Power Electronics - MOSFET Power Losses

Join Dr. Martin Ordonez and graduate student Ettore Glitz in a lesson on power losses in MOSFETs. This video briefly introduces a simplified model of a MOSFET, and then discusses both conduction and switching losses in the device.
Script: Matthiew Amyotte
Visuals/Narration: Ettore Scabeni Glitz
Production: Martin Ordonez
Audio/Video Editing: Gregor Morrison
Music: Back to the Ocean - Jesiah
Photo Credits: UBC Hover Collective, Drew Hays

[@LolodeMillau]

I am currently designing an SMPS and I thank you very much for this reminder. This is very clear and well summarized. FYI, I found a mistake @4:56 : in the Eoff formula, you must replace "tfv" by 'trv" as the voltage is rising to Vdd when the mosfet is getting off. But it is not that important as everyone sees the principle that you have well explained.

[@kabandajamir9844]

The world's best teacher thanks sir

[@ShivamChauhan-ki9jq]

Very well explained... Please keep posting ... Specifically for gate driver circuit, snubber circuit etc...

[@darrellhambley7245]

This video shows a good simplified picture of switching losses, very good for new students to understand. The next step after this is to learn what CAUSES the rise/fall times of the current and voltage so that the engineer can make meaningful estimations of losses.

[@miguelsanguinetti8384]

sry but this is not for new students in my opinion,

[@devinerhythem]

Simple and understandable explanation. eye opener

[@saurabhjha5401]

In the next lectures of loss please consider Diode Reverse recovery too.

[@johnf3326]

Thats really simply explained even for the 'amateur'! THANK YOU!

[@23lkjdfjsdlfj]

For lower frequency operation Rds(on) isn't necessarily the cause of dissipating the most power - the capacitance can be responsible for up to 300% more power loss.

[@radhasri9802]

thanks alot...for the clear explanation with neat diagrams

[@abhisekrout8522]

How would I find the values of voltage rise and fall time and the current rise and fall time? In the datasheet I found rise time and fall time. Are they voltage rise time time and fall time or current rise time and fall time???

[@RSP13]

I also wanna know

[@venkat261990]

Dear Professor, The quality of your video is incredible and Mr. Glitz has explained the concept in a lucid manner. I am aware that, it take huge effort and time for making such a video. However, if your team can make more videos on advanced topics (such as resonant converters, switch-node ringing control etc.), then it would be awesome and helpful for students.

[@Chupacabras222]

You are wrong. Vgs does NOT change immediately as you show in your charts. Because of gate capacitance Vgs cannot change instantaneously, but it is rising/dropping with some delay.

[@Cutycats9]

Yeah vgs like a capacitor charging but i think he want to ignore it to make it simple

[@muhammadazka1868]

i want to ask how did Pcon ended up being 3W?, how to calculate it? (10A^2 x 50mOhm x 0.6 Duty Cycle = 3W? HOW?

[@Ozzy3333333]

Does this apply when driving a inductive load (inductor or transformer)? As the inductance will never rise to a flat state.

[@MRdivod]

Correct me if I am wrong but don't you need (In some designs) to account for the power losses due to current flowing from source-drain via the body diode whilst the MOSFET is OFF???

[@danielchatrie3517]

So when my smps circuit have 2kw of mosfets but I can only get 800 watt from the 2kw of mosfet at 225khz and the main voltage only drop 3 voltage and the input current is 12amp 133 volt and am losing 1200 watt how is this even possible to have 60 percent losses

[@dp0813]

Good GAWD! I graduated with a B.S. in EE a few months ago but this video really helped me understand this topic better. Great job guys!

[@deivid7257]

Best video thanks from Colombia 🇨🇴 👌

[@bryanst.martin7134]

Electrons are negative charged particles. Therefore the Source of current flow. Why do academics teach reverse flow?

[@ajingolk7716]

So lets say for a mosfet with rds(on) 0.042 and a current and duty 50% = 33.6 watt ?

[@rohitseth6615]

You explained very well in just small 9 minutes video ,Excellent from INDIA

[@廖宏軒-y2r]

Sir, may I ask why the Vds wouldn't drop till the ID = IDon ?

[@johnaweiss]

4:17 When the device is turned off, does voltage rise because something is building up on one side of the switch? Or something is emptying out on the other side? Or neither? Or both?

[@cnchow323]

It is very nice explanation on all calculation and useful. do you have a template to do all calculations by inputs all necessary parameters? I like to get a copy if possible

[@vitaliyl7030]

Why does the vds start to drop only after drain current reached max?

[@wailalwajeeh6520]

Thanks a lot . Very clear explanation. Can you make a video to calculate power losses in an IGBT? Or is it the same?

