Fixing conmutation for high voltage switching with power mosfet












0












$begingroup$


and thanks for reading. I have an issue with a charge circuit for a capacitor as load. I want to switch approximately 400 V DC to charge a 1000 uf 600 V capacitor. I'm using a power mosfet for this application. I need it to charge instantly as soon as it turns on, or in a few milliseconds. The problem is that to do that i saturate the mosfet and then turn it off using a 10V signal to Gate source to drive the mosfet, it works the first time, as soon as I send the signal it charges, but the problem is that the capacitor gets damaged and all the terminals get shorted. The mosfet is a IRPF460, it is a 500V, 20 A and 0.27 ohm mosfet, i choose it because it seems to be the correct for this application. I put a 10 A fuse next to the mosfet to verify if it was being damaged by some inrush current but it wasn't because ass soon as I turn on the mosfet the fuse didn't pop and the current i measure was not above 5.5 A, and the mosfet brokedown anyway. THe only think that could be causing the problem is the conmutation therefore, the problem must be in Gate-Source or the driving part. Another think that called my atenttion is that if i apply almost 8 V to Gate-Source the capacitor charges but only to a half of the voltage with a single pulse of a button., and the mosfet does not suffer any damage. The driving signal for the mosfet will be a pulse that can go from 55 ms to 1 sec. so it has to charge the cap within these times too. I looked for snubber circuits that can handle this but the ones i found were parallel to the mosfet and would get 400 V as soon as the power supply is conected, so i would need components to deal with this and i dont have them. Even if i would get them i dont know if it would work. THis circuit will have another part to discharge the capacitor, but first i need the charge to work. I would like to know if i can implement some kind of snubber for Gate-Source or what can i do to avoid damaging the mosfet and switching the voltage needed. I think the mosfet could be leaving the safe operating area (SOA) when switching. I also tried to put a diode with a parallel resistor on gate, but no results. Please I need help. Thank you all, again.



This is my circuit:
enter image description here









share







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  • 1




    $begingroup$
    Did you examine the SOA curve of the FET?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 2




    $begingroup$
    Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
    $endgroup$
    – Blair Fonville
    6 hours ago










  • $begingroup$
    Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
    $endgroup$
    – Kevin White
    5 hours ago


















0












$begingroup$


and thanks for reading. I have an issue with a charge circuit for a capacitor as load. I want to switch approximately 400 V DC to charge a 1000 uf 600 V capacitor. I'm using a power mosfet for this application. I need it to charge instantly as soon as it turns on, or in a few milliseconds. The problem is that to do that i saturate the mosfet and then turn it off using a 10V signal to Gate source to drive the mosfet, it works the first time, as soon as I send the signal it charges, but the problem is that the capacitor gets damaged and all the terminals get shorted. The mosfet is a IRPF460, it is a 500V, 20 A and 0.27 ohm mosfet, i choose it because it seems to be the correct for this application. I put a 10 A fuse next to the mosfet to verify if it was being damaged by some inrush current but it wasn't because ass soon as I turn on the mosfet the fuse didn't pop and the current i measure was not above 5.5 A, and the mosfet brokedown anyway. THe only think that could be causing the problem is the conmutation therefore, the problem must be in Gate-Source or the driving part. Another think that called my atenttion is that if i apply almost 8 V to Gate-Source the capacitor charges but only to a half of the voltage with a single pulse of a button., and the mosfet does not suffer any damage. The driving signal for the mosfet will be a pulse that can go from 55 ms to 1 sec. so it has to charge the cap within these times too. I looked for snubber circuits that can handle this but the ones i found were parallel to the mosfet and would get 400 V as soon as the power supply is conected, so i would need components to deal with this and i dont have them. Even if i would get them i dont know if it would work. THis circuit will have another part to discharge the capacitor, but first i need the charge to work. I would like to know if i can implement some kind of snubber for Gate-Source or what can i do to avoid damaging the mosfet and switching the voltage needed. I think the mosfet could be leaving the safe operating area (SOA) when switching. I also tried to put a diode with a parallel resistor on gate, but no results. Please I need help. Thank you all, again.



