Real hero

Thank you!

I agree :-)

i agree too.

Yes I do agree with you. I haven't experienced this before. A great voice and great explaintion. 👍

You're welcome. TI powerbench is a nice tool and TI provides some very good reference designs. I also like using LT Spice for simulating the designs too. Thank you for watching. -Dr. K

Clark, you are welcome. Glad the video helped. -Dr. K

Trowan, thank you. -Dr. K

Mahmoud, thank you. Here's the link to part 2 kzfaq.info/get/bejne/r9aUeaaivp3SeoE.html. In the course I teach on power electronics, we cover diode selection in more detail. There are only a couple of options for the traditional buck converter. The first is a fast recovery diode, the second is a Schottky diode and the third is a SiC diode. Another approach not covered is to utilize a synchronous Buck converter topology. In this topology, the diode is replaced with a power switch (i.e. MOSFET). Thank you for the feedback and best wishes on your designs. -Dr K

Thank you. Glad the videos helped. Best wishes on your thesis. Keep me posted on your work. -Dr. K

Thank you.

You're welcome.

@@powerelectronicswithdr.k1017 Hi Dr K, is there any chance you could order your lectures in a playlist?, the content is awesome but I'm having a hard time knowing what lecture comes first and what comes second. thanks

@@martovify Joaquin, that is a great suggestion. Our term is ending next week and we go on summer break at MSOE. I'll plan to add playlists and organize the channel better. Interestingly, there's only a handful of videos that people come to the channel for. The buck converter is one and the LLC converter is the other. I've been thinking of doing a series on fly-back converters. Again, thank you for the suggestion. -Dr. K

Hi Jim, glad the videos are helpful. Best wishes on your summer internship. Power electronics is a great area to start your work as an engineer. Keep me posted. -Dr. K

Thank you. Texas Instruments also provides some good information for the converter/regulator you have selected. Best wishes on your design. -Dr. K

Thanh, glad this helped.

@@powerelectronicswithdr.k1017 At 13:22 you circled 30A and said "30A forward voltage". I am sure you mean 30A forward current. IMO, I wouldd not choose FET 30V rating and Diode 45V rating. Leakage inductance and parasitic capacitance would cause high voltage spike on top steady state voltage. These voltage spikes will breakdown the FET and Diode. I would choose at least 75V or 100V rating on both devices. Thanks

@@user-power_electronics Hi Thanh, yes I think I messed up in that area. With regards to sizing the FET and diode, my advice is to start at x2 times the maximum nominal value and then adjust accordingly. Over sizing to a factor of x3 or x4 increases the component costs. The sensitivity to cost in the design is something to consider. Low volume and high reliability might justify a higher rating. High volume/cost competitive applications require the designer to explore ways to keep costs down. As an example, the PCB layout and component placement can drastically impact lead inductance and parasitic capacitance. For such an application, I recommend using the vendor's datasheets, application notes and reference designs as guides. There are also integrated solutions that package the switching regulator and switch in the same package. Thank you for the feedback and best wishes on your designs. -Dr. K.

@@powerelectronicswithdr.k1017 Hi Dr. K, Thank you for your comment. I totally agree with you regarding sizing and readjust accordingly. Size and Cost always a major factor in production. Sorry, I used to work for satellite applications and always emphasize on reliability and follow mil spec strictly. Your power electronic tutorial classes are very practical combining with simple theory behind. Best Regards.

Thank you. -Dr. K

You are very welcome. Best wishes on your designs. -Dr. K

Anderson, thank you. Best wishes on your converter design. -Dr. K

Hi RGB, the answer is "it depends." Sometimes that datasheets are accounting for losses or differences. My advice is to start with the theoretical design equations and then compare to the data sheet equations. You may find that the values are close. Hope this helps and best wishes on your design. -Dr. K

Abuzar, excellent observation. My calcs assume zero voltage drop across that fly-back diode. However, the diode is one the major loss components in the entire design (see synchronous buck-converters). That node voltage at the diode, reference to ground, when the diode is conducting is about 0.5V to 0.65V depending on the diode. The input voltage is still about 15V. Therefore, the drop across the MOSFET when it is "off" is just under 15V. As you noted, it doesn't matter too much. Often we will choice a voltage rating for that MOSFET that is a factor of x1.5 to x3 of the highest voltage we would expect. Hope

