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Topic: Best most reliable way to drive a N channel mosfet with a mechanical switch (Read 1 time) previous topic - next topic

MikeLemon

Hello,

I'm looking for a way to drive a high voltage source voltage(50V) with the most reliable best possible way using this around 2 or more mosfets: IRF7530 in parallel and a simple switch that either shorts or opens the circuit to be used as an E switch.


Any ideas how how to do that?

From what I understood the challenge is to reduce that 50V don to a more suitable voltage (Up to 20V) to drive the gate and prevent the gate from floating and protecting it from noise(High curent) and other stuff.

Also How many mosfets of those kind should I put in parallel in order to drive those mosefets without active cooling or a heat sink assuming 32deg Celsius ambient and 50-55A continues current?

on the way what voltage do you have to put in the gate in order to drive around 50-55A through it?

Thanks for helping.

DVDdoug

You'll have to check the specs on the particular MOSFET.

You can use a voltage divider (2 resistors) to knock-down the gate voltage.

You shouldn't have noise problems if you use the voltage divider to pull-up the gate and then switch it to ground to turn it off.

The amount of heat generated depends on the RDS (on) resistance.*       50A is a lot and I don't think you'll find a MOSFET that will work without a heatsink.   Heatsink calculations are complicated and there is tolerance so if you put parts in parallel you can't count-on the heat being exactly-evenly distributed.  For worst-case heatsink calculations use the maximum  RDS (on) spec. 



* One way of calculating power (Watts) is Current2R.

ReverseEMF

I'm going to answer this from the assumption that you know how to actually use a MOSFET, because otherwise, the scope of this question would, frankly, just be too large to cover in one reply:

Driving a MOSFET with a switch is quite simple:


V1 can be a different voltage than 9V -- just be sure, whatever voltage you choose, that the MOSFET will be driven to saturation by that voltage on the gate. 

R1 I chose to discharge the Gate capacitance at a rate that insures the MOSFET will turn off fast enough that it doesn't linger in the Active region long enough for the MOSFET to burn up -- that's a LOT of current to be switching!  R1 can, probably, be of higher resistance, but that can't really be determined properly until the MOSFET is selected, and the Gate voltage determined.

Q1: There are MOSFETs out there that can handle 55A and more, and do it without even the need for a Heatsink.  Like, for instance, this little beauty: FDP027N08B-F102.  But, how to select the best MOSFET for the job is a huge topic.  Best to Google that and find one or more of the many sites that do an adequate job of teaching such things.

R2 merely represents whatever load you will be switching with this MOSFET.


"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

MikeLemon

Hi,

Thanks for the effort but guess you didn't quite get my question.

I was looking for a completer solution for this problem, all I have is that 50V source a mechanical switch and a mosfet now I am trying to look for a solution to drive that mosfet as reliably as possible using those parts and just a few more not using an external voltage source.

I thought of using a zener diode to create a reference or some series resistors to reduce that voltage or go full studip and add a straight up voltage special high voltage regulator just to reduce the voltage for that gate but I'm not sure what are the advantages and disadvantages of each power wise so I'm looking for an advice.


TomGeorge

Hello,

I'm looking for a way to drive a high voltage source voltage(50V) with the most reliable best possible way using this around 2 or more mosfets: IRF7530 in parallel and a simple switch that either shorts or opens the circuit to be used as an E switch.
An E switch should be a mechanical switch, not a semiconductor switch.
Why do you want to use a MOSFET?
Tom... :)
Everything runs on smoke, let the smoke out, it stops running....

MikeLemon

Due to high currents and low volume constrains and a sparking problem with fast rushing electrons and initial activation  :)

MarkT

Use 100k resistor and 100k || 12V zener diode as a voltage divider - the zener across the gate/source of the
MOSFET will protect it too.

Switch the top of the divider to your incoming 50V supply.

Note a || b means a in parallel with b, ie 100k resistor and zener in parallel for the bottom of the divider,
another 100k for the top part.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

MikeLemon

Ok sounds good, But this switch will be suseptible to vibrations that could slightly knok it open for a brief.

Do you thing adding a capacitor between the gate and source will fix it? if so with what estimated value?

And I've seen lots of design utilizing a 100R resistors in series with the Mosfet gate to prevent over current or something to the gate what do you thing about it?

Also how many mosfets of those kind should I put in parallel in order to drive those mosefets without active cooling or a heat sink assuming 32deg Celsius ambient and 50-55A continues current?

never was good with calculation those thing with the data sheet data.

MorganS

Ok sounds good, But this switch will be suseptible to vibrations that could slightly knok it open for a brief.
Sounds bad. Buy a new switch.

Do you thing adding a capacitor between the gate and source will fix it? if so with what estimated value?
Also bad. A simple capacitor will slow down the transition from 12V to 0V or 0V to 12V. So the MOSFET spends a few milliseconds in that 'linear' switching region. Since you're switching 2500 watts, your MOSFET will have to dissipate something about that size for that time. With no capacitor, the switching occurs in nanoseconds and a few nanoseconds of 2500W is not a lot of energy at all.

