Myth Busters 3 – Myth: “You must have a gate resistor”

[Boardburner2]

I couldn't figure out an easy way to do that without a current probe, I can try adding a small series

I think we had one in the kit.
If it is still here friday i will have a go.

The spike i was looking for on the transitions did not appear although poking around a robomower drive showed some spikes the usual scopes do not see.

Hi guys


 I suspect that the need for a gate resistor, might be a "folk memory", passed down by 'generations' of old farts like me, who have never had any need or reason to doubt the accepted wisdom.


Fof

[/quote]

I was wondering the same having read this thread.
Not particularly involved in design though so knowledge/experience of the detail is rusty to say the least.

[MrAl]

Hi guys

Thoroughly enjoyed this very informative thread.

Here are a couple of thoughts I've had.  In my professional career, I spent most of my time in development labs, working directly and in parallel with lots of very  clever EEs.  They, invariably, like me, tended to include a small gate resistor.  This thread has got me thinking as to where the gate resistor requirement came from.

Very many of you will never have come across the situation where it was obligatory to attach heat shunts to the legs of transistors, before applying a soldering iron.  Yes, when transistors first started to appear in commercial devices, this is what we had to do.  Ever hear of the expression "the fastest fuse on three legs"?  This is what transistors were commonly called, as they would die, just by looking at them the wrong way.  We used to joke that they were only there to protect the main fuse.

In these sort of scenarios, if I was a design engineer, I would use each and every form of insurance I could incorporate.  Don't forget that the development of transistors took quite a few years before one could (almost) guarantee many of the stated characteristics and parameters required for the design.

As FETs have always been the 'delicate' flower compared to BJT, due to potential for static damage, I suspect that the need for a gate resistor, might be a "folk memory", passed down by 'generations' of old farts like me, who have never had any need or reason to doubt the accepted wisdom.

Thanks for a great, educational read.

Fof

Hi again,

The original gate resistor requirement came from the oscillation due to there being some inductance in every mosfet source circuit.  What we are talking about here though sort of ignores that and we were focusing on what the Arduino pin can handle from a typical Uno pin for example.  If the pin is worked too hard the dissipation in the chip could rise too much, and we were also worried about burning out the pin itself due to repeated high currents.

What i found out from the scope pics is first of all that mosfet was acting like a typical mosfet when turning it on.  Also, the dissipation comes out to a formula like:
P=K/Tp

where K is some constant based on the driver and Tp is the total period and P is the total power dissipated by the microcontroller chip.

This of course simply means that the frequency is limited because as frequency goes up Tp goes down, and when Tp goes down P goes up.  The chip can only handle so much power because of it's surface area, and the local effects are probably more strict because we cant assume that the heat distributes perfectly.
I was thinking maybe a rule of thumb would be, assuming only one pin being used, one-half the power allowed for the whole chip, which i think would then be 0.5 watts because the total power of the chip is about 1 watt max probably at 25 degree C ambient.

For a very rough and crude estimate, this seems to be around 100kHz.  That should be proved though by experiment, and of course different mosfets could possibly limit this more.  From what i remember from the 1980's, that was a typical limit anyway.

The other interesting thing is that when we add a gate resistor to the Arduino we also slow the mosfet down, which would tend to increase the turn on time.  The difference is some of the power would be moved out of the chip then.

With all the variables here it is starting to look like the best way to test for resistor or no resistor is to run it in the application and see if the chip gets hot.  If the chip gets hot then either lower the frequency or add a gate resistor.  Of course if the efficiency goes down too much it could be a fight to find the right combination.

BTW another application that requires a gate resistor is when connecting mosfets in parallel to obtain a higher current and lower Rds.  One gate resistor per device is the common scheme.  That is to help with oscillations as well as help to get the mosfets all switching at nearly the same time.

[polymorph]

The cascode configuration works with BJT, JFETs, MOSFETs, and a mixture of them. It also means you can stack transistors to get a higher voltage than any individual rating.

