Dont know which mosfet driver to use.

Thread Starter

Foksie89

Joined Jul 28, 2020
3
Good day to everyone.

I am new to this forum, this is my first post.

I like to try and design things in my spare time, and my current project requires the use of MOSFETS.

I can help myself with basic electronics, but nothing advanced.

So I am using an Arduino Uno that must switch 8 IRF540 MOSFETS. Sometimes only 2, 4, 5, 6 or all 8 at once, so I assume I will need 8 separate drivers.

The Mosfets will switch up to 18V 12A each, up to a maximum of 100Hz.

If anyone can please advice me on which MOSFET drivers to use. Or what specs on the driver I should be looking out for?

Also if the MOSFETS I have will be up to the task.

I have 1N5822 Schottky Diode`s for each Mosfet as flyback diodes.

Is there any other components I should put in the circuit? Resistors caps anything?

Any help will be appreciated.
 

djsfantasi

Joined Apr 11, 2010
6,718
First, what are the specifications of the eight loads? Current drawn and voltage required.

Second, you’ll need logic level MOSFETs. IRL540 instead of IR540.

Third, are you switching the low side, switching the ground connection? That’s easier than switching the high side - meaning switching the positive voltage.
 

BobTPH

Joined Jun 5, 2013
2,505
A gate driver is not normally needed unless you are switching at high speeds, at a max of 100 Hz, none should be needed. As @djsfantasi said, though, the IRF540 will need 10V on the gate. You can either use a level shifter (singled BJT or MOSFET) or use the logic level MOSFET instead and driver the gate directly out of the Arduino, with, maybe, a 10 Ohm resistor in between.

Bob
 

Thread Starter

Foksie89

Joined Jul 28, 2020
3
First, what are the specifications of the eight loads? Current drawn and voltage required.

Second, you’ll need logic level MOSFETs. IRL540 instead of IR540.

Third, are you switching the low side, switching the ground connection? That’s easier than switching the high side - meaning switching the positive voltage.
Good day.

Sorry for the delayed response, things got hectic at work.

The load will be a maximum of 18V 12A each, but all 8 must be able to switch simultaneously.

If I use logic level mosfet, will the max current from an arduino (20mA) per pin be enough to switch the mosfet on fast enough?

I am trying to switch low side yes.
 

Thread Starter

Foksie89

Joined Jul 28, 2020
3
A gate driver is not normally needed unless you are switching at high speeds, at a max of 100 Hz, none should be needed. As @djsfantasi said, though, the IRF540 will need 10V on the gate. You can either use a level shifter (singled BJT or MOSFET) or use the logic level MOSFET instead and driver the gate directly out of the Arduino, with, maybe, a 10 Ohm resistor in between.

Bob
Good day.

Sorry for the delayed response, things got hectic at work.

Using logic level mosfets, will the arduino max current per pin of 20mA each be enough to switch all 8 mosfets on simultaneously?
 

BobTPH

Joined Jun 5, 2013
2,505
Get the total gate charge from the datasheet and divide by the current to get the time needed to turn on.



For example, the IRLZ44N has a total gate charge of 48nC. At 20 mA, this means a turn on time of 24 usec. For 8 in parallel, it will be 8 times longer, or 192 usec.

Bob
 

RPLaJeunesse

Joined Jul 29, 2018
109
If you are switching multiple 12A loads with FETs I'd look at transistors that dissipate less than a few watts each at that current. And a 100V transistor is not needed to switch an 18V load, 40V-50V is more than adequate. Many common FETs today can reach the 0.010 Ohm range and are not cost-prohibitive. Take a look at the FDP8447L, a 40V rated TO220 part rated 0.011 Ohm max at 4.5V drive, 25C. I've also had good luck with the TI CSD series FETs, having in production a boost supply producing 19V at 95W from as little as 6V input, and needing minimal heatsinking.
 

LowQCab

Joined Nov 6, 2012
39
I was going to mention some of the Logic Level FETs but RP beat me to it, however,
some "slowing down" of the FETs may be a bonus.
Are your loads Inductive, or Resistive ????
Why are you switching them at 100hz ??? is this for PWM of power ???
Slowing down the FETs can lessen or eliminate problems caused by an Inductive Load.
Just keep in mind that when you are driving a FET Gate,
you are basically driving a Capacitor,
which initially, looks like a Dead Short to the driver, until it charges up.
All you need is an appropriately sized Resistor on the Gate.
Slowing down the FET WILL make it dissipate more heat to some extent,
but it may be worth it to reduce or eliminate
Switching Noises / Flyback Spikes caused by ridiculously fast Switching Speeds.
 

BobTPH

Joined Jun 5, 2013
2,505
me2me2...


... or fast ON OFF to avoid overheating your 'FET at linear region
. . .
perhaps
I understand that. Do you understand why this is a larger problem when switching at 100KHz than at 100Hz? In fact, the power dissipation due to time in the linear region is 1000 times greater at the higher speed.

Bob
 

377Ohms

Joined Sep 9, 2015
10
Get the total gate charge from the datasheet and divide by the current to get the time needed to turn on.

For example, the IRLZ44N has a total gate charge of 48nC. At 20 mA, this means a turn on time of 24 usec. For 8 in parallel, it will be 8 times longer, or 192 usec.

Bob
Hmmm... with dimensional analysis:

1C = 1A*s: 48nC / 20mA = 48E-9A*s / 20E-3A = 2.4E-6s = 2.4us not 24us for one MOSFET gate. 8 * 2.4us = 19.2us for eight MOSFET gates in parallel, not 192us.

But I am sure you understand this turn on calculation is only a rough estimate. MOSFET gate turn on characteristics have at least five separate timing regions arising from at least that many various phenomena. This app-note discusses this in more detail while deriving equations that estimate the turn-on and turn-off switch times:

Power MOSFET Basics: Understanding Gate Charge and Using it to Assess Switching Performance - Vishay AN608a:

https://www.vishay.com/docs/73217/an608a.pdf
 

ci139

Joined Jul 11, 2016
1,677
Last edited:

BobTPH

Joined Jun 5, 2013
2,505
Hmmm... with dimensional analysis:

1C = 1A*s: 48nC / 20mA = 48E-9A*s / 20E-3A = 2.4E-6s = 2.4us not 24us for one MOSFET gate. 8 * 2.4us = 19.2us for eight MOSFET gates in parallel, not 192us.
Hah! I just did the calculation again. Not using scientific notation. The calculator stopped after nine digits when I was entering 1,000,000,000 to get nC, and thus I was off by a factor of 10.

Bob
 
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