Lost trying to calculate switching time of Power MOSFET - Page 1
Forum section: electronics / projects

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 03:16:40 am
content: Hey All,I am trying to drive IRF7749L2PbF MOSFETs with DRV8323 Gate driver, and I thought it would be nice idea to calculate switching time to estimate switching loses. However more I tried to study about it more I got confused, I understand that there are some basic parameters given in the data sheet on the MOSFET which are related to switching are:Turn on delay time - T(d(on)) : 17ns,Turn off delay time T(d(of)) : 78ns,Rise Time - T(r) : 43ns,Fall time T(f) : 39ns,Total Gate charge - Q(g) : 200nC andGate to Drain charge - Q(gd) : 71nC,I understand that total turn on time is T(on) = T(d(on)) + T(r) and same foes of T(off).I understand that there timing given in the datasheet are in specific condition and the gate drive must me able to source and sink the gate charge, DRV8323 can source 1A and sink 2A in the datasheet (section 9.2.1.2.2) of the drive the equation given to calculate I(DRIVEP) as Q(gd)/t(r) and I(DRIVEN) as Q(gd)/t(f)However on the youtube I have seen equation to be I(drive) = Q(g)/t(r) so I am lost a little.To make the matters worst for me I compared IRF7749L2PbF with AUIRF7749L2TR which has lower Q(g) and Q(gd) but has higher switching time T(d(on)), T(r), T(d(off)) and T(f), I thought the less the gate charges faster the switching time.Voltage is Aprox 48V from a battery so max 55V, Current consumption is about 60A RMS. I am designing a ESC for BLDC so yes it is an inductive load.I wish someone can help me with this, Thanks in advance

author: rx8pilot
author_rank: Super Contributor
timestamp: February 15, 2018, 06:43:56 am
content: The turn on delay time is not particularly useful spec. It tells you (roughly) how long it takes just to get to the Vgs(th) threshold where the device begins to flow current.That is the beginning of the ohmic region where the device will be dissipating heat.I watched a series of really great videos recently that cover this topic. When I get back to desk I will post the links.Like I mentioned in your other thread, it's a thick topic. For small currents, you can make mistakes and still have a working circuit. At the current levels you are after, small mistakes turn into smoke real fast making troubleshooting very difficult.Short and misplld from my mobile......

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 15, 2018, 10:07:46 am
content: Quote from: blueskull on February 15, 2018, 03:18:44 amwhy run a simulation when we can test this in real life? consider the switching circuit as black box. put a load and power source on it. measure Vin, Iin, Vout Iout. efficiency will be Pout / Pin = (Vout . Iout) / (Vin . Iin) x 100. another way is probe with DSO the worst case is ((Ttotal - Tswitch) / Ttotal) x 100. Ttotal = total off-on-off time, Tswitch = time needed from mosfet fully off to fully on + time needed from mosfet fully on to fully off. sorry i cant help with math intensive of this because too many unknown during switching on and off period, for eg is Rds linear with Vds? i dont know. but i believe they can be measured with eyeballing the DSO for V and I curve.

author: amspire
author_rank: Super Contributor
timestamp: February 15, 2018, 10:41:15 am
content: You have left out some of the most important information such as voltage rails, current and if there an inductive load. A circuit as well is even better.Lets start with the inductive load - it will be trying to maintain a constant current as the mosfet is turning off. If the drives can manage 2A and the inductor has 10A current, then that inductor current has to flow somewhere - into the inductors internal capacitance, into any external capacitors, into the mosfet drain to source capacitance and into the drain-gate capacitance. If you have over 2A flowing into the gate from the inductor, you can have the mosfet actually turning on partially as the drain voltage is rising and this can be a massive stress on the mosfet. With the dV/dT across the gate-collector capacitance, you can get very high currents.Can you give more details? I always prefer just to use some simple calculations based on the I = Cdv/dt formula to understand the circuit. It really helps in adjusting the circuit. Ideally you can run very conservative and the hand calculations will be fine. If you need to run right on the mosfet limits, then simulating the circuit may be needed.

author: danadak
author_rank: Super Contributor
timestamp: February 15, 2018, 11:28:56 am
content: Could work with this as a first order approximation, modify it for gatedrive parametrics.https://www.circuitlab.com/editor/#?id=t67g8zRegards, Dana.