This is my circuit:
enter image description here









share







New contributor




M.Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Did you examine the SOA curve of the FET?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 2




    $begingroup$
    Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
    $endgroup$
    – Blair Fonville
    6 hours ago










  • $begingroup$
    Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
    $endgroup$
    – Kevin White
    5 hours ago
















0












0








0





$begingroup$


and thanks for reading. I have an issue with a charge circuit for a capacitor as load. I want to switch approximately 400 V DC to charge a 1000 uf 600 V capacitor. I'm using a power mosfet for this application. I need it to charge instantly as soon as it turns on, or in a few milliseconds. The problem is that to do that i saturate the mosfet and then turn it off using a 10V signal to Gate source to drive the mosfet, it works the first time, as soon as I send the signal it charges, but the problem is that the capacitor gets damaged and all the terminals get shorted. The mosfet is a IRPF460, it is a 500V, 20 A and 0.27 ohm mosfet, i choose it because it seems to be the correct for this application. I put a 10 A fuse next to the mosfet to verify if it was being damaged by some inrush current but it wasn't because ass soon as I turn on the mosfet the fuse didn't pop and the current i measure was not above 5.5 A, and the mosfet brokedown anyway. THe only think that could be causing the problem is the conmutation therefore, the problem must be in Gate-Source or the driving part. Another think that called my atenttion is that if i apply almost 8 V to Gate-Source the capacitor charges but only to a half of the voltage with a single pulse of a button., and the mosfet does not suffer any damage. The driving signal for the mosfet will be a pulse that can go from 55 ms to 1 sec. so it has to charge the cap within these times too. I looked for snubber circuits that can handle this but the ones i found were parallel to the mosfet and would get 400 V as soon as the power supply is conected, so i would need components to deal with this and i dont have them. Even if i would get them i dont know if it would work. THis circuit will have another part to discharge the capacitor, but first i need the charge to work. I would like to know if i can implement some kind of snubber for Gate-Source or what can i do to avoid damaging the mosfet and switching the voltage needed. I think the mosfet could be leaving the safe operating area (SOA) when switching. I also tried to put a diode with a parallel resistor on gate, but no results. Please I need help. Thank you all, again.



This is my circuit:
enter image description here









share







New contributor




M.Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




and thanks for reading. I have an issue with a charge circuit for a capacitor as load. I want to switch approximately 400 V DC to charge a 1000 uf 600 V capacitor. I'm using a power mosfet for this application. I need it to charge instantly as soon as it turns on, or in a few milliseconds. The problem is that to do that i saturate the mosfet and then turn it off using a 10V signal to Gate source to drive the mosfet, it works the first time, as soon as I send the signal it charges, but the problem is that the capacitor gets damaged and all the terminals get shorted. The mosfet is a IRPF460, it is a 500V, 20 A and 0.27 ohm mosfet, i choose it because it seems to be the correct for this application. I put a 10 A fuse next to the mosfet to verify if it was being damaged by some inrush current but it wasn't because ass soon as I turn on the mosfet the fuse didn't pop and the current i measure was not above 5.5 A, and the mosfet brokedown anyway. THe only think that could be causing the problem is the conmutation therefore, the problem must be in Gate-Source or the driving part. Another think that called my atenttion is that if i apply almost 8 V to Gate-Source the capacitor charges but only to a half of the voltage with a single pulse of a button., and the mosfet does not suffer any damage. The driving signal for the mosfet will be a pulse that can go from 55 ms to 1 sec. so it has to charge the cap within these times too. I looked for snubber circuits that can handle this but the ones i found were parallel to the mosfet and would get 400 V as soon as the power supply is conected, so i would need components to deal with this and i dont have them. Even if i would get them i dont know if it would work. THis circuit will have another part to discharge the capacitor, but first i need the charge to work. I would like to know if i can implement some kind of snubber for Gate-Source or what can i do to avoid damaging the mosfet and switching the voltage needed. I think the mosfet could be leaving the safe operating area (SOA) when switching. I also tried to put a diode with a parallel resistor on gate, but no results. Please I need help. Thank you all, again.



This is my circuit:
enter image description here







mosfet switch-mode-power-supply power-electronics switching powermosfet





share







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asked 6 hours ago









M.BrianM.Brian

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  • 1




    $begingroup$
    Did you examine the SOA curve of the FET?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 2




    $begingroup$
    Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
    $endgroup$
    – Blair Fonville
    6 hours ago










  • $begingroup$
    Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
    $endgroup$
    – Kevin White
    5 hours ago
















  • 1




    $begingroup$
    Did you examine the SOA curve of the FET?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






  • 2




    $begingroup$
    Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
    $endgroup$
    – Blair Fonville
    6 hours ago










  • $begingroup$
    Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
    $endgroup$
    – Kevin White
    5 hours ago










1




1




$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
6 hours ago




$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
6 hours ago




2




2




$begingroup$
Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
$endgroup$
– Blair Fonville
6 hours ago