You could add another MOSFET that switches the gate separately and put a capacitor on that one's gate, since it's only switching milliwatts. But that is a poor solution for replacing the switch.

And I've seen lots of design utilizing a 100R resistors in series with the Mosfet gate to prevent over current or something to the gate what do you thing about it?
The 100R resistor will slow down the MOSFET switching speed. That's good for switches driven at kHz frequencies that end up radiating radio frequencies into adjacent electronics. For this example, no, don't do that.

Also how many mosfets of those kind should I put in parallel in order to drive those mosefets without active cooling or a heat sink assuming 32deg Celsius ambient and 50-55A continues current?

never was good with calculation those thing with the data sheet data.
Good news! Today is your day to learn!
"The problem is in the code you didn't post."

MikeLemon

Sounds bad. Buy a new switch.
Also bad. A simple capacitor will slow down the transition from 12V to 0V or 0V to 12V. So the MOSFET spends a few milliseconds in that 'linear' switching region. Since you're switching 2500 watts, your MOSFET will have to dissipate something about that size for that time. With no capacitor, the switching occurs in nanoseconds and a few nanoseconds of 2500W is not a lot of energy at all.

You could add another MOSFET that switches the gate separately and put a capacitor on that one's gate, since it's only switching milliwatts. But that is a poor solution for replacing the switch.
The 100R resistor will slow down the MOSFET switching speed. That's good for switches driven at kHz frequencies that end up radiating radio frequencies into adjacent electronics. For this example, no, don't do that.
Good news! Today is your day to learn!
Quote
Also bad. A simple capacitor will slow down the transition from 12V to 0V or 0V to 12V. So the MOSFET spends a few milliseconds in that 'linear' switching region. Since you're switching 2500 watts, your MOSFET will have to dissipate something about that size for that time. With no capacitor, the switching occurs in nanoseconds and a few nanoseconds of 2500W is not a lot of energy at all.

You could add another MOSFET that switches the gate separately and put a capacitor on that one's gate, since it's only switching milliwatts. But that is a poor solution for replacing the switch.
Ok first this happens with vibrations and it always happen that siwtches get disconnected from a little knob it is just usually for a milli seconds or less and capacitors on the circuit companshaite for that so you don't feel it, I can't change the laws of physics....

Quote
Also bad. A simple capacitor will slow down the transition from 12V to 0V or 0V to 12V. So the MOSFET spends a few milliseconds in that 'linear' switching region. Since you're switching 2500 watts, your MOSFET will have to dissipate something about that size for that time. With no capacitor, the switching occurs in nanoseconds and a few nanoseconds of 2500W is not a lot of energy at all.
You could add another MOSFET that switches the gate separately and put a capacitor on that one's gate, since it's only switching milliwatts. But that is a poor solution for replacing the switch.
And the switching rate is around 0.02millli Hz a normal human just support to short something when he want's to turn on this device it's not for some motor or DC DC converter control otherwise I'd use a mosfet driver....

Quote
The 100R resistor will slow down the MOSFET switching speed. That's good for switches driven at kHz frequencies that end up radiating radio frequencies into adjacent electronics. For this example, no, don't do that.
for the resistor it's the same answer above.

Quote
Good news! Today is your day to learn!
And that's what I'm trying to do....
 

MorganS

So you seem to have already worked out what information is required for calculating overheating: the max current and the ambient temperature.

The MOSFET has an on-resistance, called RDS(on). That's the resistance between drain (D) and source (S) terminals when driven on at a specific gate voltage or VGS. That gate voltage is almost universally 10V. Under those conditions, the MOSFET acts like an ideal resistor, so Ohm's law can be applied.

V = I*R


Given the resistance and the current, what is the voltage across the MOSFET? No, it's not 50V. You should get an answer under 1V.

Given the voltage and current through the MOSFET, how many watts of power is it dissipating?

P=I*V


Then you have to find the "thermal resistance" in the datasheet. It will be given in degrees per watt. Sometimes degrees C, more often degrees K, which makes no difference to us. This is highly variable depending on what heatsink you attached the MOSFET to. The datasheet doesn't know what heatsink you have so it will only give you one or two examples.

There's actually a whole lot of thermal resistances that can be important here. The 'junction' is the actual semiconductor element inside the MOSFET. You're interested in junction-to-ambient, since you've specified an ambient temperature. Or maybe you will be using a heatsink so you want junction-to-case. That is the thermal resistance from the junction to the outside heatsinking surface of the device.

Junction-to-ambient for a TO220 case MOSFET might be 50ºK/W. So hitting it with 2 watts of heating will heat it up by 100 degrees above ambient. Add the ambient of 32ºC and you're at 132ºC.

How hot is it allowed to get? That's in the datasheet too. Usually maximum junction temperature is 125ºC, so you've exceeded that. The thing will burn up.

The thermal resistance junction-to-case will be much lower. Maybe 2ºK/W. So putting it on a heatsink will help. Maybe you find a heatsink with 15ºC/W thermal resistance. That's actually a pretty small heatsink. Maybe $1 or $0.50 and it just clips over the TO220 case. So 2W power will heat up the heatsink to 60ºC above ambient. 92ºC total. The junction is then only 4 degrees hotter than that, so the device is safe and won't burn up with 2W power dissipation.