Hi, the resistor is not really part of the design. It is just show as a load that your converter would be connected to and fixed-resistive loads are easy. The value of the resistance you use for testing the design is Vo/Io or Vo^2/Po. Hope that helps and best wishes with your design. -Dr. K

Hi Jackson, without specifics its hard to tell what's happening. Have you simulated your design with different input voltages? Note that the equations that I present in this video are for continuous conduction mode (CCM) and it is possible to go into discontinuous conduction mode (DCM). Also, I did not discuss closed loop control for these designs. Most DC/DC converters use a controller that uses both voltage and current feedback control for varying the duty cycle. Keep me posted on your findings and thanks for watching. -Dr. K

Not necessarily, it depends on the ripple value you are aiming to get. As Dr. K says in the video, it's generally taken as 0.4 (40%), but it comes down to what you are aiming for.

Hi Caleb, great question. That 22uF is called a bypass capacitor. Here's more info from TI on the definition of bypass capacitors www.ti.com/lit/an/scba007a/scba007a.pdf. This capacitor provides for a very low input impedance power supply for the IC and most ICs recommend some form of bypass or input capacitor. It should be physically located very close to the power terminals on the IC to reduce lead inductance. Hope that helps. Best wishes on your design. -Dr. K

You are welcome. Best wishes on your designs. -Dr. K

No. The switching loss will be constant regardless of the duty cycle. It will increase if you increase your switching frequency. You will need to use the maximum current throught the MOSFET and the maximum voltage Vds across the MOSFET. However the conduction loss will be proportional to the duty cycle and is D*Ids^2*rds. Hope that helps. -Dr.K

@@powerelectronicswithdr.k1017 Vds in this case will be my input voltage Vin and the maximum current will be input current?

Hi Erol, excellent observations. Yes, a P channel MOSFET could be used in the design and this would make the switching control electronics easier. However, N channel devices typically have lower Rds and therefore lower losses as compared to a P channel device. Therefore, the additional cost/complexity of a high-side gate driver for the N channel device can be justified in the design. Have you watched my video on high-side gate drivers. The audio is terrible, but the content is good. Gate drivers and control of these topologies is also a critical part of the design that I don't cover in this series. Thanks for watching and best wishes on your design! -Dr. K

@@powerelectronicswithdr.k1017 Can you share the link of the video on high-side gate driver ?

@@tarekbenhacine8743 HI Tarek, this might help. Sorry about the audio in this one. I've never had the time to re-edit. However, it should help with understanding high-side gate driving and gate driving in general kzfaq.info/get/bejne/nL18osKD2afToYk.html

You are welcome. Best wishes on your design. -Dr. K

Pedro, as long as you keep the ripple current (delta-I) less than the average the full rated output current, you will maintain CCM. For example, if Io = 10A and you design for a 25% ripple voltage, delta-I will equal 2.5A and you will maintain CCM. The boundary between DCM and CCM is when the average output current equals the ripple current. Entering DCM isn't always the worse thing to happen and some controller designs allow for both CCM and DCM modes. Hope this helps. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 Understood, thank you very much!

Hi and you are welcome. These calculations work for all basic buck converter designs. However, once we start looking at specific applications other requirements come into action. These include galvanic isolation between the bulk DC power source and the output power, sizing of the components for transient response, etc, regulation and control of the output current/voltage, soft-starting, OCP, etc. There are some great design tools provided by vendors such as TI and others that will help walk through setting up some of the initial stages. Best wishes on your design.

@@powerelectronicswithdr.k1017 Thank you sir for your reply.Sir actually I am designing a DC-DC converter for PMSG based wind energy system.