$begingroup$
Please edit your post to fix all the typos. And add a link to the datasheet - there’s no such thing as IRPF460.
$endgroup$
– Blair Fonville
6 hours ago












$begingroup$
Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
$endgroup$
– Kevin White
5 hours ago






$begingroup$
Charging a capacitor like this will result in half of the charging energy being dissipated within the FET. There will be extremely large currents and energy dissipation that will destroy any reasonable size device. You need to either use an appropriate size series resistor that can handle the energy or better approaches use a series inductor and diode to recover that energy and put it back in the capacitor.
$endgroup$
– Kevin White
5 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Analysis



Cap Specs not given so a typical part

e.g. 1mF @600V ESR=92[mΩ] @ 10kHz 20°C using this Cap, Kemet ALC70(1)102FP600



FET RdsOn= 270 mΩ so out of 270+92 total The FET will draw 75% of the power and energy.



The cap Ec= 1/2CV² = 1/2 * 0.001F * 400²V = 80J so the cap will dissipate 20% or 20J while charging up to 80J. so the FET must transfer and dissipate 75% of 100J or 75J.



The worst case FET Safe Operating Area (SOA) must be observed.
enter image description here
Yet the FET can only handle about 900 mJ at 92uS but with RdsONC= 270mΩC=270us the SOA curve points to about 500 mJ vs a requirement to transfer to dissipate 75J.



So a much bigger FET is needed with lower RdsOn in the 10 mΩ range, I suspect. I doubt if the supply or Cap can handle a steady diet of these pulses, so back to the drawing board. The term "instantly" needs to be specified and relaxed with a current limiter.



Short Circuit current on the Cap is about 4000 Amps at 400 V.



"Houston, I think we have a problem"





share











$endgroup$





















    0












    $begingroup$

    Sunnyskyguy-ee75 gives you a really good answer regarding the power problem. Ultimately, I believe you will need to consider the problem you are trying to solve. Either generate a lot of heat by charging the cap quickly with a high current (Warning caps will self heat, Al electrolytic in particular, and can be destroyed by too much current). Or increase the charging time and generate less heat.



    Maybe a non-linear solution is best:




    • Pulse the MOSFET (you'll need a catch diode for the parasitic
      inductance).

    • Take this a little further and make a buck DC/DC converter by adding
      one inductor and diode to your circuit. A fixed duty cycle or fixed
      peak current are both simple switch control methods that will charge
      the cap. The high voltage is put across the inductor and not the FET. Bonus is that most power in the inductor goes to charge the cap too rather than being wasted as heat.


    Linear solutions:




    • Power resistor in series with MOSFET to limit the charging current. This is still a trade-off between power in the FET and charging time. The resistor provides another knob to turn so you can balance power and charge time.

    • Massive MOSFETs with lots of heat sinking configured as a current source. Feedback circuit to control the current flow by modulating their Vgs. This means a current sense resistor between the FET's source and the negative power terminal. Opamp compares the sense resistor voltage to a reference voltage and drives the MOSFET gate. This can be a difficult circuit to stabilize with a large FET. Step changes on the reference voltage will excite instability.


    Power MOSFETs in brief:



    Most power MOSFETs are designed to act as switches (in switching power converters for example). They can stand off the rated Vds when off. When turned on the NFET pulls its drain down to its source quickly, typically faster than 1 us.



    The power MOSFET is designed to be the lowest impedance looking into the node. In your situation the capacitor is the lowest (AC) impedance.



    There are MOSFETs called Linear FETs that are intended more for this type of operation. A linear FET has an expanded SOA, a lower gm, and typically a higher Ron than other similar switching power FETs. IXYS (now Littelfuse) has a selection here: N-Channel Linear Power MOSFETs.






    share|improve this answer











    $endgroup$













    • $begingroup$
      These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
      $endgroup$
      – Sunnyskyguy EE75
      3 hours ago













    Your Answer





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    2 Answers
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    2 Answers
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    3












    $begingroup$

    Analysis



    Cap Specs not given so a typical part

    e.g. 1mF @600V ESR=92[mΩ] @ 10kHz 20°C using this Cap, Kemet ALC70(1)102FP600



    FET RdsOn= 270 mΩ so out of 270+92 total The FET will draw 75% of the power and energy.



    The cap Ec= 1/2CV² = 1/2 * 0.001F * 400²V = 80J so the cap will dissipate 20% or 20J while charging up to 80J. so the FET must transfer and dissipate 75% of 100J or 75J.



