Fixing conmutation for high voltage switching with power mosfet
$begingroup$
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:
mosfet switch-mode-power-supply power-electronics switching powermosfet
asked 2 hours ago
M.BrianM.Brian
62
New contributor
M.Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
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:
mosfet switch-mode-power-supply power-electronics switching powermosfet
asked 2 hours ago
M.BrianM.Brian
62
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
2 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
1 hour 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
1 hour ago
add a comment |
$begingroup$
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:
mosfet switch-mode-power-supply power-electronics switching powermosfet
asked 2 hours ago
M.BrianM.Brian
62
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:
mosfet switch-mode-power-supply power-electronics switching powermosfet
mosfet switch-mode-power-supply power-electronics switching powermosfet
asked 2 hours ago
M.BrianM.Brian
62
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.
asked 2 hours ago
M.BrianM.Brian
62
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.
asked 2 hours ago
M.BrianM.Brian
62
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.
asked 2 hours ago
M.BrianM.Brian
62
asked 2 hours ago
M.BrianM.Brian
62
62
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.
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.
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Check out our Code of Conduct.
1
$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
2 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
1 hour 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
1 hour ago
add a comment |
1
$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
2 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
1 hour 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
1 hour ago
1
1
$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
2 hours ago
$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
2 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
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"
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
$endgroup$
add a comment |
$begingroup$
@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). - That 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.
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.
answered 4 mins ago
jherboldjherbold
813
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
2
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$begingroup$
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.
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"
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
$endgroup$
add a comment |
$begingroup$
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.
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"
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
$endgroup$
add a comment |
$begingroup$
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.
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"
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
$endgroup$
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.
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"
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
edited 1 hour ago
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
answered 1 hour ago
Sunnyskyguy EE75Sunnyskyguy EE75
68.4k22598
68.4k22598
add a comment |
add a comment |
$begingroup$
@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). - That 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.
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.
answered 4 mins ago
jherboldjherbold
813
$endgroup$
add a comment |
$begingroup$
@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). - That 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.
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.
answered 4 mins ago
jherboldjherbold
813
$endgroup$
add a comment |
$begingroup$
@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). - That 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.
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.
answered 4 mins ago
jherboldjherbold
813
$endgroup$
@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). - That 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.
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.
answered 4 mins ago
jherboldjherbold
813
answered 4 mins ago
jherboldjherbold
813
answered 4 mins ago
jherboldjherbold
813
answered 4 mins ago
jherboldjherbold
813
813
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1
$begingroup$
Did you examine the SOA curve of the FET?
$endgroup$
– Sunnyskyguy EE75
2 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
1 hour 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
1 hour ago