[MarkT]

The resistor stack circuit is a signal level shifter, it cannot pull the output down to ground
without requiring the top device to handle the full voltage (which makes the other devices
redundant).  

C1 is wrong.  The ratio between R6 and R7 sets the voltage gain, the output would be expected to
stay near 1000V, and you'd probably add a capacitor chain to the resistor chain to ensure the
base voltages are evenly spaced during power-up transients.

Alternatively it may be done that way purely to spread the heat dissipation between several devices,
in which case I'm talking out of my, erm, drain terminal...

[polymorph]

That'll be news to a lot of EEs.

I suggest building it in Spice and trying it out.

[stuart0]

The cascode configuration works with BJT, JFETs, MOSFETs, and a mixture of them. It also means you can stack transistors to get a higher voltage than any individual rating.

Some interesting looking circuits there polymorph. I'm not trying to be rude or anything, but perhaps you could start another thread to discuss them, as they seem quite a bit off topic in this one.

I understand that the cascode configuration somewhat isolates the driving device from the Miller effect, but that's the only tenuous connection I can see to this topic.

[stuart0]

Hi again,The original gate resistor requirement came from the oscillation due to there being some inductance in every mosfet source circuit.

Yes this was my understanding as well MrAl. I've certainly seen this oscillation effect occur, and it can quite drastically increase the switching losses. Reducing the source inductance (the length and layout of wiring from source to common ground point with gate drive) and/or increasing the gate series resistance is the cure.

Regarding the other issue of transient pin power dissipation and the danger of overloading the pin without the resistor, I very much doubt that this is an issue with small mosfets as per the OP's example (2n7000).

Of course it depends on what voltage you're switching, but the typical value of total gate charge for that mosfet is only around 1nC. So we have about 1nC * 5V = 5nJ of energy loss per switching cycle. Even at 100 kHz that only amounts to 0.0005 Watts of additional dissipation in the micro controller.

Regarding the instantaneous power dissipation that may occur while switching, consider the pin output characteristics (taken from 328P datasheet) below. Note that the output resistance is about 25 ohms (20mA at 0.5V drop). Also note the slight curvature of the characteristics, showing increasing resistance with increased voltage drop as is typical for a mos output.

So we could estimate that the maximum instantaneous power dissipation would be somewhat less than 100mA * 2.5 V = 0.25 Watts, albeit very briefly.

Personally I would be very surprised if the power dissipation (instantaneous or otherwise) would be an issue. Even for larger mosfets I suspect that the need for an external driver (to achieve snappier switching and reduce switching losses in the mosfet) would arise well before damaging power losses occurred in the MPU pins.

Having said that however. Not everyone has an oscilloscope to detect unwanted oscillations, or the ability to layout the gate drive to avoid it, so guidelines like adding the series gate resistance really aren't a bad idea in any case.

[MrAl]

Yes this was my understanding as well MrAl. I've certainly seen this oscillation effect occur, and it can quite drastically increase the switching losses. Reducing the source inductance (the length and layout of wiring from source to common ground point with gate drive) and/or increasing the gate series resistance is the cure.

Regarding the other issue of transient pin power dissipation and the danger of overloading the pin without the resistor, I very much doubt that this is an issue with small mosfets as per the OP's example (2n7000).

Of course it depends on what voltage you're switching, but the typical value of total gate charge for that mosfet is only around 1nC. So we have about 1nC * 5V = 5nJ of energy loss per switching cycle. Even at 100 kHz that only amounts to 0.0005 Watts of additional dissipation in the micro controller.

Regarding the instantaneous power dissipation that may occur while switching, consider the pin output characteristics (taken from 328P datasheet) below. Note that the output resistance is about 25 ohms (20mA at 0.5V drop). Also note the slight curvature of the characteristics, showing increasing resistance with increased voltage drop as is typical for a mos output.