    The worst case FET Safe Operating Area (SOA) must be observed.
    enter image description here
    Yet the FET can only handle about 900 mJ at 92uS but with RdsONC= 270mΩC=270us the SOA curve points to about 500 mJ vs a requirement to transfer to dissipate 75J.



    So a much bigger FET is needed with lower RdsOn in the 10 mΩ range, I suspect. I doubt if the supply or Cap can handle a steady diet of these pulses, so back to the drawing board. The term "instantly" needs to be specified and relaxed with a current limiter.



    Short Circuit current on the Cap is about 4000 Amps at 400 V.



    "Houston, I think we have a problem"





    share











    $endgroup$


















      3












      $begingroup$

      Analysis



      Cap Specs not given so a typical part

      e.g. 1mF @600V ESR=92[mΩ] @ 10kHz 20°C using this Cap, Kemet ALC70(1)102FP600



      FET RdsOn= 270 mΩ so out of 270+92 total The FET will draw 75% of the power and energy.



      The cap Ec= 1/2CV² = 1/2 * 0.001F * 400²V = 80J so the cap will dissipate 20% or 20J while charging up to 80J. so the FET must transfer and dissipate 75% of 100J or 75J.



      The worst case FET Safe Operating Area (SOA) must be observed.
      enter image description here
      Yet the FET can only handle about 900 mJ at 92uS but with RdsONC= 270mΩC=270us the SOA curve points to about 500 mJ vs a requirement to transfer to dissipate 75J.



      So a much bigger FET is needed with lower RdsOn in the 10 mΩ range, I suspect. I doubt if the supply or Cap can handle a steady diet of these pulses, so back to the drawing board. The term "instantly" needs to be specified and relaxed with a current limiter.



      Short Circuit current on the Cap is about 4000 Amps at 400 V.



      "Houston, I think we have a problem"





      share











      $endgroup$
















        3












        3








        3





        $begingroup$

        Analysis



        Cap Specs not given so a typical part

        e.g. 1mF @600V ESR=92[mΩ] @ 10kHz 20°C using this Cap, Kemet ALC70(1)102FP600



        FET RdsOn= 270 mΩ so out of 270+92 total The FET will draw 75% of the power and energy.



        The cap Ec= 1/2CV² = 1/2 * 0.001F * 400²V = 80J so the cap will dissipate 20% or 20J while charging up to 80J. so the FET must transfer and dissipate 75% of 100J or 75J.



        The worst case FET Safe Operating Area (SOA) must be observed.
        enter image description here
        Yet the FET can only handle about 900 mJ at 92uS but with RdsONC= 270mΩC=270us the SOA curve points to about 500 mJ vs a requirement to transfer to dissipate 75J.



        So a much bigger FET is needed with lower RdsOn in the 10 mΩ range, I suspect. I doubt if the supply or Cap can handle a steady diet of these pulses, so back to the drawing board. The term "instantly" needs to be specified and relaxed with a current limiter.



        Short Circuit current on the Cap is about 4000 Amps at 400 V.



        "Houston, I think we have a problem"





        share











        $endgroup$



        Analysis



        Cap Specs not given so a typical part

        e.g. 1mF @600V ESR=92[mΩ] @ 10kHz 20°C using this Cap, Kemet ALC70(1)102FP600



        FET RdsOn= 270 mΩ so out of 270+92 total The FET will draw 75% of the power and energy.



        The cap Ec= 1/2CV² = 1/2 * 0.001F * 400²V = 80J so the cap will dissipate 20% or 20J while charging up to 80J. so the FET must transfer and dissipate 75% of 100J or 75J.



        The worst case FET Safe Operating Area (SOA) must be observed.
        enter image description here
        Yet the FET can only handle about 900 mJ at 92uS but with RdsONC= 270mΩC=270us the SOA curve points to about 500 mJ vs a requirement to transfer to dissipate 75J.



        So a much bigger FET is needed with lower RdsOn in the 10 mΩ range, I suspect. I doubt if the supply or Cap can handle a steady diet of these pulses, so back to the drawing board. The term "instantly" needs to be specified and relaxed with a current limiter.



        Short Circuit current on the Cap is about 4000 Amps at 400 V.



