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.
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.
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.
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.
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?
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.
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
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?
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?
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???
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???
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.
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
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!?
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
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.
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
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.
@@abeditani8293 I was going to suggest these 2 projects: www.codrey.com/electronic-circuits/simple-electronic-dc-load/ hackaday.com/2017/02/28/beefy-100-amp-electronic-load-uses-two-mosfets/ but I came across your other comment where you have already made a similar device. All I can suggest to you is to do the following: 1. Use more MOSFETs in parallel (as many as 10-15) 2. Use the thermal paste used for computer CPUs, such as MX-2 3. Place a copper pad under the MOSFETs which would take the heat away significantly faster than the aluminum itself. Basicallly a small square flat block of copper on top of a VERY large aluminum heatsink with many fins 4. Place a few large fans on the heatsink to keep its temperature down under maximum load. Just look at the CPU heatsinks and fans, consider that they take away over 100W of heat, to give you an idea of how much heatsink and fan surface you need for sufficient cooling 5. Consider even water cooling if necessary, either by immersing a portion of the heatsink in a large bucket of water, or by using pipes to run the water through your heatsink. 1500W is a lot of power, almost as much as the water heater which takes an hour and a half to heat up 50 liters of water. 6. Since your dummy load likely doesn't need high speed MOSFETS, all you have to worry about is their voltage, current and power rating, and connecting their gates in parallel shouldn't affect the behavior of the circuit, since they use practically no current at DC or very slow changing drive voltages.
@@edinfific2576 yes . I don't want the mosfet to be high speed but I want it fast respond in the gate Soo it can responded very quickly from the feed back . But I have too much problem to control mosfet in voltage above 60v with 4amp . It just get exploded before even get temperature rise to 60c. I think that I need special specific mosfet for this subject
@@abeditani8293 Devil is in the detsils. I had once tried using a MOSFET on 200+ volts DC from solar panels, and I used a 16V zener diode to protect the gate from surges. The gate was still damaged and I for months I didn't know why. Later I realized that a zener diode of less than 1W power rating has a very poor surge/transient response, and it doesn't conduct quickly enough. Another issue I have discovered accidentally in a darasheet just 2-3 days ago is that a MOSFET's Vds rating is significantly reduced when you apply a reverse bias to the gate. Same thing with your problem: there are some details you have missed in the datasheets or in your circuit. Just like some people have suggested, the power applied to your MOSFET is too high for it and your cooling may also be poor. Without seeing the actual circuit loading it, and knowing the exact MOSFET model, I can't help you much further.
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.
@@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.
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 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
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!
@@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 ?
Dear Sir , i have a 12v battery .After using it , it's voltage is decreasing(which is normal) . but my relay needs constant 12v to switch . how can i get constant 12v for relay while the battery voltage less than 12 v from the same battery ? Actually A LM7812 needs =>12V . that's why i am confused ...
A voltage regulator needs a voltage higher than the output voltage to work properly. A Lm7812 needs at least 1V higher or 13V input (check datasheet for minimum voltage input).
You can use a circuit booster (based on a circuit voltage doubler DC to DC) where any "clock" can be used (NE555, Schmith Hex trigger, ... ) to raise the voltage at least at the desirable voltage, and then a TL431 (a shunt regulator, instead of a regulator in series ) to get the final voltage. (While a LM7812 takes a toll of 1V, the TL431 may take no voltage at all, since it is in parallel with the load, as a sharply defined voltage Zener diode would do.)
What a great video! It covers not only the power losses of MOSFETs but also the basic operating principles of them. Thank you!
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.
sry but this is not for new students in my opinion,
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!
Excellent simple explanation.
Gave an in-depth understanding.
Thank you so much.
Please keep posting
Thats really simply explained even for the 'amateur'! THANK YOU!
Thank you for this film. You make things easier to understand.
Excellent, all my doubts got cleared in a video
so flipping simple and elegant. THANK YOU
Thank You! This helped so much with my senior design project!