[@BrijeshKumar_electrical]

Very good lecture.... Nice

[@mayais]

Can you explain how the Cgs and the Cds affects the calculation??

[@leandroalgenton]

it affects the Ton and Toff calculation, the Cgs and Cds are the capacitance between the gate-source and drain-source, The components and voltage level that you will switch the gate, will interact directy with the capacitance (Cgs) and consequently the time to reach the threshold saturation. In that case, as quick you reach this threshold smaller wil be the energy dissipated.

[@ahmedazaiz2327]

An excellent course on Mosfet losses. Thank you very much.

[@yingpang6366]

This is tooooo over simplified and erroneous

[@RSP13]

Explain, please.

[@teoprotoulis3419]

What a great video! It covers not only the power losses of MOSFETs but also the basic operating principles of them. Thank you!

[@ajingolk7716]

Why rise time or fall time 1/2 ?

[@giorgiofocco6746]

Schöne Grüße an Prof. Tieste.

[@Rizemize123]

When the MOSFET turns on at 3:00 , why the I_D(on) increases *linearly?*

[@alexnoggle1874]

Very handy information.

[@rokibsumon2160]

I have a 650v ups and its use IRLB4132 mosfet. But I can not find this mosfet in local maket please tell me a alternative mosfet of IRLB4132.

[@AbdulWahab1994]

Its explained in really simple way I think my professor should quit teaching anymore.

[@oliverwijesooriya2067]

very excellent explanation

[@andrwsk23]

Thank you! I was struggling with the differences

[@ccv3237]

Excellent explanation. Congratulations, and thank you for sharing.

[@magloirentaba8142]

You're a life saver 🙏

[@abhilasbiswas2085]

Thank you.

[@chethanraju6647]

Excellent, all my doubts got cleared in a video

[@caioborgonha3900]

This vídeo was so helpful, thank you for sharing this knowledge!

[@damny0utoobe]

Excellent explanation of power losses in a mosfet

[@thuyavansathiamurthi3592]

so flipping simple and elegant. THANK YOU

[@mjeezyca]

Thank you for presenting this it is very helpful

[@ManojKumar-jw5ys]

thank you sir

[@snehasheri9353]

Wht is the solution to turn off the load like fan immediately ,when gate voltage is made negative

[@billy936]

If I had a single pulse for .1 seconds, would the power calculation be the same? Could I still just treat it like a duty cycle, D = .1/1 for the calculation?

[@NawtieBoy96]

excellent video . Can I know why a Mosfet has a body diode ? and what does it do ?

[@avicohen3035]

Thank you for this film. You make things easier to understand.

[@nabilahkhoirul458]

Very good and funny videos bring a great sense of entertainment!

[@nachiketadeshmukh8444]

The conduction loss expression may be incorrect in my understanding. Due to a duty ratio of 0.6, the RMS current through the MOSFET should be less than 10 A.

[@meletiosmastrodemos6300]

No, the expression is correct, and you are also correct that the current is less than 10 amps, which is shown in the equation. The RMS current is I*SQRT(Duty Cycle). Since the power loss is calculated by Irms^2*R, Irms^2 = [I*SQRT(Duty Cycle)]^2, which simplifies to I^2*D, so the conduction power loss is I^2*D*Rds,on

[@ashritham1165]

Thank you, It was a very informative video and Impressive presentation. Sir, can I know which software you have used for presentation. The presentation slides where awesome.

[@kabandajamir9844]

So nice

[@clementugbaga6866]

what is the reference book of this lecture

[@jaiminpatel1576]

Nice

[@TamilTechnocrat]

small correction at 4:54 last expression in the RHS ==> (trv + tfi)/2 (corrected) (it is the rise time of the voltage but not the fall time)

[@rameshp7135]

Very nice explanations.....

[@gayathrisony2456]

Great explanation thankyou sir

[@muhammadbilal2000]

Beautiful and to the point. Thank you

[@EngineeringEE]

Thank you so much for the video. Wonderfully described.

[@humbertocuevas5395]

Saludos cordiales atte.desde cd. Juárez Chihuahua México.

[@devarajan2291]

Excellent simple explanation. Gave an in-depth understanding. Thank you so much. Please keep posting

[@asafabraham7206]

Simple and understandable explanation. Thank you

[@wordsoccer747]

not correct for switching loss

[@kabandajamir9844]

The world's best teacher

[@thinarangana7316]

Well explained!! Thank you so much.

[@user-pf3tq9wr4v]

v helpful! thanks

[@Antagon666]

Feels like Dr. Martin Ordonez is trying to steal my soul with that look. Otherwise great video!

[@rocksonrong8401]

bro??

[@Antagon666]

@@rocksonrong8401 It's a proper stare right in to your soul. I mean the first rule of filmography, don't stare into the camera the whole time, looks really unnatural and unnerving.

