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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

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.
I realy don't understand how delaying the transaction from going On to OFF can damage the mosfet?

All I want is to put a small little capacitor in parallel with the mechanical switch to prevent any sudden disconnections from shutting the system down...

wvmarle

I realy don't understand how delaying the transaction from going On to OFF can damage the mosfet?
Heat.

55A at 1.15mΩ = 3.5W (not much, though a small heat sink is a good idea).

Partly on MOSFET, say resistance of 1Ω, current will now drop to about 25A (as you have an approx. 1Ω load in series) but you're dissipating 625W. That's above the max rating of the rather impressive 375W of this MOSFET.

Quote
All I want is to put a small little capacitor in parallel with the mechanical switch to prevent any sudden disconnections from shutting the system down...
Get a proper switch. You're talking about milliseconds here, you won't be able to bridge much more than that. That's in the realm of mechanical failure, not human intervention.

Also this beast of a MOSFET has a gate capacitance to match: roughly 24 nF (based on gate charge value). You'll have to use quite small resistor values for the gate to make it switch fast enough, as you probably want to go <1 µs switching. A typical 10k pull-down will take in the tune of 500 µs to switch off that MOSFET, most of that time it's partly open. Not good.

What should work quite OK is a 100Ω pull-down resistor (switching it off in a few µs), combined with a 220Ω resistor in series with your switch. This voltage divider gives you some 15.6V at the gate, a good number. Now you do run into another problem: that 100Ω resistor gets to dissipate 2.4W, and the 220Ω one 5.4W. That's a lot of heat to deal with.

Now this 100Ω resistor also means you need a 22µF cap just to keep your MOSFET from switching off if your switch is off for 1 ms. But that also means you have a great risk of smoke coming out of the MOSFET being kept long time in partly open state...

Anyway, these numbers suggest you need a MOSFET driver circuit. A circuit that can switch between the two voltages, handle the current spike needed to charge/discharge the gate, while not leaking as much current as those resistors would.
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.

MikeLemon

This just got soooooo complicated now...

So you say this schematic can't work at all with that with those specs for a human to simply switch a system on or off?:


allanhurst

Use a resistive/capacitive load on your switch to take out bounce and  then a schmidt trigger  to achieve a  fast MOSFET switch time.... I take it a few mS delay is no problem?

I'll knock up a circuit ( couple of BJT's etc) if interested.

Allan

wvmarle

Do annotate your parts, makes referencing them a lot easier.

There appears to be a connection missing between two MOSFETs.

Don't put resistors towards the gate.

No need for a zener if you wire the switch with series resistor - that's a voltage divider, works just as well.

That 100k pull-down is WAY too big as I explained already. Same for the 2x 100k around the switch. A recipe for smoke.

That 100 nF cap is not doing anything useful other than helping to kill your MOSFET by delaying the off switching - in your schematic I estimate it's taking about 40ms to switch off those MOSFETs  based on 3x gate capacitance + another 100 nF. That's slow, very slow, and you risk very quick heat build-up. A flaky switch causing many very short off times may cause enough heat to have them smoke, catch fire, explode, or show other undesired behaviour.

Your schematic suggest you have 12V available (for the LED) - you said you only have 50V.
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.

MorganS

Thanks quite helpful!

Is it possible to Isolate the live side with that n channel mosfet?

Yes, but you need a gate drive voltage about 10V HIGHER than the 50V input to hold the MOSFET in the ON condition.
"The problem is in the code you didn't post."


MikeLemon

Use a resistive/capacitive load on your switch to take out bounce and  then a schmidt trigger  to achieve a  fast MOSFET switch time.... I take it a few mS delay is no problem?

I'll knock up a circuit ( couple of BJT's etc) if interested.

Allan
That just makes it so much more complicated for such a simple application....

so what do you recommend from start to finish in terms of components(or even IC's) if I may ask?

Just not planning to add more than a SOP-8 IC to that system can't turn this into a PC mother board...

TomGeorge

Hi,

  • What is  the LOAD?
  • What is the application?
  • What are you using at the moment to turn the LOAD ON and OFF?
  • What is the 50V supply?
  • Can you please tell us your electronics, hardware experience?
  • Is your E switch an Emergency Stop button?



Thanks.. Tom.. :)
Everything runs on smoke, let the smoke out, it stops running....

raschemmel

Quote
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.
First of all, "E-Switch" doesn't mean anything more than "Electronic Switch"
Electronic Switch



Quote
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?
Is there a reason why you have yet to answer the repeated question : WHAT IS YOUR ELECTRONICS BACKGROUND/EXPERIENCE ?"




Let's take an example:
Let I = 60A (worst case)
Let RDS= 9mOhms (0.009 Ohms)

 EVolts = IAmps x ROhms
  E = 60A x 0.0015 Ohms
  E = 0.09 V

 PWatts = IAmps x EVolts
 P = 60A x 0.09V
 P = 5.4 W
 
The IRFS7530 has a Junction to case max resistance of 0.40 degrees Celsius/per watt.
 Tcase  0.40 degrees Celsius/per watt x 5.4 W
 Tcase = 2.16 + Ambient

IRFS7530 DATASHEET

The junction operating temperature range is :
TJ = -55 to 175 (celsius)

All of this looks good , but only applies to the test switching time:

Quote
Pulse width  400μs; duty cycle  2%.
.

I saw nothing in your replies that even remotely suggests you have any plans to
to use "switching" (ie: switching frequency).
Everything in your replies suggests you are simply planning to use the device in the ON CONTINUOUSLY
mode. (ie; 100 % duty cycle).

If you can reduce the time in the linear range to < 400uS then maybe the mosfet will be ok.

Allanhurst's recommendation of a Schmitt trigger is  excellent advice.

If you ask me, if you are not inclined to use any heat sink or cooling, you should use a contractor with
a snubber network.


 
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MikeLemon

So First off all I don't have much experience with mosfets and that's it,

About the temperature calculations you made some data simply doesn't make any sense as that I'm sure a single mosfet of that kind at 60A continues simply can't handle that current and stay that cool.

and for the third time I just want to connect a mechanical switch to close and open the circuit from the mosfet array and thats it,  how hard is it to understand?

wvmarle

and for the third time I just want to connect a mechanical switch to close and open the circuit from the mosfet array and thats it,  how hard is it to understand?
Apparently about as hard as it is to answer questions posed to you. Answers to which are key to how you may be able to electronically switch your 50+ A load, which as you should have figured by now is not trivial.

Still wondering where the Arduino is in this whole thing.
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.

MorganS

Quote
The IRFS7530 has a Junction to case max resistance of 0.40 degrees Celsius/per watt.
 Tcase  0.40 degrees Celsius/per watt x 5.4 W
 Tcase = 2.16 + Ambient
Yes but junction-to-ambient is 40 C/W in the datasheet.

It needs a heatsink.
"The problem is in the code you didn't post."

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