Power Electronics with Dr. K
Power Electronics with Dr. K
Welcome to my channel. I am a professor at Milwaukee School of Engineering and teach electrical engineering courses. Here you will find some of my lecture topics for courses that I teach in Power Electronics course.
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Very valueable video! But I think if we use MOS to replace Diode, the power loss of rectifier MOS should be lower than another MOS. Because Buck converter need to make a dead time, so its body diode will conduct firstly, and then MOS conduct. I guess there is no switching loss in this MOS.
Yes, good observation. Often that diode is replaced with a MOSFET. This is called a synchronous converter. It has lower loss (improved efficiency) but requires a more complex controller. There is still switching loss in the MOSFETs, but if the frequency is low enough, the overall loss will be better. There are converters such as resonant converters that utilize zero-voltage, zero-current (ZVZC) conditions and have even better efficiency. Best wishes on your designs. -Dr. K
@@powerelectronicswithdr.k1017 thanks for your kind response! Sync buck do need a complex controller, that's why I mention the dead-time. During the dead-time, the body diode of MOS will conduct firstly. That will cause an extra power loss: Vdiode✖️Io✖️（dead-time/T）, but body diode dosen't have a ton and toff time, it exists automatically, so I think the rectifier MOS will cause a body diode power loss, but has almost no switch loss.
Excellent sir! Thanks for sharing!
why a Schottky diode?
Lower loss when the diode conducts. The other selection would be a fast recovery diode. Many converters are also synchronous and use a MOSFET instead of a diode. Great question and best wishes on your design.
Great video! Could you also show how the equations deduced or put links? Really appreciate it.
Hi! The links for the equations can be found in the video description. Best wishes. -Dr. K
Does bootstrapping limits switching frequency ?
Great question which applies to high-side driving of typically a P-channel MOSFET. The answer is yes. However, the switching frequency range is relatively broad and will partially determine the value of boot-strap capacitor used. As an example, it is impossible to never switch because the boot-strap capacitor requires charge and this capacitor charges up when one side is switched to ground either through the driver or through the low-side switch. Once charged, the boost-capacitor can then be used to drive the MOSFET and acts like a floating voltage supply. However, the capacitor will eventually discharge over time and the result is Vgs will slowly decrease to an in operable point. If the switching frequency is to high, the boot-strap capacitor never fully charges up or takes too long to charge. Often the high-side gate drive datasheet will provide information on selecting the boot-strap capacitor based on your switching frequency. Best wishes on your design. -Dr. K
@@powerelectronicswithdr.k1017 thanks for perfect explanation. one more question about gate drive. Gate drive require high current for a moment ,(say 500 mA), due to Cgs. If we use bootstrap, does bootstrap capacitor can deliver such a high current?
I think the transistors you chose are all of the same negative type. is it correct ?
Yes, that is correct. I do not show the gate-driver circuitry or control for this. That would be another video. N-channel MOSFETs require special circuitry for high-side driving. There are h-bridge topologies in which the two high-side switches are P-channel MOSFETs. This makes for easier switching. Best wishes on your design. - Dr. K
What about the output is the Vin ~= Vout ?
Thank you so much. If I want to build a variable output buck convertor, should I do the calculations on the highest output voltage and choose the components please? Thanks.
Hi Ziadfawzi. Great question and the answer is that it depends. For example, will you try to maintain a constant output power at all voltage ranges? If so, then the output current will be maximum at the smaller output voltage values and minimum at the larger output voltage values. It would be a good practice too look at both extremes and see which values will "stress" your components the most. Sorry it is not as easy as just the highest output voltage. Best wishes on your design. -Dr. K
Hello! how do we choose the resistance? please let me know
Hi Jviccii, the resistance is not part of the design of the Buck Converter. It is an easy way (although not always accurate) to implement a load on the power supply. You can set the value to Vo/Io. However, please take caution as not all loads are purely resistive. Great question and best wishes on your design. -Dr. K
In switching loasses formula instead of ton and toff there should be rise and fall time of mosfet, since it's loss during switching .. not conduction loss is it correct?