[@prashantupadhyay5110]

Why there is diode connected parallel to MOSFET

[@LolodeMillau]

It is called the mosfet "body diode". By design, all mosfets has a reverse body diode that makes the mosfet drive current in the opposite direction even if the mosfet is OFF.

[@vaibhavjagtap7772]

Good explanation 👍

[@Hhtfvhrcbgrejkkjhffvbhb]

We chek gate and source voltage (high frequency) by led? When we reparing smps module

[@tanvirfakir2356]

very good explaination

[@bocheng3911]

Where is the 1/2 come from at 3:44, Eon =ID*vDD*t*1/2?

[@leandroalgenton]

It is because of the energy dissipation area (red triangle), which is like to calculate de triangle area : Area= (base*height)/2. The ID*VDD is the "height" , the Ton is the "base" and the *1/2 is divided by 2.

[@engrsyedowaisathar5556]

I love the video excellent job ... Excellent and to the point ... typically professional ....

[@theengineer9910]

great explaination !

[@niraimathir7198]

u can make more videos on advanced topics like multilevel inverter then it would be awesome and helpful for students.

[@chehar]

Under DC conditions i.e. duty cycle = 1, the power losses I^2*Rds(on) is much lower than say Vds*I. Ohms law doesn't seem to work here. Power is not equal under supposedly interchangeable equations? what am i missing!?

[@EdwinFairchild]

yes they are, you messed up your math somewhere. Vds x I = 10Amps x 50mOhms = 500mVolts ... so then 500mVolts x 10Amps is 5W @ 100% duty cycle. Like wise with the other equation 10^2 x 50mOhms also equals 5W

[@MMaheshThakur]

thanks u so much

[@lolog450]

Excellent video!!!!! Congratulations....

[@RedlabsTV]

Perfect thank you!

[@rajatsachan9301]

dear sir , your video helped me a lot . can you please makes videos on igbt, and converters dc-dc , dc-ac.

[@txdare1830]

Cool!

[@pawanchandrapandey3106]

its very helpful.. thanks

[@ahsanbaig74]

you are awesome and the way you explained is extremely excellent that anyone can understand

[@zaidhamid1622]

شكرا جزيلا

[@bl8713]

Beautiful, accurate explanations.

[@abeditani8293]

Thanks . This helpful information . But I am using my mosfet like irfp260 or irfp90n20d as a DC test dummy load in DC region of the mosfet . And that make my mosfet like crazy crazy very very extremely hot . And I try my best to not over low the 100C in best case . Even if the mosfet Handel 300W . In DC region only handle 100W !!!!!!!! And that a big Lost power

[@Antagon666]

What's the current, do you have a heatsink ?

[@abeditani8293]

@@Antagon666 5A at 20v . So I have a 100W pure heating up inside the mosfet in the DC region . I need to reach 300W demand of ( 60v at 5A ) My mosfet get dead in 40v 5A even if I have big heathink

[@Antagon666]

Well that's non intended way of using a MOSFET. In order to dissipate 300W of heat, you would need to keep it's package temperature at 25C. And there is no freaking way to cool 300W with the to247 package, because the contact area is simply too small with the heatsink. And the power rating decreases with temperature.. at 100°C, the maximum sustainable power dissipation drops to 100W. It has nothing to do with DC rating of MOSFET, it really is 300W for really short amount of time, until it heats up. So the solution ? 1)Use transistor with a much bigger package size 2) use at least 5 irfp260 in parallel, to spread the power dissipation over a larger area!

[@abeditani8293]

@@Antagon666 yes your right . There no way to be exactly to dissipated more than 180w with TO247 package what ever mosfet or IGBT model you use . The area is not enough to transfer this big heating . Can I install like 5 of them all is same one input gate with out problem or corruption with the input mosfet control signal ?

[@gandhi770]

Thank You! This helped so much with my senior design project!

[@RSP13]

AMAZING CONTENT! Thank you

[@GeomicroSolution]

Nice video

[@humbertocuevas5395]

Saludos cordiales atte

[@karthikeyan111991]

one of the best, I have seen. keep it going.thanks

[@alg21343]

Many thanks. it really helps.

[@mahoneytechnologies657]

Very Nice and Clear explanation - The illustrations are super, presentation speed just right for me. Well Done! Are your discussions available in text form? Does UBC have an Online course given in this format?

[@emililie2244]

Interesting and very clear. Thanks We also need to add the power lost through the diode to the inductive load.

[@snnwstt]

I think their intention is only to consider the case when the current I_DS is positive, for which the intrinsic diode does not conduct.

[@alg21343]

many thanks.

[@RogerLuo1995]

Very good video!