You're welcome. Glad these provide value. -Dr. K

Hi Sagar, yes sort of. You actually want to look at the maximum. But for sizing the inductor one often will look at the effective or RMS current which will be slighlty higher than the average. The inductor current spec is used to keep the design from melting off the coil wire lacquer and shorting the coil. Some vendors also spec the voltage across the inductor, which can be important. Note that you will often size slightly higher to place a safety factor in your design. Hope this helps. Best wishes on your design. -Dr K

Hi Kosi, if the output was restricted to be between 3-4V, then one would estimate delta-Vo as 1V and the average value of Vo is 3.5V. That's a relatively high ripple current if one is trying to provide a regulated output voltage. Hope this helps. Best wishes on your design. -DrK

Hi Daniel, great questions. This video only covers the basic concepts of the switching buck converter. Control of the output voltage is almost always done by using voltage and sometimes current feedback. This enables a constant output voltage over a wide range of input voltages. Use of an application specific IC is almost always recommended. For example, the AP63205 is a wide input buck converting regulator with a fixed 5.0V output. That said, you still need to size the inductor. If you plan to keep the design in continuous conduction mode (CCM), then size the inductor by checking the output ripple values at both extremes of the range. The upper range of the input voltage should result in a larger value of inductance if you want to design for a worst-case ripple current. With regards to output current, many of the controller ICs have current feedback functionality and will reduce the output voltage to limit the output current. Hope this helps. Best wishes on your design. -Dr. K

Hi Tanbir, the spec on the output ripple voltage wasn't provided for this example and therefore I made an assumption and used a percentage of the nominal output voltage. When you make that assumption, use the full-load value for the output current as this will be your worse-case scenario. The capacitor has to provide charge (i.e. output current) when the switch is off. Design is very iterative and you may find that you will either select a size that is slightly higher (more cost/size but lower output voltage ripple) or slightly lower (less cost/size and higher voltage ripple). Thanks for watching and best wishes on your design. -Dr. K

You're welcome. I'm glad it was helpful. -Dr. K

Hi Chibuzo, I'm planning on doing a video on PWM control of the buck converter this weekend and will keep you posted. -Dr. K

The peak value of current yes, but not the average value. Both values are important when selecting the diode. The reason the average value of current through the diode is less, is because the diode is only conducting when the switch is open. This flyback state is (1-D)% of the time where D is the duty cycle. Therefore the average current through the diode is (1-D)*Io. The peak current is the same for both devices. Hope this helps and thanks for the question. -Dr. K

@@powerelectronicswithdr.k1017 thank you.

Hi Farid, I assume you mean the amplitude of the pulse driving Vgs? There is a bit of an impact, but I am assuming we are in the Ohmic region of operation for the MOSFET (i.e. the MOSFET is fully on). Therefore, we really only care about the pulse duration or the PWM duty cycle as a percentage. Best wishes on your design. -Dr. K

Hi abdalah, the capacitor equations are in part 2 of this design example. -Dr. K

You are welcome.

Hi, the value of delta-V is often determined by the application. Some applications can work well under a higher ripple voltage and other applications require tighter regulation. Yes, if the output is 20V and the specified delta-V is 1%, then the output should range between 19.9V to 20.1V or delta-V = 0.2V. Hope this helps. Best wishes on your design. -Dr K

@@powerelectronicswithdr.k1017 thank you so much sir. Have a great day. Thanks for the video and reply.

That's a great idea and I'll put that on the list of videos to make. In the meantime, here's a paper that describes the design process "Designing an LLC Resonant Half-Bridge Power Converter," by Hong Huang, Reproduced from 2010 Texas Instruments Power Supply Design Seminar SEM1900, SLUP263, Texas Instruments, 2010. The design of an LLC resonant converter is not as straightforward as the buck converter. There's many different options to select. Thanks for watching. Dr. K

Hi TK, great question. R is really not part of the design. It is external to the design of the buck converter and the answer is that it depends on the application. This video is a first iteration of the design and we assumed that R = Vout^2/R. This assumes the load is resistive in nature. If the load is not resistive, then you will need to model with the necessary reactive elements. Note that these reactive elements will only really matter for transient effects such as turn-on, turn-off or when the load drastically changes. At steady-state reactive load elements, such as series inductance, will have zero impact the DC output. Hope this helps, -Dr. K

@@powerelectronicswithdr.k1017 So Sir, is it okay if I insert the R value with the said method ( Vout/Pout ) ? Because my lecturer only gave us the input voltage and output voltage to design the whole converter. Other videos they all have R value and ripple factor as headstart. There is one person he first did a minimum value for L , then he suddenly calculate another L value by making the value 25% larger. Then I have no idea why he did that....