For manual switching, you don't need to consider switching losses unless you do something stupid like putting a capacitor on the gate to slow it down.
"The problem is in the code you didn't post."

TomGeorge

Hi,

Quote
Due to high currents and low volume constrains and a sparking problem with fast rushing electrons and initial activation  :)
How do you switch the load ON and OFF normally?

E switch MUST disconnect/isolate the positive or live side of the load.

If you have a short to ground, or a possible fatal electrocution, disconnecting the gnd or low side will not ISOLATE the load.

never was good with calculation those thing with the data sheet data.
Can you please tell us your electronics,  hardware experience?
Have you consulted any suppliers/advisers in industrial safety equipment?
Tom.... :)
Everything runs on smoke, let the smoke out, it stops running....

wvmarle

Mmm... Surprised no-one looked up the IRF7530's data sheet as it turns out this MOSFET has a maximum VDS of 20V, and a maximum drain current of 5.4A, making it utterly unsuitable for this 50V, 50-55A load. It's a dual MOSFET package even.

Interestingly this IRF model is logic level, fully open at VGS = 4.5V.

A quick search on Digikey turned up the IRFB3607PBF as very suitable alternative. Max 75V, 80A. On resistance 9mΩ @ 45A / 10V VGS so you still have to deal with 27W of heat in that MOSFET (so add a big heat sink).

What kind of load are you switching, really? If it's a motor you need some special measures to not blow up your MOSFET upon switching.

Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

Coding Badly

An E switch should be a mechanical switch, not a semiconductor switch.
Depends on what the definition of "E" is.


MikeLemon

So you seem to have already worked out what information is required for calculating overheating: the max current and the ambient temperature.

The MOSFET has an on-resistance, called RDS(on). That's the resistance between drain (D) and source (S) terminals when driven on at a specific gate voltage or VGS. That gate voltage is almost universally 10V. Under those conditions, the MOSFET acts like an ideal resistor, so Ohm's law can be applied.

V = I*R



Given the resistance and the current, what is the voltage across the MOSFET? No, it's not 50V. You should get an answer under 1V.

Given the voltage and current through the MOSFET, how many watts of power is it dissipating?

P=I*V



Then you have to find the "thermal resistance" in the datasheet. It will be given in degrees per watt. Sometimes degrees C, more often degrees K, which makes no difference to us. This is highly variable depending on what heatsink you attached the MOSFET to. The datasheet doesn't know what heatsink you have so it will only give you one or two examples.

There's actually a whole lot of thermal resistances that can be important here. The 'junction' is the actual semiconductor element inside the MOSFET. You're interested in junction-to-ambient, since you've specified an ambient temperature. Or maybe you will be using a heatsink so you want junction-to-case. That is the thermal resistance from the junction to the outside heatsinking surface of the device.

Junction-to-ambient for a TO220 case MOSFET might be 50ºK/W. So hitting it with 2 watts of heating will heat it up by 100 degrees above ambient. Add the ambient of 32ºC and you're at 132ºC.

How hot is it allowed to get? That's in the datasheet too. Usually maximum junction temperature is 125ºC, so you've exceeded that. The thing will burn up.

The thermal resistance junction-to-case will be much lower. Maybe 2ºK/W. So putting it on a heatsink will help. Maybe you find a heatsink with 15ºC/W thermal resistance. That's actually a pretty small heatsink. Maybe $1 or $0.50 and it just clips over the TO220 case. So 2W power will heat up the heatsink to 60ºC above ambient. 92ºC total. The junction is then only 4 degrees hotter than that, so the device is safe and won't burn up with 2W power dissipation.

For manual switching, you don't need to consider switching losses unless you do something stupid like putting a capacitor on the gate to slow it down.
Thanks quite helpful!

Hi,
How do you switch the load ON and OFF normally?

E switch MUST disconnect/isolate the positive or live side of the load.

If you have a short to ground, or a possible fatal electrocution, disconnecting the gnd or low side will not ISOLATE the load.
Can you please tell us your electronics,  hardware experience?
Have you consulted any suppliers/advisers in industrial safety equipment?
Tom.... :)
Is it possible to Isolate the live side with that n channel mosfet?

Mmm... Surprised no-one looked up the IRF7530's data sheet as it turns out this MOSFET has a maximum VDS of 20V, and a maximum drain current of 5.4A, making it utterly unsuitable for this 50V, 50-55A load. It's a dual MOSFET package even.

Interestingly this IRF model is logic level, fully open at VGS = 4.5V.

A quick search on Digikey turned up the IRFB3607PBF as very suitable alternative. Max 75V, 80A. On resistance 9mΩ @ 45A / 10V VGS so you still have to deal with 27W of heat in that MOSFET (so add a big heat sink).

What kind of load are you switching, really? If it's a motor you need some special measures to not blow up your MOSFET upon switching.


My bad I meant "IRFS7530"

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