So we could estimate that the maximum instantaneous power dissipation would be somewhat less than 100mA * 2.5 V = 0.25 Watts, albeit very briefly.

Personally I would be very surprised if the power dissipation (instantaneous or otherwise) would be an issue. Even for larger mosfets I suspect that the need for an external driver (to achieve snappier switching and reduce switching losses in the mosfet) would arise well before damaging power losses occurred in the MPU pins.

Having said that however. Not everyone has an oscilloscope to detect unwanted oscillations, or the ability to layout the gate drive to avoid it, so guidelines like adding the series gate resistance really aren't a bad idea in any case.

Hi,

Now that you mention it, i dont know what the internal power dissipation would be for a direct short to ground for a long time period.  If the pin current is limited then the power cant be that high.  For example, with 50ma output with 5v the internal dissipation can only be 0.050*5=250mw also.  If 100ma, then 500mw of course.  That would be with an infinitely large capacitor or equivalent gate capacitance.  I guess i would have to test the pin to see what the max current could ever be assuming a 5v drop and constant short to ground.  A 50 ohm resistor would immediately limit this to 100ma if in fact the pin could actually put out that much current.  If it cant, then maybe the resistor does not help for this particular issue.

[krupski]

MOSFET designers do a lot to reduce the Miller capacitance in devices, its not that important at low voltages.

The plateau is where the channel is formed, basically - the charge in the gate mirrors the charges forming
the gate.  Most of the channel area is protected from the drain voltage since the substrate is electrically
connected to the source.

There is, of course, still some drain-gate capacitance, but the total charge on the gate is much larger
than the charge on this capacitance in a good device.

I did some experiments switching 3A with a drain voltage of 0.5V, 9V and about 25V on an STP3020L
logic-level StripFET, and the fall in drain voltage (blue) happens basically before the bulk of the gate
charge.  The gate drive is an Arduino pin (so about 30 or 40 ohm source impedance)





Note the horizontal scale is 25ns/div, vertical is 1V/div for yellow (gate voltage), 5V/div for drain(blue).
The time taken to charge the plateau is extended for larger drain voltages, which suggests some sort
of delayed Miller effect, but note the plateau is present even at a 0.5V drain voltage, when the Miller
effect is for all purposes absent.

Note also the sudden onset of drain voltage falling at the moment the plateau starts - clearly this
is when a channel first forms (the inversion layer)

[ Note that due to the bench power supply and long leads I didn't get meaningful switch-off waveforms, and thinking about it I suspect the change of charge in the drift region is probably responsible for the delayed second part of the plateau - that change is proportional to the drain delta-V... ]

Those traces are very interesting. Notice in the first trace that the capacitance is actually feeding forward a small charge into the drain (the tiny positive bump).

As far as a "delayed Miller effect", I disagree. In the second and third image, you see that the Arduino drive pin is held at an almost constant voltage (the Miller effect "fighting" the Arduino pin trying to go high), then when the mosfet is mostly saturated, the more shallow slope is due to ONLY the gate capacitance slowing down the rise time of the Arduino pin.

Notice that it takes almost 250 nS for the Arduino pin to reach equilibrium. I'm assuming (since you didn't mention it) that you used no resistor between the pin and the mosfet gate, correct?

All in all, very interesting data!

[krupski]

As FETs have always been the 'delicate' flower compared to BJT, due to potential for static damage,

That's only during handling. Look at how tiny SMT mosfets are used to switch 10 amperes of more on PC motherboards (the multi-phase CPU power supply).  It's amazing that those little parts drive current through toroids wrapped with #14 or #12 wire!

[krupski]

We used to joke that they were only there to protect the main fuse.

That's a common joke: "The transistor will always blow to protect the fuse". 

[TomGeorge]

Hi,
[soapbox]
I find in servicing equipment with MOSFET/IGBT catastrophic failure.

If a gate resistor is fitted then most of the  time the driving circuit is protected.
Repair, replace MOSFET usually fixes problem.