@@TeeTarik Yes, you really do need an either a maximum output current or output power specification. If you have the output power, then you can estimate the output current as Po/Vo. Again, R is not part of the design it represents the load your converter would be connected to. The first iteration of the design assumes a resistive load and also note it is the output voltage squared divided by the output power. It is possible to start the design with a minimum L value and I have seen the design process started with finding minimum L that keeps you in continuous conduction mode (CCM). Design is not a linear process and often can take numerous iterations as you work your way through the selection process. Best wishes. -Dr. K

Hello Robert, the equation you have listed is the smallest value inductor that will maintain CCM for the buck converter. I think somewhere in the video, its stated that we are designing for complete CCM and our ripple current for this example is 40% Therefore, if your requirements are okay with entering DCM mode, by all means use a smaller inductor, but note that this will also impact the sizing on your filter output capacitor. With regards to your second question, the equation for the inductor sizing is a modification of equation 5 from the Texas Instruments app www.ti.com/lit/an/slva477b/slva477b.pdf. The TI Power Topologies Handbook has some information too.

@@powerelectronicswithdr.k1017 Thank you sir.

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 Muhammet, use the output for computing the ripple voltage. If your output is 1%, then use 0.15V as your ripple. In most designs there will also be a capacitor on the input to. Also, the duty cycle will be controlled to regulate the output voltage and the output current. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 first of all I hope your health is good during pandemic. okay, current is 2A and output voltage is between 12-15 V. So I’ll use 0.15 as a ripple voltage. But the thing is 1% output will cause any problems ? I guess 1% is good. I’m sorry but hopefully it’s my last question:) Thank you Dr. K

@@yukselle. Hi Muhammet, all is good here. Thank you for asking. Hope you are doing well. The answer to your question is "It depends." The output specifications really depend on the application that your converter is being used for. Some applications are more tolerant of higher voltage variations that include ripple (higher frequency variations) and load/line regulation (variations due to the changes in the output load and the input line voltage value). Sorry, but I don't have a good answer for you on this one. -Dr K

@@powerelectronicswithdr.k1017 I’m doing good sir, thank you. Thank you for your answers sir, have a good day. Kindest Regards.

Hi Sanjay, I would look at both cases. However, using Dmax (i.e. the xase when Vin = 120Vdc) will result in lower ouput ripple current when your input voltage is 300Vdc. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 ok sir 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

Hi Ekow, great question. One interpretation of the buck converter is that it is a DC-chopper followed by a low pass 2nd order filter. The LC circuit acts as a second order low-pass filter that smooths out the chopped DC waveform. High frequency components of the chopped DC waveform will be blocked by the series inductor (and shunted by the parallel capacitor) and the DC will pass right through the series inductor. Hope this helps. -Dr K

Hi Moni, yes, but depending on the application you may want to look at other topologies. The power and voltage values you list prompt me to ask the following questions. What's the application? What type of efficiency are you looking to obtain? Is isolation required? These high-voltage, high-current applications require a different level of design considerations than what's covered in this video. Hope this helps answer your question. Best wishes. -Dr. K

@@powerelectronicswithdr.k1017 yaa yes.i understood but for your questions is there any documents is available?if any means can plz suggest or share me...becoz,I want to explore that one

@@monimonish302 I would look at research on some of the high power density converters. One place to start is by looking at some of the reference designs provided by the electronic vendors such as Wolfspeed, ON and Infinion. All are in the SiC and GAN space for power electronics. Best wishes. -Dr. K

@@powerelectronicswithdr.k1017 thank you...

Hi! The links for the equations can be found in the video description. Best wishes. -Dr. K

Ram, great observation and yes. Here's a link to a video where I discuss this loss kzfaq.info/get/bejne/aNuHe86GptmpkYU.html . Best wishes on your design. -Dr. K

Hi Shankar, these topologies will work at 10A. However, you will need to resize the inductor, diode and MOSFET accordingly. Note that once you start increasing the output power, other topologies, such as the resonant converter, become more advantages due to the improved efficiencies. Design is all about the trade-offs between cost, performance, and complexity. Best wishes on your design. -Dr. K

Hi Srinivas, great question and the answer is "it depends." It would be great to switch at higher frequencies. This would decrease the inductor and capacitor size, but this comes at the expense of increased switching loss for the MOSFET and the reverse recovery losses of the diode. Switching in the 1kHz to 20kHz creates possible audible noise. Finally, if the frequency is too high, there EMI/EMC standards that come into play. Sorry that I don't have a good answer for you and it's a great question. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 Thanks a lot for your reply. And kindly requesting you is post more videos on KZfaq it's help full so many people like me...