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 03:36:03 pm
content: Hey,Sorry for leaving out the details.Voltage is Aprox 48V from a battery so max 55V, Current consumption is about 60A RMS. I am designing a ESC for BLDC so yes it is an inductive load.Kind Regards,Sparsh

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 03:59:43 pm
content: Is it possible to get an estimate for the worst switching time with those MOSFETs with those Gate Drives, so I can get a estimate on the worst switching power loss?Also can someone please exaplain why the MOSFET with lower gate charge has higher switching time?

author: rx8pilot
author_rank: Super Contributor
timestamp: February 15, 2018, 04:41:44 pm
content: Quote from: satyamfifa on February 15, 2018, 03:59:43 pmYou are still giving too much weight to the data sheet terms - td(on) does not describe much.The total time spent in the linear region and the integrated power dissipation is based on many factors and td(on) is merely an interesting note.

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 04:59:33 pm
content: Hey,Thanks for your reply, I do understand what you are saying, however according to you own post Q(g) should be a determining factor on the T(r), and currently I don't understand why a MOSFET with lower Q(g) and Q(gd) about have higher T(r) and T(d(on)), it's highly counter intuitive to what I know so far.And I am asking this because I would rather use AURIRF7749 instead of IRF7749 in my design to lower the switching loss, and maybe even some other MOSFET with higher RDS(on) with lower switching loss.Kind Regards,Sparsh

author: Benta
author_rank: Super Contributor
timestamp: February 15, 2018, 05:10:43 pm
content: Datasheets are tricky. I haven't read yours, but here's a couple of ideasYou are trying to compare two similar devices, but you really need to check out the test setups in the datasheets.Eg, what is the generator impedance feeding the gates? How is rise time defined (10%...90% or 20%...80%)? What is the load impedance? And so on. Reducing a comparison to just a couple of figures won't work.And remember: some suppliers excel in "specmanship".Cheers.

author: rx8pilot
author_rank: Super Contributor
timestamp: February 15, 2018, 05:26:41 pm
content: Quote from: Benta on February 15, 2018, 05:10:43 pmAint that the truth - many specs are also calculated as estimates and never tested practically.I have spent a ton of money to measure my actual circuits to see what they actually do. In my case, it is usually faster to just build and test as opposed to trying to anticipate all the factors and build an accurate simulation. Please keep in mind, even if my posts look smart - I myself am not claiming to be a MOSFET genius in any way.It is true that Qg is a big factor in determining how fast you can transit the linear region of your device - but there are other factors that can impact that. There are also other loss factors like Coss and other capacitances.

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 05:29:10 pm
content: Everything in the datasheet test setup seems similar, Load and Impedance is not defined, rise time is deffined the same way in both (10%....90%), also gate resistance R(g) is same in both the MOSFETs.Sorry I am not familiar with my spacemanships

author: rx8pilot
author_rank: Super Contributor
timestamp: February 15, 2018, 05:39:12 pm
content: 'specmanship' is the art of designing specs that make your device look more appealing.Keep in mind (with any datasheet) they are marketing tools and the companies have marketing people do the final review. Where a number can be legitimately stretched....it just may be.Many of the specs are extremely hard to measure - so they are estimated or calculated. Much of what you see is a 'best case' scenario that is hard to achieve in the real world.

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 05:48:43 pm
content: So is it good to assume that the AUIRF version, which is the Automotive qualified chip is more "truth telling" than the normal IRF version?Sorry I am looking for some quick answers here, as neither I have equipment nor time to actually test the MOSFETs.....

