings said they were up to the job.
    The relay you linked is very similar to the one I linked up there and while it can indeed switch north of 100VDC (according to the datasheet), the current allowed even at 35VDC will only be around 1,5A thanks to the square law of current...
    The snubber seem like an interesting solution indeed but not very safe since the current will not be "galvanically" interrupted because of that...


    • by  David Ashton
      AB - points taken but the truth lies somewhere in between I think.  Your linked relay you will note has two current ratings (NO/NC - ie normally open (closing contacts) and normally closed (opening contacts).  The contacts can break a lot less than they can make.  Temperature also plays a factor.  But I'd question whether the ratings go down as much as you say.   I tried to find some documentation on this but it is very hard.  None of the relays you, I or Rick cited have ratings at different voltages.  My understanding has always been that up to about 50VDC you don't have to downrate much (but I'm open to correction...) And 500A is a LOT of current at any voltage.

      Seems like you have a choice....use relays and worry about current capacity, or use solid state switches (yours or commercial) and be prepared for heatsinks and cooling.  My strong advice would be to switch smaller powers even if it means more loads and maybe driving several loads from one pin, you're less likely to got too near the limits of your switches, of whatever type.

      Can I be rude and ask what this is for?  It does sound fairly hairy stuff.

      • by  A B

        Hi David,


        Sure :  the load will be used to discharge some battery packs (a few kWh each) and thermally characterize them. (dT vs. I)

        We would have used inverter and inject in the grid ideally but cost is too high for that...

        Regarding derating, in the first page of the datasheet I linked, you can find a DC load breaking capacity curve with load limit 1 & load limit 2. In both cases, @40V, you get a little over 1A at the load limit 1 or around 3,5A at load limit 2.

        While you're pretty much asymptotical (infinite) at 12V...

        About the heatsink and cooling in solid state, I just don't see where they lie since the On-state conduction loss will be that of a few Ws and virtually no switching /linear loss...
        • by  David Ashton (edited)

          AB - sorry, being in Australia means I sleep different times from the rest of the world.

          Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

          Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  David Ashton
    AB - points taken but the truth lies somewhere in between I think.  Your linked relay you will note has two current ratings (NO/NC - ie normally open (closing contacts) and normally closed (opening contacts).  The contacts can break a lot less than they can make.  Temperature also plays a factor.  But I'd question whether the ratings go down as much as you say.   I tried to find some documentation on this but it is very hard.  None of the relays you, I or Rick cited have ratings at different voltages.  My understanding has always been that up to about 50VDC you don't have to downrate much (but I'm open to correction...) And 500A is a LOT of current at any voltage.

    Seems like you have a choice....use relays and worry about current capacity, or use solid state switches (yours or commercial) and be prepared for heatsinks and cooling.  My strong advice would be to switch smaller powers even if it means more loads and maybe driving several loads from one pin, you're less likely to got too near the limits of your switches, of whatever type.

    Can I be rude and ask what this is for?  It does sound fairly hairy stuff.

    • by  A B

      Hi David,


      Sure :  the load will be used to discharge some battery packs (a few kWh each) and thermally characterize them. (dT vs. I)

      We would have used inverter and inject in the grid ideally but cost is too high for that...

      Regarding derating, in the first page of the datasheet I linked, you can find a DC load breaking capacity curve with load limit 1 & load limit 2. In both cases, @40V, you get a little over 1A at the load limit 1 or around 3,5A at load limit 2.

      While you're pretty much asymptotical (infinite) at 12V...

      About the heatsink and cooling in solid state, I just don't see where they lie since the On-state conduction loss will be that of a few Ws and virtually no switching /linear loss...
      • by  David Ashton (edited)

        AB - sorry, being in Australia means I sleep different times from the rest of the world.

        Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

        Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  A B

    Hi David,


    Sure :  the load will be used to discharge some battery packs (a few kWh each) and thermally characterize them. (dT vs. I)

    We would have used inverter and inject in the grid ideally but cost is too high for that...

    Regarding derating, in the first page of the datasheet I linked, you can find a DC load breaking capacity curve with load limit 1 & load limit 2. In both cases, @40V, you get a little over 1A at the load limit 1 or around 3,5A at load limit 2.

    While you're pretty much asymptotical (infinite) at 12V...

    About the heatsink and cooling in solid state, I just don't see where they lie since the On-state conduction loss will be that of a few Ws and virtually no switching /linear loss...
    • by  David Ashton (edited)

      AB - sorry, being in Australia means I sleep different times from the rest of the world.

      Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

      Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  David Ashton (edited)

    AB - sorry, being in Australia means I sleep different times from the rest of the world.

    Re the curve, I do indeed stand corrected.  The only comment I can make on that is to suggest a proper contactor (A big mains relay), but they are usually intended for AC use and probably would not be characterised for DC, and would have the same limitations as these smaller relays.

    Heatsinking - solid state relays do not have ~ zero voltage drop like relays.  The Solid state relay suggested by Rick Curl  has a datasheet here which states that the voltage drop is <1.5V (so take that as a maximum) and max current is 40A.  So worse case power dissipation will be 1.5V x 40A = 60W.  In practice it would probably be a fair bit less than this but you'd be looking at 30-40W I'd say.  They don't give any thermal information but it does seem to have a metal base and holes for attaching to a heatsink.  You could probably find a finned heatsink that would fit these, and use a fan powered by your batteries (the load would be inconsequential compared to the power you're switching).  You'd probably have to get 12V fan and use a regulator (switching preferably) to generate the 12V from your 36V batteries.  These SSRs would comfortably take your 30A load per resistor and you could probably develop a module controlling 4 or 5 KW loads this way.

  • by  A B
    Oh and yes, an SSPC is basically what I'm trying to do with my bad design MOSFET design :-)

    Only cheaper... :-)

  • by  Elizabeth Simon

    One suggestion to allow the use of less expensive relays would be to use a hybrid circuit where you first turn on the MOSFET  but then immediately also turn on a mechanical relay that is in parallel. once the mechanical relay contacts close the current goes through the relay and the MOSFET can be turned off (or not). When you want to open the circuit, you first turn the MOSFET on (if you previously turned it off) then turn off the mechanical relay. When the relay has opened, you turn off the MOSFET.

    The trick bit is that you either need two inputs for each relay or some sort of RC circuit to make sure that the MOSFET stays on long enough for the relay to open.  The advantage is that you aren't switching the entire voltage with the relay  so you can use less expensive relays and at the same time, you don't need the heatsink  because you only dissipate the power in the MOSFET for a short time.

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