Because in that case we get duty cycle value ranging from Dmax to Dmin. Then for hose cases we get two values of L and C. Then how to choose a value such that L and C value fulfills the requirements output for the entire input voltage range??

Hi Siddhi, great observation in that there will be a wide range of duty cycles. Typical one will select the max value for L and C. but this is not a hard and fast rule. One approach is to look at min-max variations of I/O. For example, you can do a worse case analysis and analyze with maximum output current and then look at the min/max ranges of your input voltage and output voltages. Then see how well the design meets the requirements. Design is an iterative process and the material presented here just provides a first order estimate for your component values based on nominal I/O values. Further work is required to check your design over a wide range of input voltages and output loads. Also dynamic control of the duty cycle to account for these variations is another topic. I would also suggest using a simulation tool such as LTSpice to help with that analysis. Best wishes on your design. -Dr. K

Hi Vivek, at the 8:50 section I discuss how to size the capacitor. The value of the capacitor is the ripple current/(8*switching_freq*ripple_voltage). This only provides the value of the capacitance to meet the ripple current and ripple voltage requirements. Selecting the type of capacitor is highly dependent on the output voltage value, the amount of ripple current, output power etc. There's a link in the description to an application note on selecting MLCC types of capacitors, assuming that's in the power range for your application. Best wishes. -Dr. K

@@powerelectronicswithdr.k1017 Yes exactly, how did you get the value 0.2v as ripple voltage, never mind, I am bit struggling to figure it out. could you explain me?

@@nustanature2655 It was an estimate. Typically the ripple voltage requirement would be provided as part of the design specification. It was an estimate that I assumed. Decreasing the value of the capacitor will increase the ripple voltage at rated load and increasing the value of the capacitance will decrease the ripple voltage. Some times you have to makes some estimates in the design process. Start with 1% or 0.1% and see how your design is impacted. -Dr. K

@@powerelectronicswithdr.k1017 Thank so much professor I greatful to you.:)

This is just an assumption (assumes load is static) used to do the analysis. The output capacitor does a pretty good job of filtering and keeping the output voltage near constant for a static load. Unfortunately, I do not discuss the control of converters in my videos and assume a steady state condition. Dynamic loads that vary rapidly place another level of design constraints on the topology. Thank you for watching and best wishes on your design. -Dr. K

Hi Johan, it was a design requirement that used for this example. For CCM, you can often go up to 40% current ripple. In this example I used 20%. Hope this helps and best wishes on your design. -Dr K

Hi Necip. The answer to your question is yes, but I am really glad you asked as this often when the learning happens. I'm just going though the algebraic steps faster than I shouldin that section. If you go back to kzfaq.info/get/bejne/f9affJl2t9SZqoE.html, I compute D = 1/3 as it is equal to Vin/Vout or 5V/15V .Therefore 1-D = 2/3. Sorry for the confusion and hoping we cleared things up. Keep me posted on your design and thank you for watching. -Dr. K

@@powerelectronicswithdr.k1017 oh yes now I saw. Thanks your quick reply.

Hi, I developed those graphics using MS Visio, MS Power point and some of the simulations were done in LTspice.

Hi Max, CCM means that the current flowing through the inductor is always non-zero. In Discontinuous Conduction Mode (DCM), the current flowing through the inductor will go to zero during a portion of the time when the high-side switch is open. In this situation, the output capacitor has to provide additional energy to keep the output current to the load. Hope this helps and thank you for asking. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 This is the best video on youtube explaining SMPS buck converters to start experimenting and designing practically, watched hundreds of videos and never understood anything how to actually make a buck and boost converters. It would be amazing if you can someday make a tutorial in the same manner of how to design practically of the input EMI filter section where it is before the bridge rectification. Thanks!

@@ShopperPlug great suggestion. I thought of adding the EMI choke into my course this year and would also do a video. Thank you.

