Does driving MOSFET gate create heat? - Page 1
Forum section: electronics / beginners

author: fonograph
author_rank: Frequent Contributor
timestamp: June 04, 2017, 11:48:48 am
content: In the book I read that while reactance acts similiar to resistance,it doesnt dissipate energy like dumping current into resistor would.Corrent me if I am wrong,mosfet gate is practicly a capacitor.Lets say I have mosfet that is being driven by AC with 1W power,will that mosfet gate absorb that 1W and turn it all into heat?Notice I ask purely about gate... source to drain current in our example is zero,this is purely about heat generation inside gate by AC current from gate driver.

author: Rerouter
author_rank: Super Contributor
timestamp: June 04, 2017, 12:03:32 pm
content: the capacitor in this case has an ESR of some value, as with all capacitors, this is both the lead, the bond wire, and the die traces, these will dissipate a resistive component of the power meaning yes, some heat will be generated. even if it is ideally a capacitor, parasitics come out to play.

author: fonograph
author_rank: Frequent Contributor
timestamp: June 04, 2017, 12:09:07 pm
content: How many % of the power gets converted to heat on average? Does the gate ESR,leads and trace resistance turn 100% of thepower into heat if we drive the gate with contimuous sinewave or its less than 100%?

author: danadak
author_rank: Super Contributor
timestamp: June 04, 2017, 12:22:28 pm
content: Most definitely CV2f Pdiss consumed in gate -http://www.eetimes.com/document.asp?doc_id=1278970http://pdfserv.maximintegrated.com/en/an/AN1832.pdfhttp://ww1.microchip.com/downloads/en/AppNotes/00799b.pdfRegards, Dana.

author: rs20
author_rank: Super Contributor
timestamp: June 04, 2017, 12:26:11 pm
content: Asking about the percentage is a truly bizzare question to ask, and I suspect you're barking up the wrong tree thinking this way. There may indeed be a "typical" figure, but I suspect no actual engineer has any idea what that percentage is, other than "negligibly small". Thinking in terms of a percentage is pointless if you don't know what the magnitude of either value is; if you have a specific circuit in mind, it's easy to calculate the dissipation in the gate directly in absolute terms -- see Dana's links for answers to useful questions.

author: Rerouter
author_rank: Super Contributor
timestamp: June 04, 2017, 12:26:42 pm
content: http://www.ti.com/lit/an/slyt664/slyt664.pdfpage 24the majority of your power will be dissipated in what is driving the mosfet gate, not the gate itself, as the signal source will have some esr of its own, and its esr is generally a higher number,if you know the 2 values, you could determine the ratio,If you somehow had a magical signal source with lets say 1 fempto ohm esr driving the gate, then yes almost the entirety of the power would be dissipated by the mosfet gate, for all of 10 nanoseconds as the bond wire to the gate would vaporize from the amount of current needed to reach 1W of signal power.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 04, 2017, 12:35:29 pm
content: Most power MOSFETs will have a specification for the intrinsic internal gate resistance, called Rg, which can be used to determine the power dissipated in the gate structure.

author: danadak
author_rank: Super Contributor
timestamp: June 04, 2017, 12:52:08 pm
content: Basic Capacitance power consumption -http://web.mit.edu/6.012/www/SP07-L13.pdfhttp://www.ti.com/lit/an/scaa035b/scaa035b.pdfRegards, Dana.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 04, 2017, 01:47:31 pm
content: Quote from: rs20 on June 04, 2017, 12:26:11 pmAbout 10%.What's wrong with that?In the end, loss is of course 100%, because the capacitive energy is charged and discharged through equivalent resistance. This is the amount into and out of a typical gate, not counting the driver, and corresponds to the amount of power you could save with a clamped inductive ("lossless") driver.The loss is smaller when you go slower, because the equivalent circuit is basically ESR + C. At higher frequencies (towards the practical cutoff of the transistor in question -- for power MOSFETs, usually 10s of MHz), it should become apparent that it's not just ESR, but distributed ESR, with a diffusion component: the Q drops to a modest value and stays there over some frequency range, while the resistance and capacitance both fall with rising frequency (the signal isn't getting as deep into the gate structure, so you're seeing less of its resistance and capacitance).It occurs to me that I've not seen a plot or equivalent circuit of this, but it should be the case. It would be interesting to measure.So, if you want low losses, you need to switch more slowly, or get transistors with lower Rg (gate spreading resistance).Conversely, if you want to drive power MOSFETs very fast (for fast switching, or use as a linear RF amp), you need to contend with the input impedance being quite low, and only moderately capacitive.RF transistors are made with far lower losses, so that they are largely capacitive until a few 100 MHz.Tim