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 15, 2018, 06:10:35 pm
content: Quote from: satyamfifa on February 15, 2018, 05:48:43 pmthis is going to be fun, since nobody is expert including the datasheet manufacturer... just take the rise and fall time figure, scale with your working voltage vs the test voltage in the datasheet, if you want to be optimistic, divide that into half, then decide how long it will be turned on. punch in the figures in my formula above then you get your answer quick... but if you want to get academic, it will take a little while.

author: rx8pilot
author_rank: Super Contributor
timestamp: February 15, 2018, 06:19:33 pm
content: Quote from: satyamfifa on February 15, 2018, 05:48:43 pmThat was my biggest surprise when I started doing power electronics - test and measurement is very hard and very expensive. I learned that I could rough in a circuit that initially worked but then later broke and I rarely had any idea why.The more I studied the circuit behavior and the devices involved, the more I learned about what can go wrong. Then I learned that seeing those wrongs in a test is quite a challenge. All the elements in a circuit - including parasitics - add up to a very complex interaction. You can be violating the SOA of the MOSFET and never know it since it is tough to measure accurately. That can easily be a problem where your circuit stops working after some amount of time.Power electronics is very challenging.Quote from: Mechatrommer on February 15, 2018, 06:10:35 pmYes - at some point you just have to make your best guesses. If you cannot test it - it becomes a big risk. It is easy to get a circuit to power up once. Vastly more challenging to get a reliable circuit with reasonable performance. Especially a 60A 48v switching power converter. That is a challenge for someone very skilled and experienced in the field.

author: satyamfifa
author_rank: Regular Contributor
timestamp: February 15, 2018, 06:30:56 pm
content: Okay I did some comparison of IRF7749-AUIRF7749 with IRF7759-AUIRF7759 and IRF7769-AUIRF7769, and I found that the discrepancy does not exist in those other version infact even the gate charge is same in other version, therefore everything is similar in AU and IRF version.The only thing I found exceptional about AUIRF7749 is that the Gate Voltage V(gs) is rated at 60 V instead of usual +-20 in all the versions both AU and IRF.So I think I might be onto something here

author: jmelson
author_rank: Super Contributor
timestamp: February 15, 2018, 09:13:21 pm
content: The numbers out of a datasheet for the transistor are TOTALLY meaningless. The gate driver probably contributes more than the FET itself. So, you really have to evaluate the driver, the gate resistor PLUS the FET as a system to get any meaningful knowledge in advance of testing it out.Some gate drivers have really weak drive, that will give slow rise/fall times and long turn-on and turn-off plateaus due to the Miller effect.Some gate drivers can deliver amps of gate current both charging and discharging, and that can give you much faster turn-on/off, to the point that the dI/dt can cause circuit malfunctions. A problem I ran into in a motor drive with half-bridge switches was that when the high-side transistor shut off after building up strong current flowing out, the inductance caused a reverse voltage across the low-side transistor, which took out the gate driver. I had to put an ultra-fast diode and an RC snubber circuit on the node between the two transistors to prevent his.Jon

author: danadak
author_rank: Super Contributor
timestamp: February 15, 2018, 09:31:46 pm
content: Some ref material on MOSFET switching -https://www.infineon.com/dgdl/an-937.pdf?fileId=5546d462533600a40153559ea1481181https://www.infineon.com/dgdl/Infineon+-+Application+Note+-+Power+MOSFETs+-+OptiMOS+P-channel.pdf?fileId=db3a304341e0aed001420380cc13101bhttps://toshiba.semicon-storage.com/info/docget.jsp?did=13416https://www.fairchildsemi.com/application-notes/AN/AN-7502.pdfhttps://www.vishay.com/docs/73217/an608a.pdfhttp://www.ti.com/lit/an/slva716/slva716.pdfhttps://www.microsemi.com/document-portal/doc_view/14692-mosfet-tutorialhttp://www.fujielectric.com/products/semiconductor/model/powermosfets/application/box/pdf/MOSFET_E_140926_01.pdfhttps://www.silabs.com/documents/public/application-notes/AN1009-driving-mosfet-and-igbt-switches-using-the-si828x.pdfhttps://pdfserv.maximintegrated.com/en/an/AN6307.pdfBedtime reading......Regards, Dana.