        "Houston, I think we have a problem"






        share













        share


        share








        edited 5 hours ago

























        answered 5 hours ago









        Sunnyskyguy EE75Sunnyskyguy EE75

        68.4k22598




        68.4k22598

























            0












            $begingroup$

            Sunnyskyguy-ee75 gives you a really good answer regarding the power problem. Ultimately, I believe you will need to consider the problem you are trying to solve. Either generate a lot of heat by charging the cap quickly with a high current (Warning caps will self heat, Al electrolytic in particular, and can be destroyed by too much current). Or increase the charging time and generate less heat.



            Maybe a non-linear solution is best:




            • Pulse the MOSFET (you'll need a catch diode for the parasitic
              inductance).

            • Take this a little further and make a buck DC/DC converter by adding
              one inductor and diode to your circuit. A fixed duty cycle or fixed
              peak current are both simple switch control methods that will charge
              the cap. The high voltage is put across the inductor and not the FET. Bonus is that most power in the inductor goes to charge the cap too rather than being wasted as heat.


            Linear solutions:




            • Power resistor in series with MOSFET to limit the charging current. This is still a trade-off between power in the FET and charging time. The resistor provides another knob to turn so you can balance power and charge time.

            • Massive MOSFETs with lots of heat sinking configured as a current source. Feedback circuit to control the current flow by modulating their Vgs. This means a current sense resistor between the FET's source and the negative power terminal. Opamp compares the sense resistor voltage to a reference voltage and drives the MOSFET gate. This can be a difficult circuit to stabilize with a large FET. Step changes on the reference voltage will excite instability.


            Power MOSFETs in brief:



            Most power MOSFETs are designed to act as switches (in switching power converters for example). They can stand off the rated Vds when off. When turned on the NFET pulls its drain down to its source quickly, typically faster than 1 us.



            The power MOSFET is designed to be the lowest impedance looking into the node. In your situation the capacitor is the lowest (AC) impedance.



            There are MOSFETs called Linear FETs that are intended more for this type of operation. A linear FET has an expanded SOA, a lower gm, and typically a higher Ron than other similar switching power FETs. IXYS (now Littelfuse) has a selection here: N-Channel Linear Power MOSFETs.






            share|improve this answer











            $endgroup$













            • $begingroup$
              These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
              $endgroup$
              – Sunnyskyguy EE75
              3 hours ago


















            0












            $begingroup$

            Sunnyskyguy-ee75 gives you a really good answer regarding the power problem. Ultimately, I believe you will need to consider the problem you are trying to solve. Either generate a lot of heat by charging the cap quickly with a high current (Warning caps will self heat, Al electrolytic in particular, and can be destroyed by too much current). Or increase the charging time and generate less heat.



            Maybe a non-linear solution is best:




            • Pulse the MOSFET (you'll need a catch diode for the parasitic
              inductance).

            • Take this a little further and make a buck DC/DC converter by adding
              one inductor and diode to your circuit. A fixed duty cycle or fixed
              peak current are both simple switch control methods that will charge
              the cap. The high voltage is put across the inductor and not the FET. Bonus is that most power in the inductor goes to charge the cap too rather than being wasted as heat.


            Linear solutions:




            • Power resistor in series with MOSFET to limit the charging current. This is still a trade-off between power in the FET and charging time. The resistor provides another knob to turn so you can balance power and charge time.

            • Massive MOSFETs with lots of heat sinking configured as a current source. Feedback circuit to control the current flow by modulating their Vgs. This means a current sense resistor between the FET's source and the negative power terminal. Opamp compares the sense resistor voltage to a reference voltage and drives the MOSFET gate. This can be a difficult circuit to stabilize with a large FET. Step changes on the reference voltage will excite instability.


            Power MOSFETs in brief:



            Most power MOSFETs are designed to act as switches (in switching power converters for example). They can stand off the rated Vds when off. When turned on the NFET pulls its drain down to its source quickly, typically faster than 1 us.



            The power MOSFET is designed to be the lowest impedance looking into the node. In your situation the capacitor is the lowest (AC) impedance.



            There are MOSFETs called Linear FETs that are intended more for this type of operation. A linear FET has an expanded SOA, a lower gm, and typically a higher Ron than other similar switching power FETs. IXYS (now Littelfuse) has a selection here: N-Channel Linear Power MOSFETs.






            share|improve this answer











            $endgroup$













            • $begingroup$
              These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
              $endgroup$
              – Sunnyskyguy EE75
              3 hours ago
















            0












            0








            0





            $begingroup$

            Sunnyskyguy-ee75 gives you a really good answer regarding the power problem. Ultimately, I believe you will need to consider the problem you are trying to solve. Either generate a lot of heat by charging the cap quickly with a high current (Warning caps will self heat, Al electrolytic in particular, and can be destroyed by too much current). Or increase the charging time and generate less heat.