@@powerelectronicswithdr.k1017 Thank You

Hi Francis, great question. If you go just about 39 seconds earlier kzfaq.info/get/bejne/f9affJl2t9SZqoE.html, I mention that it was an assumption. For your design, if an output ripple isn't specified, then you will need to provide one. Note that the smaller the ripple voltage, the greater the value of the output capacitor. Hope that helps and best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 Thank you very much.

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

Hi, thank you for the question. I’m assuming a 20% ripple current. The percentage of ripple current through the inductor can be as great as 40% at rated load for CCM. Best wishes on your design. -Dr K

Hi Moni, I've had a number of requests to do a flyback converter and will plan for that over the summer. Thank you for the suggestion. -Dr. K

Tqz fo the consideration.and one more thing in this video u done calculations for CCM in case of DCM what is the calculation to design buck,boost,buckboos,flyback etc

Hi, any standards used would depend on the application of the power supply. For example, electronics parts manufacturers often specify temperature ranges for parts. Also, many ICs have a commercial grade and automotive graded part. The component you select depends on the application the converter will be used in. Hope this helps. -Dr. K

@@powerelectronicswithdr.k1017 Thank you very much for your clear answer. Your answer helps me fulfill the requirements of ABET at the level of graduation projects

Great question. The value of 40% is low enough to keep the design in continuous conduction mode (CCM). Selecting a higher percentage might run the risk of DCM and also creates larger current swings. Moving in the other direction, a design goal of 20% or 10% would greatly increase the size of the inductor, perhaps unnecessarily.

Hi Gidgy, great question. Yes you could most certainly use a p-type enhancement MOSFET for the high side switch. And as you are probably aware (based on your question) that driving a high-side p-type MOSTET is relatively easy as compared to an n-type. However, n-type devices are often more efficient. An n-type device usually (not always) has a lower Rds and the Cgs and Cgd are lower too for the same current/voltage ratings. This enables a faster turn-on/turn-off time, lower conduction loss and lower switching loss. Its often a design trade-off between cost, functionality, performance, etc. Perhaps I should create a video on high-side gate drivers? Great question and I hope this answer clears things up. Best wishes on your design. -Dr. K

@@powerelectronicswithdr.k1017 Thanks for the quick response. Yes this certainly cleared things up. And about your question, the more content the better :)

Ciss = Cgd + Cgs, and is the input capacitance of the device. Cgd is the capacitance between the gate and drain terminal of the device. Rg is the gate resistance that is internal to the device. You should be able to find Ciss and Rg on the datasheet for the MOSFET you plan to use in your design. Cgd will need to be estimated. Hope this helps and best wishes on your design. -Dr. K.

@@powerelectronicswithdr.k1017 Thank you doctor i really appreciate your giving ❤️❤️❤️

Rendy, by "interverter," I assume you mean DC-AC. The answer to your question would be no. Here's my video on PWM-based single-phase inverter kzfaq.info/get/bejne/r7qAY9uotZrSpX0.html.

Eric, yes. Often specifications will include temperature ranges. In this video we did not focus on heat transfer for this design. There's another video where I cover investigating thermal considerations and show students how to do a simple static analysis. Best wishes on your design. -Dr. K

Ashish, thank you for watching. -Dr K

Hi. Great question. The schematic used more often than not depends on the controller used in the design. The datasheet for the controller will often have one or two reference schematics. There will be more detail such as couple capacitors, components for setting the switching frequency, sizing the voltage divider network for feedback, current limiting, etc. I would start my design by selecting a controller that will meet your specifications, if possible. Best wishes on your design. -Dr. K

Practical mosfet gate triggering circuit is very critical for your circuit to operate the mosfet in switching mode.

Yes, absolutely. This video focuses on the MOSFET high-side switch (which sometimes is integrated into the controller), the flyback diode for a non-synchronous controller, the inductor and the bulk output capacitance. -Dr. K

@@powerelectronicswithdr.k1017 Yes, Sir; Just now lab tested this Buck Converter using the high-side switch the MOSFET technology and circuit is working fine.

Akash, lol. I have been around. Thanks for watching. -Dr. K

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.