Hi Gan-rc2im, yes you are correct. I am using these terms liberally. On the datasheet, the rise time is actually the amount of time it takes to turn the MOSFET "off". This is the time duration where Vds rises from near 0 V to the supply value Vdc. The fall time is the duration it takes to turn the MOSFET "on". The is the duration where Vds drops from Vdc to near 0V. Please note that the datasheets use very specific testing parameters and your application of rise and fall times will vary. Great observation and thank you for watching. -Dr. K
Hello Dr.K, your videos about buck converter is helping full for me when i need reference in my master's thesis. Thank you!!! But can you explain where the pictures and graphs about buck converter came from? 😄
Hi, I developed those graphics using MS Visio, MS Power point and some of the simulations were done in LTspice.
The intro music was annoying, at best ...
Lol... You do know you can skip past the introduction. Have a great day MrSummitville and best wishes on your designs. - Dr. K
@@powerelectronicswithdr.k1017 LOL - You do know, that adding crap noise to your video does not make it better? Have a nice day...
I want to design a full bridge DC-DC converter. Can you explain how to choose a MOSFET for that?
Hello sir, the Vdc is 50 volt, but the v(out+, out-) is 100 volt, is it correct?
Yes, that would be correct. The units would be 100 Vpp (volts peak-to-peak) or 50 Vp (volts peak).
Very good presentation of MIOSFET turn on characteristics. Really good that you show how charge control calculations are done in a design.
thank you :)
You’re welcome.
in the formula some words I can't understand. Would you show clear text?
So nice thanks sir
Very understandable
I love your clip
Thank you for good technology
it depends on the duty cycle D <3
oh Jesus, you have a high level to explain power electronic, im so grateful that i see your video, thank u so much
Great video Dr. K, thank you very much😃
14:20 I'm glad you said AVERAGE current. I see so many people making the mistake of saying dQ/dt = PEAK current which is completely wrong. Since the gate is a capacitor resistor network, the peak current can be estimated by Q/t * 5 (five time constants).
Why can we assume Io is constant while the voltage accross the capacitor is not constant?
Often one assumes the load is resistive and because the output voltage is relatively constant ( with the exception of the voltage ripple), the average output current is constant. The output capacitor is therefore used to store and provide charge to help maintain that constant voltage assumption. Thank you for the question and best wishes on your design. -Dr. K
It was wonderful lecture
Thank you.
Hi Dr. K, i've a question about the voltage regulation when the buck converter has no load. how do you regulate the voltage output when the output is floating? which means the capacitor won't be able to discharge itself.
Dear Professor, @12:42 should not we observe the Vd(t) negative? and also @16:21 the Vds is 2/3 Rds when it is conducting, right?
Pablo, YES! Great catch. The voltage should be in the range of -0.5V to -0.65V. Luckily we have a low forward voltage drop and it will not impact the results drastically. Also, you are correct, the average value of Vds during that period will be (Io/D)* Rds and that is 2/3*Rds. Best wishes on our design. -Dr. K
@@powerelectronicswithdr.k1017 thank you very much
Hi Dr.K! Thank you for the video! Everything else was very clear but I have one question. Can you tell me how did you get average current on secondary side of the transformer as Is = (pi*I0)/2?
That's actually the peak value of the current on the secondary side. This comes from the computation of the average value of a fully rectified sine wave. The average value of |Apk*sin(wt)| is 2*Apk/pi, where Apk is the peak value. In this case it Is. Solving for Is using the average, one gets Io*pi/2. If only 1/2 wave rectification is used, then the equation would change to Io*pi. Hope this helps. Best wishes on your design. -Dr. K
@@powerelectronicswithdr.k1017Thank you Dr. K! That makes more sense. I think you mispoke in the video by mistake and I took it literally😅😂. Thank you for your lecture too! Have a great day!
Very basic knowledge but explained in a very confusing way. Where is the reference for MOSFET switching in High Side Switching Mode? Do you have a MOSFET driver supplying pulses at the gates of M1? The back -BEM will do what? M1 will never switch unless there is a driver IC or bootstrap for M1. Yes it is adviseable to run the pulses at 90% duty cycle. But your way of explanation is vague, incoplete and not description. Instead of building an understanding, you are confusing more. I disliked your video.
Please do not confuse my audience. You only switch with a 50% duty cycle for a resonant converter. You are correct in that gate drivers are required. This video is not about MOSFET gate drivers. This an explanation of how resonant converter functions and has soft switching. Best wishes on your designs.
Hey Dr. K! Loved the video. Is there a textbook that talks about these considerations that you would suggest. Do not need detail, just a good source about power electronics and thermal considerations.