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 16, 2018, 08:22:28 am
content: Quote from: jmelson on February 15, 2018, 09:13:21 pmnot to be too pedantic, its not totally meaningless. they can be a good estimate for rough figure such as gate charging/switching time, maximum current etc, SOA and derating curve also can be usefull as long as you dont work near the curve boundary... imho...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 16, 2018, 07:00:09 pm
content: Incidentally, there are two outliers I know of regarding rise time:- A lot of Fairchild (now On Semi) parts were (are?) measured at 50 or 25 ohm drive impedance. That is, a 50 ohm function generator, or terminated into 50 ohms (50 || 50 = 25 ohms). This gives these parts a strikingly long rise/fall time spec (surely this was NOT written by marketing?!). Yes, they still go fast when you drive them harder.- Some TI NexFETs have been rated with "0 ohm" drivers. Yes, I asked, and no, they don't know how.Most choose a drive impedance typical for use, usually in the 1 to 10 ohm range for larger, power-ey transistors.Tim

author: jmelson
author_rank: Super Contributor
timestamp: February 16, 2018, 08:46:09 pm
content: Quote from: Mechatrommer on February 16, 2018, 08:22:28 amJust plain quoted turn-on/turn-off times for a FET are of little use, I don't know why they even specify this without specifying the gate driver. That is what I was referring to in the "meaningless" comment, SPECIFICALLY in reference to switching time. The actual gate charge numbers are very useful, when COMBINED with currents available from the gate driver, to calculate the turn-on and turn-off times. And, of course, OTHER parameters, like on resistance, gate threshold, max currents at various conditions are all QUITE useful.Jon