Excellent explanation of power losses in a mosfet
one of the best, I have seen. keep it going.thanks
An excellent course on Mosfet losses. Thank you very much.
Excellent video!!!!! Congratulations....
Thank you so much for the video. Wonderfully described.
you are awesome and the way you explained is extremely excellent that anyone can understand
Beautiful, accurate explanations.
Simple and understandable explanation. Thank you
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.
I love the video excellent job ... Excellent and to the point ... typically professional ....
Excellent explanation. Congratulations, and thank you for sharing.
Beautiful and to the point. Thank you
Simple and understandable explanation. eye opener
thanks alot...for the clear explanation with neat diagrams
This vídeo was so helpful, thank you for sharing this knowledge!
You explained very well in just small 9 minutes video ,Excellent from INDIA
Well explained!! Thank you so much.
Very well explained...
Please keep posting ...
Specifically for gate driver circuit, snubber circuit etc...
The world's best teacher thanks sir
AMAZING CONTENT! Thank you
Thank you! I was struggling with the differences
Thank you for presenting this it is very helpful
Many thanks. it really helps.
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.
great explaination !
Best video thanks from
Colombia 🇨🇴 👌
Very good and funny videos bring a great sense of entertainment!
Very good video!
very excellent explanation
The world's best teacher
Thanks a lot . Very clear explanation.
Can you make a video to calculate power losses in an IGBT? Or is it the same?
Interesting and very clear. Thanks
We also need to add the power lost through the diode to the inductive load.
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.
Very nice explanations.....
Perfect thank you!
Great explanation thankyou sir
its very helpful.. thanks
Very good lecture.... Nice
Very handy information.
very good explaination
We chek gate and source voltage (high frequency) by led?
When we reparing smps module
many thanks.
Good explanation 👍
Thank you.
Thnk u so much.. u actually helped me out.. 👍👍
anjali nosaria nice vedio.
#PowerElectronicsAPracticalApproach
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.
excellent video . Can I know why a Mosfet has a body diode ? and what does it do ?
u can make more videos on advanced topics like multilevel inverter then it would be awesome and helpful for students.
v helpful! thanks
thank you sir
So nice
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?
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.
Super explanation with quiet neat example sir.
For irf3205 the duty cycle given is 2%, which means D=0.2 or d=2.
Excuse me, but I have the same question. Have you found the answer to yours?
@@mostafasadek2943 Hello, did you find the answer?
@@ADGGMB I believe it's 0.02. 100% corresponds to D=1. Similarly a 2% means D=0.02.
2% duty cycle means D=0.02
Saludos cordiales atte.desde cd. Juárez Chihuahua México.
Wht is the solution to turn off the load like fan immediately ,when gate voltage is made negative
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
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.
Does this apply when driving a inductive load (inductor or transformer)? As the inductance will never rise to a flat state.
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?
thanks u so much
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?
When the MOSFET turns on at 3:00 , why the I_D(on) increases *linearly?*
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???
Nice video
Cool!
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???
I also wanna know
So lets say for a mosfet with rds(on) 0.042 and a current and duty 50% = 33.6 watt ?
Can you explain how the Cgs and the Cds affects the calculation??
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.
Why does the vds start to drop only after drain current reached max?
Sir, may I ask why the Vds wouldn't drop till the ID = IDon ?
شكرا جزيلا
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?
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)
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
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!?
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
Why rise time or fall time 1/2 ?
nice one
May Allah bless you both for more and more
what is the reference book of this lecture
Nice
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.
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
Where is the 1/2 come from at 3:44, Eon =ID*vDD*t*1/2?
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.
Is there a way to made a powerful DC load like 1500w test dummy load . Need something to test voltage from 0 up to 100V and current from 0 up to 50A
Hello, Abed, greetings from Bosnia!
Are you trying to make a MOSFET-based dummy DC load? Or what are you trying to acomplish? Please be more specific.
@@edinfific2576 mosfet based dummy load Dc 1500w power .
Can test 100v 50amp with out getting the mosfets exploded .