Hello Dr. K! I came across your channel a while ago and the material is very good. Thank you for your videos. You are a good teacher. I have a question about the consumption of MOSFETs. If I am not wrong: - The total consumption is equal to % of time in saturation multiplied by the power due to RDS(ON), added to the power due to the two switching (on / off) multiplied by the frequency. (Proportional part of the time in saturation, plus two switchings, every period. The absolute power loss of ecah switch is independent of the frequency). - All the power consumed by the MOSFET is transformed into heat. If that is correct, measuring the temperature that the transistor reaches and dividing by the thermal resistance R-JC, we should have more or less the same number, is that correct? I am doing tests with several MOSFETs to see it experimentally, and I see a very strange temperature graph: For a pwm of 10 KHz, pulses between 1% and 3% cause the temperature to rise excessively, and then it gradually drops to the calculated theoretical value. I've measured that temperature for 10, 20, 30 .... 250, and then 500, 750, 1000... 4095, with a 12 bits pwm. Lot of points. For an IRF540, for example, the temperature reached with a duty of 2 us is 14ºC, and with 20 us it is 9ºC (above ambient temperature). This is normal? Why can it happen? If you find it strange and want me to send you specific information about the tests, I would be happy to do so. (Sorry for my English, I'm Spanish, I hope you will understad that I wanted to explain...) Thank you so much!
You have very good material in you channel. Thank you for your videos.
This vidio is so valuable. Please edit and make this better(interupt audio). Thanks bytheway.
You are right. I do need to re-edit that video. It is one of the more popular ones and the audio terrible. My summer project. Thank you for the feedback. -Dr. K
Hi Professor, I tried to design a buck converter using IRF3205 NMosfet for an input of 24V and Output of 7.3V, 40A. But i couldnot drive the mosfet from driver circuit. Control pulse from microcontroller, tried with PC817,MCT2E and IR2101. Can you please suggest me a driver circuit design. Thank You.
The IR2101 has both a high-side and low-side gate driver. Are you doing a synchronous buck converter? Also the IRF3205 N-channel MOSFET needs about 10V for the gate-to-source pin. What is the voltage you are using on the Vcc pin for the IR2101. This needs to be anywhere from 10-15V. I would probably use a regulator specifically designed for buck converting that has a built in gate-driver. Infineon does make switching regulators for buck converters. Best wishes on your design. - Dr. K
Hi Professor, Thank You for the response. I have tried Power Mosfet driver IR2101 for high side Buck Converter with Vcc between 12 to 15V. But Mosfet is not going to off state. Hence the required output voltage is not observed across the load. @@powerelectronicswithdr.k1017
Prof. could you please make a step by step video on a SiC MOSFET. Please!!!
Thanks Power Electronics 🙏🌹
what is the meaning of Ts ,can you please tell me?
Ts is the total switching period. For example, if the switching frequency is 10kHz, then Ts = 0.10ms. The switches are only activated for a portion of that period (i.e. the duty cycle D) and we can control the voltage by adjusting the duty cycle. Hope this helps and best wishes on your design. -Dr. K
🙏 Thanks 🙏👍
You are welcome. Best wishes on your design.
Good morning sir I want to calculate Switching loss of the MOSFET for the spwm inverter
The switching loss can be estimated if you know the max turn-on and turn-off power values. It would then then be 0.5*(p_max_on*t_on + p_max_off*t_off)*fswx, where fswx is the frequency of your sawtooth carrier wave. The tricky part will be estimating the turn on and off times and the max power values.
Thanks 😊👍🙏💯
One thing to keep in mind with ceramic capacitors is the reduction in capacitance with voltage. The lower the voltage rating on the capacitor, the lower the capacitance when you approach the rated voltage. Awesome video
What a great video.
Thank you so much Dr. K. I have been looking for this explanation for hours now. You are the first person to explain this clearly and concisely.
The music at the back is disturbing inspite of a good explanation
Qu'est-ce que c'est que ce tintamarre d'enfer en introduction particulièrement pénible et lequel couvre votre voix ? Déjà qu'il n'est pas simple d'entendre vos paroles et de vous comprendre, il y aurait intérêt à supprimer ce bordel ambiant, parfaitement inutile et désagréable. Merci !
Thank U very much