author: amspire
author_rank: Super Contributor
timestamp: February 17, 2018, 12:52:18 am
content: Quote from: satyamfifa on February 15, 2018, 03:16:40 amA lot of the specifications you have quoted are typical under one set of conditions. They can be useful if they happen to exactly match your circuit, but it is always worthwhile to get back to some basics.First, there is the ON channel resistance which depends on the gate voltage. For 60A switching, you will need at least 10V gate drive. It is typically 1.1 milliohms for this device and it increases as the device gets hotter. Doubling it would probably be a good initial design approximation. The Rds power loss is just simple I2R calculations.The next loss is due to the charge in the drain to gate and drain to source capacitance. Every time the mosfet switches on, this charge is dissipated in the mosfet. Don't over-complicate things - just add the two capacitances together and don't worry about the fact that the gate is not at 0 volts.Looking at the curves, at 55v, the total drain capacitance is a bit under 1800pf - that is close enough.The energy in this capacitor = 1/2CV2= 2.7uJ. So if you were running at 100KHz, that would be 0.27W of power.The gate charge from the graphs with a 10V gate drive and 48V on the drain is about 210nC. The gate resistance is 1.1ohms. This charge has to be added and removed every cycle, so at 100KHz, the average current is100KHz x 210nC x 2 = 42mA (The times 2 factor is for the charge and discharge)So the power lost by the gate resistance at 100KHz is 42mA x 1.1 = 46.2mWNow we get to the hardest factor - the losses due to the mosfet still conducting while the drain voltage is rising or falling. Also the rise time changes with the current. If you really wanted to maximize mosfet efficiency, you would drive the gate with +/- 10V with a 10A capable driver. Practically, a driver like the DRV8323 with a 1A source and 2A sink and a 0 - 10V? voltage is more common. This is an interesting driver - it has programmable drive currents and it actually acts as a constant current source/sink.As we mentioned, at 55V, the drain capacitance is about 1800pF. That effects the rise time and it is dependant on the inductor current. At 1A in the inductor, rise time is = C.dV/I = 100nS. At 60A, this calculation drops to below 2nS. So at the peak current, you can forget it. At low currents, it is relevant.Now the gate charge is 210nC and the driver IC can sink this at 2A. The gate resistance can discharge the gate at almost 5A, so the 2A current limiting of the driver dominates here. To discharge 210nC at 2A takes 210nC/2 = 105nS. This number is misleading since the mosfet is still on at 5V, so during the 10V to 5V discharge time, the mosfet has not started to turn off significantly.The mosfet only starts to turn off at 60A when the gate voltage is about 4.5V and it is essentially off by about 3.5V. These numbers change with temperature, so I would say on = 5V, off = 2V. The 0V gate capacitance is about 12000Pf, but this falls with voltage, I am just going to use 10000pF, at 2A. The time taken to discharge the gate by 3V (between 5 and 2V) at a current of 2A is 10,000pf x 3V/2A is 15ns. That is a much better figure.However, the driver also has to cope with the current from the drain-gate capacitance which is something like 800pF. If we calculate the risetime at 2A, you get dT = 800pF x 55V/2A = 22ns.The way the maths works out, you have to add these together to get the final mosfet turn off time of 37ns.The risetime will depend on other issues, like the internal capacitance of the inductor, the properties of the ferrite, etc, but as long as the drain risetime is significantly over 37ns, then the mosfet will not be on significantly while the drain voltage is rising. If at 60A, the inductor can force the drain to rise in 20ns, then the mosfet will still be partially on while the voltage is rising. This can really start to add to the power losses, and it can also be a reliability issue for the mosfet.This is probably the point at which you would build the circuit and check the risetime. If it is 50ns or more, then the mosfet will probably run extremely efficiently. If it is 25nS at full current, the mosfet is probably getting very hot and you may have to consider a more powerful gate driver. Perhaps also using a negative voltage for turning off the mosfet.I hope that gets you started. I haven't used any equations other then basic resistance and capacitance ones. If you start this way, then when you see a design guide equations, you can actually work out how it is derived which means you can understand it.Richard.Edit: I ignored the output rise and fall times for the driver IC, but it is important. If the gate is charged to 10V and the mosfet doesn't not start to turn off till 5V, the critical thing is that the gate driver output is fully on by the time the gate reaches 5V. I haven't gone through the turn on losses. Hopefully, you can work through that yourself.Also, I forgot to calculate the gate resistive losses due to the gate-drain capacitance. At a guess, it could be 70mW at 100KHz.Finally, there can be other significant losses of power. If there is ringing on the drain when the mosfet turns off, then if the drain voltage exceeds 60V, the mosfet will start conducting like a zener diode and it will absorb this energy from the inductor. The mosfets can be used this way, but it is not efficient design. With a good layout and good Schottky diodes, hopefully this never happens.

author: satyamfifa
author_rank: Regular Contributor
timestamp: June 07, 2018, 02:56:18 pm
content: Dear Sparsh Bhonwal,We are sorry for our late answer.You observed:IRF7749 has higher gate charge 71 nC compared to AURIF7749 of 46 nC.IRF7749 has higher Ciss 12320 pF compared to AURIF7749 of 10655 pF.IRF7749 has higher Coss 1810 pF compared to AURIF7749 of 1627 pF.IRF7749 has lower tr,tf 43ns,36ns compared to AURIF7749 of 149ns,88ns.But there also is another parameter.IRF7749 has lower Rg_internal 1.1 Ohms compared to AURIF7749 of 1.5 Ohms.Furthermore, as chip parameters vary, then if they go through the same test yielding different results, the test works like a filter which decides which chip becomes an IRF7749, which chip becomes an AURIRF7749 and which chip goes to the thrash bin.Best regards,Infineon TechnologiesWerner Schmitt, Support Engineer