author: chris_leyson
author_rank: Super Contributor
timestamp: June 04, 2017, 03:23:58 pm
content: I recently had to redesign a cascade buck switcher running at 250kHz that was only 65% efficient, 48V to 15V and then 15V to 5V. The design engineer had used a 50A device for the the low side switching mosfet in the 48V to 15V converter where a 2A device would have been sufficient. The 1.5nF gate capacitance caused the controller driver stage to overheat and on a hot day the controller would shut down when it reached its thermal limit of 150C. Some of the controller chips would shut down permanently and not recover. The designer had used a 50A device thinking it would reduce conduction loss and obviously hadn't considered the required drive power. Driving a mosfet gate can cause overheating and it cost the company quite a lot in service recalls. Changed the cascade design to a flybuck 85% efficient and 20C cooler.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 04, 2017, 04:28:24 pm
content: Quote from: chris_leyson on June 04, 2017, 03:23:58 pmLikely the other 15 percentage points were in the drain energy (capacitance). Every time that thing switches on, it discharges all ~50V in one big gulp. You can't measure this directly, because the capacitance is in parallel at the transistor -- the current spike doesn't go out the source pin, it stays inside. But the power is there, oh yes it is...Also, 1.5nF might be the small signal capacitance, but the large signal equivalent is usually triple that, give or take. A better estimate is Qg(tot) / Vg(on). So, 50nC at 10V is 5nF equivalent!Another good reason to use external gate resistors: keep the driver cool. You usually spec a driver for the current it needs to deliver, no more (because cost, and sometimes because switching loss or quiescent current in the driver itself). Typical drivers are rated by peak current into a quasi-static load (a large capacitor), so a 2A (at 10V) driver is about 5 ohms. To deliver that current into your transistor, you can't add any external resistance.This power dissipation limits the Qg * Fsw you can run a particular driver at.You can get a bigger driver and add an external resistor, to get the same total Rg for example, while shifting the heat to a safer and/or cheaper location (resistors). (Whether this is actually cheaper, or more compact, or better than just using a driver with better thermal management, who knows -- there are lots of options.)Oh, and you can use a quite small driver (a few 100mA, say) paired with a complementary pair of BJTs (like PBSS303NX/PX or any number of other similar types), to get a really low equivalent driver resistance (not quite Rdriver / hFE, but better than a 10x improvement for sure) on the cheap. I should do that from time to time...Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: June 04, 2017, 05:59:26 pm
content: I am not sure if I understand this correctly,it seems simple but the numbers I get are so beyond what I expected that I think I must not understand it properly.I am looking at transistor datasheet,total switching energy is 50uJ,maximum freqency is 100MHz.Well,50 micro joule 100 million times per second is 5000 watts.Thats ridiculous! That transistor needs 5000 W gate driver just to run at its maximum speed,note,this is single discrete mosfet.Is it really 5000W or am I reading it wrong.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 04, 2017, 06:09:11 pm
content: Quote from: fonograph on June 04, 2017, 05:59:26 pmYou are, indeed, reading the datasheet wrong. The 50uJ switching energy is not lost in the gate, it is from charging/discharging the output capacitance (mainly Cds, some contribution from Crss) and will apply to a given drain-source voltage (usually somewhere between 50% and 80% of max rated Vds). So this energy scales with the voltage being switched by the MOSFET and can be eliminated almost entirely through quasi- or full resonant transition switching.Power dissipated by the internal gate resistance will be much, much lower; it is, essentially, an I²R loss where R is very small and I is the average gate current, which is also very small (yes, peak current can be quite high, but it only lasts for about as long as it takes to slew the gate up or down, which is usually a tiny fraction of the pulse period).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 04, 2017, 09:29:09 pm
content: What transistor?MOSFETs are not specified by "maximum frequency". Applications are too diverse for such an arbitrary limit.If you've added up the switching delay times, that's a kind of maximum frequency, but only at the specified conditions, and yes, doing that continuously will more than exceed the power ratings.This is one of the differentiators about power switching devices: they are rated for far more V and I than they could possibly dissipate continuously.(The other factors are the very low Rds(on), high capacitances, and relatively high Rg, which make power switching devices suitable for lower bandwidths (on the order of ~100MHz, versus RF transistors in the 1GHz+), and switching frequencies much lower still: at least 1/10th the linear bandwidth, or lowish MHz.)Indeed, the switching transient dissipates a huge amount of power (~kW for a ~100W size converter), which is one reason why the switching frequency has to be low compared to the switching speed.Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: June 05, 2017, 04:55:08 am
content: So,if I understand it correctly,you need less power to drive the gate of mosfet when the source - drain voltage and current is zero.That means switching mosfet gate while the mosfet is disconected from source,just laying on table is easy but once its connected to source in some real circuit with current going through it,then the switching is much harder/power hungry.Is that right?

author: danadak
author_rank: Super Contributor
timestamp: June 05, 2017, 11:04:40 am
content: If a MOSFET drain and source are grounded, you will be driving theCgd and Cgs capacitance's. So no, there is power dissipated by thegenerator driving the gate.Regards, Dana.

author: tatus1969
author_rank: Super Contributor
timestamp: June 05, 2017, 11:36:11 am
content: Here is a picture showing all the capacitors are involved in the switching process:Charging a capacitor requires energy: E = 1/2 * C * V^2.When the gate capacitor, Cgs, of a MOSFET is charged, this energy comes from the driver supply. That's easy, but what happens when the transistor is turned off again? The driver shorts out the charged gate capacitor. The energy that is stored in the gate is entirely dissipated into heat during this process.That means, all of the energy that is used to charge Cgs is lost in the end and dissipated into heat, distributed between driver, gate resistance, and PCB trace resistances, depending on their values.Once you know the dissipated energy per on-off cycle, and the switching frequency, you can calculate the corresponding power loss: P = E * fYou can repeat this method to the other capacitors, Cgd and Cds.[Note that this is a bit over simplified but should explain the idea... For example, charging the gate creates additional losses in the driver.]