            Maybe a non-linear solution is best:




            • Pulse the MOSFET (you'll need a catch diode for the parasitic
              inductance).

            • Take this a little further and make a buck DC/DC converter by adding
              one inductor and diode to your circuit. A fixed duty cycle or fixed
              peak current are both simple switch control methods that will charge
              the cap. The high voltage is put across the inductor and not the FET. Bonus is that most power in the inductor goes to charge the cap too rather than being wasted as heat.


            Linear solutions:




            • Power resistor in series with MOSFET to limit the charging current. This is still a trade-off between power in the FET and charging time. The resistor provides another knob to turn so you can balance power and charge time.

            • Massive MOSFETs with lots of heat sinking configured as a current source. Feedback circuit to control the current flow by modulating their Vgs. This means a current sense resistor between the FET's source and the negative power terminal. Opamp compares the sense resistor voltage to a reference voltage and drives the MOSFET gate. This can be a difficult circuit to stabilize with a large FET. Step changes on the reference voltage will excite instability.


            Power MOSFETs in brief:



            Most power MOSFETs are designed to act as switches (in switching power converters for example). They can stand off the rated Vds when off. When turned on the NFET pulls its drain down to its source quickly, typically faster than 1 us.



            The power MOSFET is designed to be the lowest impedance looking into the node. In your situation the capacitor is the lowest (AC) impedance.



            There are MOSFETs called Linear FETs that are intended more for this type of operation. A linear FET has an expanded SOA, a lower gm, and typically a higher Ron than other similar switching power FETs. IXYS (now Littelfuse) has a selection here: N-Channel Linear Power MOSFETs.






            share|improve this answer











            $endgroup$



            Sunnyskyguy-ee75 gives you a really good answer regarding the power problem. Ultimately, I believe you will need to consider the problem you are trying to solve. Either generate a lot of heat by charging the cap quickly with a high current (Warning caps will self heat, Al electrolytic in particular, and can be destroyed by too much current). Or increase the charging time and generate less heat.



            Maybe a non-linear solution is best:




            • Pulse the MOSFET (you'll need a catch diode for the parasitic
              inductance).

            • Take this a little further and make a buck DC/DC converter by adding
              one inductor and diode to your circuit. A fixed duty cycle or fixed
              peak current are both simple switch control methods that will charge
              the cap. The high voltage is put across the inductor and not the FET. Bonus is that most power in the inductor goes to charge the cap too rather than being wasted as heat.


            Linear solutions:




            • Power resistor in series with MOSFET to limit the charging current. This is still a trade-off between power in the FET and charging time. The resistor provides another knob to turn so you can balance power and charge time.

            • Massive MOSFETs with lots of heat sinking configured as a current source. Feedback circuit to control the current flow by modulating their Vgs. This means a current sense resistor between the FET's source and the negative power terminal. Opamp compares the sense resistor voltage to a reference voltage and drives the MOSFET gate. This can be a difficult circuit to stabilize with a large FET. Step changes on the reference voltage will excite instability.


            Power MOSFETs in brief:



            Most power MOSFETs are designed to act as switches (in switching power converters for example). They can stand off the rated Vds when off. When turned on the NFET pulls its drain down to its source quickly, typically faster than 1 us.



            The power MOSFET is designed to be the lowest impedance looking into the node. In your situation the capacitor is the lowest (AC) impedance.



            There are MOSFETs called Linear FETs that are intended more for this type of operation. A linear FET has an expanded SOA, a lower gm, and typically a higher Ron than other similar switching power FETs. IXYS (now Littelfuse) has a selection here: N-Channel Linear Power MOSFETs.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 3 hours ago

























            answered 4 hours ago









            jherboldjherbold

            1113




            1113












            • $begingroup$
              These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
              $endgroup$
              – Sunnyskyguy EE75
              3 hours ago




















            • $begingroup$
              These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
              $endgroup$
              – Sunnyskyguy EE75
              3 hours ago


















            $begingroup$
            These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
            $endgroup$
            – Sunnyskyguy EE75
            3 hours ago






            $begingroup$
            These are all very good ideas but Brian needs to give life to this project with real specs on source impedance and max charge dump time. available charger spec link.. One could also slow charge and switch between 2 precharged caps as well. but the problem lacks definition and purpose. Do you agree?
            $endgroup$
            – Sunnyskyguy EE75
            3 hours ago












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