@@abeditani8293 I was going to suggest these 2 projects:
www.codrey.com/electronic-circuits/simple-electronic-dc-load/
hackaday.com/2017/02/28/beefy-100-amp-electronic-load-uses-two-mosfets/
but I came across your other comment where you have already made a similar device.
All I can suggest to you is to do the following:
1. Use more MOSFETs in parallel (as many as 10-15)
2. Use the thermal paste used for computer CPUs, such as MX-2
3. Place a copper pad under the MOSFETs which would take the heat away significantly faster than the aluminum itself. Basicallly a small square flat block of copper on top of a VERY large aluminum heatsink with many fins
4. Place a few large fans on the heatsink to keep its temperature down under maximum load. Just look at the CPU heatsinks and fans, consider that they take away over 100W of heat, to give you an idea of how much heatsink and fan surface you need for sufficient cooling
5. Consider even water cooling if necessary, either by immersing a portion of the heatsink in a large bucket of water, or by using pipes to run the water through your heatsink. 1500W is a lot of power, almost as much as the water heater which takes an hour and a half to heat up 50 liters of water.
6. Since your dummy load likely doesn't need high speed MOSFETS, all you have to worry about is their voltage, current and power rating, and connecting their gates in parallel shouldn't affect the behavior of the circuit, since they use practically no current at DC or very slow changing drive voltages.
@@edinfific2576 yes . I don't want the mosfet to be high speed but I want it fast respond in the gate Soo it can responded very quickly from the feed back . But I have too much problem to control mosfet in voltage above 60v with 4amp . It just get exploded before even get temperature rise to 60c.
I think that I need special specific mosfet for this subject
@@abeditani8293 Devil is in the detsils.
I had once tried using a MOSFET on 200+ volts DC from solar panels, and I used a 16V zener diode to protect the gate from surges. The gate was still damaged and I for months I didn't know why. Later I realized that a zener diode of less than 1W power rating has a very poor surge/transient response, and it doesn't conduct quickly enough.
Another issue I have discovered accidentally in a darasheet just 2-3 days ago is that a MOSFET's Vds rating is significantly reduced when you apply a reverse bias to the gate.
Same thing with your problem: there are some details you have missed in the datasheets or in your circuit.
Just like some people have suggested, the power applied to your MOSFET is too high for it and your cooling may also be poor.
Without seeing the actual circuit loading it, and knowing the exact MOSFET model, I can't help you much further.
Saludos cordiales atte
Why there is diode connected parallel to MOSFET
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.
dear sir , your video helped me a lot . can you please makes videos on igbt, and converters dc-dc , dc-ac.
But in datasheet, only rise and fall time of current is mentioned. Then how should I calculate switching loss in this case?
Need little help.
In IRF3205 datasheet the given is the rise/fall time of voltage VDS so maybe we neglect rise/fall time of current in calculation of switching losses.
Feels like Dr. Martin Ordonez is trying to steal my soul with that look.
Otherwise great video!
bro??
@@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.
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
What's the current, do you have a heatsink ?
@@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
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!
@@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 ?
3:40 how it's possible I D rising up without VDS going down?
Dear Sir , i have a 12v battery .After using it , it's voltage is decreasing(which is normal) . but my relay needs constant 12v to switch . how can i get constant 12v for relay while the battery voltage less than 12 v from the same battery ? Actually A LM7812 needs =>12V . that's why i am confused ...
www.electronics-lab.com/project/5v-to-12v-step-up-dc-dc-converter/
A voltage regulator needs a voltage higher than the output voltage to work properly. A Lm7812 needs at least 1V higher or 13V input (check datasheet for minimum voltage input).
You can use a circuit booster (based on a circuit voltage doubler DC to DC) where any "clock" can be used (NE555, Schmith Hex trigger, ... ) to raise the voltage at least at the desirable voltage, and then a TL431 (a shunt regulator, instead of a regulator in series ) to get the final voltage. (While a LM7812 takes a toll of 1V, the TL431 may take no voltage at all, since it is in parallel with the load, as a sharply defined voltage Zener diode would do.)