author: fonograph
author_rank: Frequent Contributor
timestamp: June 05, 2017, 02:47:14 pm
content: I get it,there is parasitic resistance in gate and you need current to drive the capacitance between gate and source/drain,but back to my question,does it take less energy to switch mosfet when its not conducting current?Becose my initial fear that after calculating the switching energy it came to be 5000 watts,then Magic Smoker posted this"You are, indeed, reading the datasheet wrong. The 50uJ switching energy is not lost in the gate, it is from charging/discharging the output capacitance (mainly Cds, some contribution from Crss) and will apply to a given drain-source voltage (usually somewhere between 50% and 80% of max rated Vds). So this energy scales with the voltage being switched by the MOSFET and can be eliminated almost entirely through quasi- or full resonant transition switching."he says this switching energy scales with voltage being switched,if the voltage/current at source > drain is zero,like in the "resonant switching",like in my example when I described a bare mosfet laying on bench table,then it should eliminate that crazy 5000 watt gate driver requirment,is that correct?I understand I will still need decent power in the gate driver becose of cgs and cgd but it will be alot less than 5000 watts,atleast I hope so,can you confirm it?

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 05, 2017, 03:29:06 pm
content: Quote from: fonograph on June 05, 2017, 02:47:14 pmAt this point I am no longer sure what you are asking, but it certainly seems like you are confusing the switching energy spec, which is for the drain-source transitions, with the gate charge spec.Generally speaking, the amount of energy lost in in the gate-source transition period (ie - going from on to off or off to on) is not dependent on the drain-source current, whereas the energy lost in the drain-source switching eventisdependent on both the current through the drain-source junction, as well as the voltage across it (once again, this applies during "hard-switched" conditions - ie, when there is an overlap of voltage and current during the transition period). If drain-source current is zero - either because of resonant conditions (ie - "soft switching") or because no load is connected then there will still be a loss from charging/discharging Cds and this loss still scales with Vds. Hopefully this answers whatever you are asking, though this new twist from mentioning 100MHz makes me wonder what daft thing you are attempting here... you definitely aren't going to be using a common power MOSFET at 100MHz; not hard-switched, anyway.When a datasheet gives a spec for "switching energy" in uJ or mJ or whatever, it is referring to the drain-source junction, not the gate-source.

author: fonograph
author_rank: Frequent Contributor
timestamp: June 05, 2017, 03:47:03 pm
content: its simple,I ask if it takes less energy to switch transistor that isnt conducting current at the moment

author: tatus1969
author_rank: Super Contributor
timestamp: June 05, 2017, 10:52:04 pm
content: Quote from: fonograph on June 05, 2017, 03:47:03 pmare you still talking about gate driving power only? and if yes, why? the major loss contributors are conduction loss from switched current while the transistor is on, and switching loss during the transition from fully on to fully off and vice versa. gate loss comes last in most cases. what switching frequency do you have in mind?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 05, 2017, 11:39:11 pm
content: Gate drive power does not depend upon load current, but it does depend on load voltage. The load voltage will not be charged up across Coss without some load current, so it's not completely independent, but that's the extent of it.Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: June 06, 2017, 06:20:50 am
content: Does that mean switching the gate takes less power when the load voltage is zero?Imagine these three switching scenarios : load voltage = zeroload voltage = gate voltage ( load is same voltage as gate requires to be ON,6V )load voltage = 100VWhich of these will take least energy to switch the gate?I read here that switching disconected transistor that is laying on table,you will need extra current becose of cgs and cgd,but if load is same voltage like gate that is fully opened,something like 6 volts,doesnt that mean that in that case you would need less gate drive power becose you will eliminate the capacitance between gate and source/drain?

author: tatus1969
author_rank: Super Contributor
timestamp: June 06, 2017, 09:06:30 am
content: can you describe more of what you are trying to achieve? is it just to understand the theory? if yes, look for the term Miller Capacitance, that origins from BJTs, but it is valid for MOSFETs as well. It explains the relationship of changing drain voltage and Cgd.

author: MagicSmoker
author_rank: Super Contributor
timestamp: June 06, 2017, 12:11:56 pm
content: Quote from: fonograph on June 06, 2017, 06:20:50 amI realize there is a potential language barrier here, but fercryinoutloud this has been answered several times already!The reverse transfer capacitance, Crss (essentially the same as Cgd), couples charge between the drain and gate and that charge will scale with drain-source voltage. At high drain-source voltages it can even come to dominate the gate drive requirement - that is, overshadow Ciss (aka Cgs), which is usually 10x-100x larger in effective capacitance value.So, yes, it takes more power to drive the gate as drain-source voltage goes up.Quote from: fonograph on June 06, 2017, 06:20:50 amYou should be able to answer this yourself by now.