If NO gate resistor fitted then driving circuit damaged.(sometimes guts blown out of driving circuit)
Replacing MOSFET not worth trying as driver circuit is usually a controller of unknown origin/program, unless its a driver IC that has not had the part number rubbed off.

No brainer fit a gate resistor everytime.
Or a dedicated Driver IC(leave the part number on it)
[/soapbox]

Tom....

[JohnRob]

I just finished with this topic in another thread.  My reason for suggesting a series output to gate resistor was to protect the  µP for damage if the Fet were to be miswired.  Or if the fet failed which usually turns the fet into a lump of resistive material.

These questions are from novice folks who need some extra safety if it can be had.  Now those that know how to design higher powered devices will not be here asking for gate resistor advice.

So the answer also depends on the audience.

[deeec]

I read the whole thread and I'm chiming in as a Physicist like BillO.

I, too, had a gut instinct that for my application with MOSFETs there was absolutely no need for the gate resistor.

In my view, the research conducted here has shown where the lines are.  But this can be extended just for the thinking.

In physics, we define what current actually is - a rate of flow of charge in Couloumbs per unit time in seconds.  This is where our contentions lie, and the first derivative with respect to time i.e. dI/dt.

Now I believe the datasheet current limits are for steady state constant current i.e. DC over a resistive load.

There is something that was briefly mentioned in this discussion but not considered further.  The size of the interconnect leading to the pin from within the die is the limiting factor.  It is a resistor.  The die has been delidded and scanned and we can measure the actual width and read up on the depth and measure the length of the wire.

Without going to too much trouble I assumed the width at 0.3 microns due to the 130nm chip process, the depth at 0.48 microns from this article's aspect ratio of 1.6: https://web.stanford.edu/class/ee311/NOTES/Interconnect%20Scaling.pdf.  I assume an interconnect internally to be 0.5mm long at most.  Someone with the inclination can be more precise.

Inputting these values into a calculator at: https://www.allaboutcircuits.com/tools/trace-resistance-calculator/ results in a resistance of 64.8 Ohms within the Atmega 328p chip itself (at room temperature.  I got the 0.3 micron number from here: https://en.wikichip.org/wiki/130_nm_lithography_process#130_nm_Microprocessors
Interconnects are often copper but sometimes gold.

We will be in the right ballpark.  Now we add the resistance from the chip leg to the arduino pin.  But we're already into too many assumptions.  I just want a ballpark number and 5 Volts divided by 64.8 Ohms equals 77 mA.  This is pretty close to the 88mA that has been seen experimentally.

I leave it to others to be more precise than this.  My MOSFET gates will charge in about 20us so my duty cycle will be low.  My average current at the frequency I use will be about 34mA which is in spec according to the data sheet and your collective experiments.

Now what gives rise to an amperage limit within the Arduino?  HEAT!  This is why when people want to overclock processors they cool them with liquid nitrogen.  Increasing the clock rate means more electron flow which means more current, which means more heat.

I think the interconnects are a limiting resistance and would allow a steady state of 88mA if the temperature were kept reasonable.  It could handle more under active cooling or submerged.

But at least we have an estimate of where this resistance and hence current limit comes from.  I played with the values and substituted gold for copper.  It's a bit of a wild goose chase.  Obviously - if lightning hit the chip for an attosecond it would fry but the average current over a whole second would not be much.  The greater hysteresis from rapid heating and cooling would reduce the life of the chip, but not too greatly - just test the temperature rise and build in cooling.  Stress test it after that.

We can understand why a chip manufacturer would want to be conservative with their current ratings.  But if you are a product designer, and you want to make a million units of something it is worth your time to get scientific proof on the actual limits in your use case.

[Smajdalf]

The MOSFET driver resistance is important IMHO - according to the typical characteristics the total pin resistance changes with supply voltage and load current. A simple bonding write resistance should be constant under those conditions.