Hello Jobs, thank you for commenting and let me respond to a couple of your statements. First, this video is for the buck converter. I do have a video for the buck-boost converter and here is the link kzfaq.info/get/bejne/rMdlqKiA2a3IhWw.html. Second, this topology will work for either a P-channel or N-channel MOSFET in the high-side. Both types of devices can be used as a high-side power switch. P-channel MOSFETS are easier to switch in a high-side configuration. However, N-channels are often used with a high-side gate driver as n-channel devices are typically faster and more efficient. Second, I'm not sure where there is a short circuit in the topology. One must always be careful to designate ground, reference nodes, and current return paths. If you notice, the symbol on my diagram is a reference node. This might be a ground location, but that is not always true. The output node is listed as Vo and is referenced between that node and the reference node. For the buck-boost topology, the output node will be at a lower potential than the reference node and the input will be at a higher potential than the reference node. Finally, there are many situations in which the source is "floating" and the reference is might not be galvanically connected to a physically grounded (to earth) node. Such is the case with battery powered devices. I hope this helps eliminate confusion and thank you for watching. Best wishes on your designs. -Dr. K

@@powerelectronicswithdr.k1017ok I understand and yes the two can use for high side switching . I have a question on the buck booster converter firstly I have a 700amp battery that I test my circuit with but the buck booster converter duty cycle at full showing 298 vdc and still look to go higher but when I turn the duty cycle down to get the desired voltage the amperage is is low and when I connected a low to it the voltage just cutting out so I don’t understand why is the however I am thinking to make push pull buck booster instead because I think I will get better results and most of the current because the one switching with with 7000uf not giving out the current I need for charging from solar panel as saw on the data sheet

@@Dc_tech386 Hi Jobs, I'm not sure I would use this type of configuration at those power levels. What is the application and what are you trying to accomplish?

Ok the buck booster i made is a papular buck booster 2 mosfet and fast switching diodes I use the tl494 Ic with feedback however the two 100 amp mosfet handling capacitor the capacitor but the capacitor or charging up to 200 voltage and I need 12 from my panel of 600watt am not getting the wattage from the panel when I connected the battery to the 200v the voltage drop don’t to 1 voltage I use 7000uf of capacitor and a big inductor with 100 turn and the wattage no staying only small load like bulb work 30 watt but only more load that that the capacitor drop out and the mosfet getting fire hot I do have a 12 volt drive to gate at 6amp so the gate is driving good I use 6 fet and the load still dropping the voltage so I try the push pull converter and is work good today am getting 14 volt at 29 amp so that a good circuit but you think it’s not good enough because the buck booster refused to keep up the booster converter work good too and the buck converter but not the fly back converter

Hi Sean, actually the duty cycle will be "about" 1/3. While 5/15 is exactly 1/3, the duty cycle will be close, but not exactly this value. The tolerances and additional losses are not taken into account when making this estimate. In addition, the losses will vary with temp, time and loading. I'll stand with my wording that the duty cycle will be "about" 33%. Best wishes on your design. -Dr. K

You are welcome. Best wishes on your design.

finally how do you define R load?

Hi Tam, your input values lead me to believe that you are trying to design an AC to DC inverter. 120V and 208V are common AC RMS values for commercial split-phase and 3-phase systems. My first question is, is this a correct assumption? If the answer to that is yes, then the next question would be offline or isolated? The answers to these questions change your approach. -Dr K

@@tamaica7770 R is just used to emulate the load. Select R for your initial design as Vo^2/Po or Po/Io^2. Keep in mind this value of R is just used to start your design. Loads are rarely purely resistive. -Dr. K

Power Electronics with Dr. K i’m trying to understand and design an AC 120/208V to DC 48V converter. First I used full bridge rectifier, then I used Buck converter. So can we input ac voltage range from 120-208V to get 48v DC or we have to choose either 120V or 208V? Also, i don’t have knowledge about offline or isolated design.

@@tamaica7770 That is called an "off-line" converter because it is directly powered off the AC commercial line. Your DC output from your FWB converter will be around165 to 300VDC because you need to use the peak value of the AC voltage, not the RMS value. You can find the duty cycle as it will be Vo/Vin. Therefore if you use D = 48/165 if the input was about 120VACrms. It will be a little be greater as there are losses that we have not accounted for. I would recommend LT Spice for simulating your design. It's a great too. Best wishes. -Dr. K