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'[PIC]: PIC driving a mosFET...driver chip or a FET'
2003\04\23@150802 by Micro Eng

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This application is simply a power switch, so there is no high frequency
switching going on.  In some reference designs I've seen both a mosFET
driver being used, and in other cases a smaller FET used to drive the gate
on the power device.  So what the better solution, or is there
one...ie....either one will work, so becomes more of a cost issue...whats
cheaper to manufacture? I know the PIC can't drive it direct, that of course
would be the simpilest and cheapest

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2003\04\23@151622 by Mccauley, Daniel H

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If you are only using a FET as a switch, a PIC should very easily be able to
drive any logic-level gate type FET
with no problems.

Remember, the gate of the FET is basically a capacitance.  You basically
charge that capacitance up to some gate voltage
which turns on the FET.  For low frequency switching like a basic power
switch, a PIC output is more than enough to drive
the gate of a logic level FET. Even for frequencies approaching a few kHz
you shouldn't have a problem driving a logic
level FET.

The only time when you need additional transistor drivers, MOSFET drivers
(i.e. TC442x etc...) is when you are switching at
high frequency and especially with FETs with a high gate capacitance.  For
example, if you are switching a FET at say 200-300kHz with a 5V output
pulse, you are charging / discharging that gate capacitance relatively fast
and the PIC won't be able to supply the current needed to switch that FET.
Thats when you need additional driver circuitry.

However, for a simple power switch, you don't need anything but a resistor
in series with the gate.  Just make sure its a logic level gate FET.

D



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2003\04\23@153146 by Gregory A. Pruden

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Hi Daniel,
I have this type of circuit on a breadboard on my desk.  I am using an
IRLZ34 and when using a series resistor, I tried 270R, 330R, 470R and each
made the MOSFET quite hot in comparison to using it without the series
resistor.  What is the purpose of the series resistor and what was I doing
wrong?
Thanks,
Gregory

{Original Message removed}

2003\04\23@155159 by erholm (QAC)

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Gregory A. Pruden wrote:
> I am using an IRLZ34 and when using a series resistor...
> made the MOSFET quite hot in comparison to using it without the series
> resistor.

Are you switching with a "high" freq ?
Say, 10-100kHz and up ?

The MOSFET probably dissipates the most heat
when beeing someware in between hard "off" and hard "on".

Having a resistor in series with the gate, will make the
charging time of the gate capacitance to be longer
then when driving it directly, and also the time the
MOSFET stays between hard "on" and "off". So you want the
FET to switch as fast as possible, which is does
when driven directly without the resistor (or a "low" value res).

Well, sounds logical to me, at least...

> What is the purpose of the series resistor

To protect whatever is driving the MOSFET from high currents. I'd
expect larger MOSFET's to have larger gate capacitance, and you could
reach a point when the current spikes gets to large (for e.g. a PIC port).

> and what was I doing wrong?

Depends on what you *intended* to be doing... :-)

Regards
Jan-Erik Soderholm.

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2003\04\23@161826 by Gregory A. Pruden

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Hi Jan-erik,
I am switching at 80khz.  So it sounds that without the resistor there is a
risk of over loading the pic but with the resistor there is a chance of
overheating the mosfet. :)

I actually checked the archives  before I breadboarded the circuit and found
both posts advocating and not the series resistor.

Thanks,
Gregory

> {Original Message removed}

2003\04\23@162729 by Madhu Annapragada

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In addition to what Jan-Erik said, the gate resistor plays another crucial
role in switching applications; When you switch the load between Vdrain to
Vsource, you induce a spike in the gate due to the gate-source capacitance
and the severity of the spike is proportional to DV/DT of the drain-source
voltage. To reduce the DV/DT, you can reduce the switching time by using a
gate series resistor. Sometimes in a Half-bridge you want your turn off to
be fast but the turn on to be a little slow in order to slow down dv/dt
induced gate spikes. In such cases you use a gate resistor in parallel with
a diode so that the diode is conducting when the gate is turning off and the
gate capacitor is charging up during turn-on via the gate series resistor.
Madhu

{Original Message removed}

2003\04\23@170737 by Micro Eng

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So what happens when you slowly drive the gate..does the conductance
increase with the gate voltage such that you reach the final on mode? In
other words, a soft start circuit.  If I wanted to minimize the spikes on
the output (such as capacitors charging) then is this the proper method?


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2003\04\23@171607 by Olin Lathrop

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> I am switching at 80khz.  So it sounds that without the resistor there
> is a risk of over loading the pic but with the resistor there is a
> chance of overheating the mosfet. :)

That means you only have 6.25uS per transition.  No wonder the FET was
getting hot.  It was spending most of its time in the region between full
on and full off where it will dissipate power.

At this frequency, you need a much faster (higher current) gate drive than
what a PIC pin can provide.  The FET should be in transition only a small
portion of the time, let's say under 1uS.  For sake of example, let's say
the total gate charge is 50nC.  50nC / 500nS = 100mA.  That's a lot more
than you can expect from a PIC pin, let alone one with a resistor in
series with it.


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2003\04\23@194511 by David VanHorn

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>
>At this frequency, you need a much faster (higher current) gate drive than
>what a PIC pin can provide.  The FET should be in transition only a small
>portion of the time, let's say under 1uS.  For sake of example, let's say
>the total gate charge is 50nC.  50nC / 500nS = 100mA.  That's a lot more
>than you can expect from a PIC pin, let alone one with a resistor in
>series with it.

I just hit this same issue with an AVR that's driving a mosfet at 250kHz as a constant current source. Same problem, ended up using a driver chip for $0.50 or so to solve the problem.  The driver's a fet also, but much smaller gate capacitance, so I can yank it up and down quickly.  If you can't switch them fast, they do get hot.

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2003\04\23@214728 by Gregory A. Pruden

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Thanks for the replies, excuse the newbie questions, and apologies for
hi-jacking this thread.  I hope the original poster also got their questions
answered.  I originally had used 100Khz but saw some distortion in the DS
waveform and backed it off to 80Khz where it looked happier :).  The
datasheet says typical 25nC but the plotted data later in the datasheet
shows 12nC or less for the Vds (13.8v) that I am using.  I am using this
circuit to manage the current to a T.E.C. and which is typically 3A.

I could back off the frequency more if you think I am pushing the pic pin
although I tuned the LC ripple filter for 80Khz.

The datasheet is here:
http://www.irf.com/product-info/datasheets/data/irlz34ns.pdf

Any suggestions would be greatly appreciated.

Gregory





> {Original Message removed}

2003\04\24@071915 by Mccauley, Daniel H

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The problem in your specific case is that you are attempting to turn on your
FET at 80kHz I think you said with different size gate resistors.  80kHz is
relatively fast and you may not be charging the gate capacitance fast enough
to "fully" turn
on the FET with the PIC output.  Therefore, the FET is operating in its
linear region which means it isn't "fully" on and the Drain-Source
Resistance is still quite high which is why your FET is heating up.
PowerDiss = Rds * Ids.

You'll need to add another driving circuit for your FET.

D



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