It's really valuable videos, a true lesson about components selection , losses calculation based on requirements; a practical application demonstration rarely found. wish all the best
Really nice breakdown of the losses, thank you ! I think there is another reason to replace that diode with a mosfet, and that is the capacitance of the V30K45 diode : it is a whopping 4 nF, which needs to be charged 0->15 V right during that switch-on time triangle. Quick calculation suggests another 250 mW or so loss because of that diode 1/2CV^2*f losses at 500 kHz. (And the diode is also more expensive than that nice 60 ct mosfet) Thanks again, I enjoyed your videos.
I really enjoyed these two videos. It’s been a long time since I was doing electronics stuff but getting back to it. I did notice a few things that I think may be incorrect. Like 5 minutes into the video a term is written 4A squared. But it should be 2A squared. Further on the correct squared value 4 is still used so it didn’t influence the result, it’s more a notational slip I guess. Then after about 13 minutes you opened a new sheet to calculate the switching loss. But if I’m correct here you left out the current so in my opinion the loss should be double of what was calculated there. Nevertheless it was really instructive and helping me to get up to speed again.
Hi Peter, the (2A)^2 = 4A^2. The square is referring to the units and not the value as an A^2*Ohm = W. I probably should have been more clear on that notation. Actually the loss in the inductor is slightly greater as there a loss associated with the ripple current. That is there is an AC RMS voltage associated with the ripple current that I ignored. On the switching loss calculations you referenced, the calculations are correct. However, I should have included the units when switching to the next page. The switching loss is (1/2)(15V)(2A)(ton+toff)(fs). This would be 15W*(ton+toff)*fs. I did the calculations on those first three terms when jumping to the next page and should have place the units in that equation. Sorry for the confusion. Thanks for watching! Best wishes on your designs. -Dr K
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
Hi Mohit, hope the videos are useful. Please reference part 1 of this series for the capacitor sizing ruclips.net/video/IpoI6ERn5zM/видео.html, Best wishes on your design. -Dr. K
@@mohitarora8754 Yes, there is loss in the capacitor and it's based on (Irms)^2(Resr). The Irms should be about Delta_I/(2*sqrt(3)). For this design, I recall talking about using MLCCs which have an ESR that is often a magnitude lower than electrolytic caps. Yes, there is some loss, but its smaller compared to the loss in the fly-back diode and the MOSFET, at least for this application, which is not always the case. If you compute this power loss for a 10mOhm ESR, you will see it is pretty small. Great questions! - Dr. K
Dr K i have a COB LED which says 10W ,12 to 14 volts , I am confused if constant current or voltage is required for the LED - Thank you so much in Advance.
Hi, the answer to your question is "it depends." You will need to look at the COB LED specifications. LED drivers have similar characteristics to buck converters in that they provide a lower output DC voltage, perhaps around 12-14 volts from a higher input voltage. When LEDs are connected in series, the nominal output voltage is approximated by the forward voltage drop times the number of LEDs connected in series. Once the LEDs are on, the intensity is controlled by the current flowing through the LEDs. Each LED has a slightly different forward voltage drop, and often times we use current to control the duty cycle of the converter. The control of the converter (voltage, current or both) is not something that I've covered in my videos. Keep me posted on your design. Lighting systems are a great application for power electronics. -Dr. K
@@powerelectronicswithdr.k1017 Thank you Dr K for the explanation , unfortunately i cannot find any details of this COB strip other than 10W 12 to 14 volts, i am still searching for the details , i will update you soon- Thank you again for your time !
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.
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
Hi Yaser, it is (2A)^2 which is 4A^2. Sorry for any confusion on that. Yes, the current is 2A. Also note that there is an RMS component to the loss to, but that is typically much lower. Best wishes on your design. -Dr. K
@@m.hassanbaig6885 Some of it depends on where your neutral/reference point is for the DC side and if you need galvanic coupling between your AC source and the DC load.
Sorry, but I do not have one yet. This is an important topic and there is very little information on RUclips covering the reverse recovery loss and reverse recovery time of a diode. Thank you for the suggestion. I'll see if I can pull something together. -Dr. K
Sir, when you calculate the switching loss the turn on and off current is not Io right? if the ripple is very huge I think can't not just use Io times Vds right?
That's correct. Using Io for the current is an average. If you know the ripple, then the switching loss for turn-off can be estimated at Vds*(Io+Iripple/2)*t_off*fs the switching loss at turn_on can be estimated as Vds*(Io-Iripplie/2)*t_on*fs. If t_on and toff are the approximately the same then a good estimate for the total switching loss is Vds*Io*(t_on+t_off)*fs. This all assumes continuous conduction mode (CCM). Hope this helps and best wishes with your design. -Dr. K
@@powerelectronicswithdr.k1017 Thank because most text book usually use average value so they don't tell why so I'm confused, and I have one question the 1/2 value is because the triangle, the vds and Ids the area alway's triangle? I see some text book use 1/6 but I don't know why , If I want to more accurate how should I do
@@rj8528 Hi, the 1/6 term comes only when the load being switched is purely resistive. For a resistive load, the instantaneous power p(t) = v(t)i(t) and v(t) = -(Vin/t_on)*t + Vin and i(t) = (Io/t_on)*t during the turn-on interval. Therefore p(t) = -(Vin*Io/t_on^2)*t^2 + (Vin*Io/t_on))*t from t=0 to t=t_on. The curve of the instantaneous power for a resistive load is an upside-down parabola with a max height of Vin*In/4. The arear (energy-dissipated) is the integral of p(t) and this equals (1/6)Vin*Io*t_on. When the load is inductive, the voltage Vin gets clamped to a constant value because the fly-back diode is still conducting while the MOSFET is turning on. This diode clamps the source node at near 0-volts relative to ground until the diode becomes reverse biased again. In this case p(t) = (Vin*Io/t_on))*t and integrating results in (1/2)Vin*Io*t_on. One of my videos illustrates the turn-on and turn-off voltages/currents for a MOSFET. Hope this helps. -Dr. K
@@powerelectronicswithdr.k1017 if the load is resistor the area I need use 1/6. Normally use 1/2 because the real situation doesn't have pure resistor load ?
It's really valuable videos, a true lesson about components selection , losses calculation based on requirements; a practical application demonstration rarely found. wish all the best
Really nice breakdown of the losses, thank you !
I think there is another reason to replace that diode with a mosfet, and that is the capacitance of the V30K45 diode : it is a whopping 4 nF, which needs to be charged 0->15 V right during that switch-on time triangle. Quick calculation suggests another 250 mW or so loss because of that diode 1/2CV^2*f losses at 500 kHz.
(And the diode is also more expensive than that nice 60 ct mosfet)
Thanks again, I enjoyed your videos.
You are correct. Also, depending on the application, the synchronous buck converter options are often the better route.
Great video. Thanks a lot of sharing the complete design procedure.
Neehar, you are welcome. Best wishes on your design. -Dr. K
I really enjoyed these two videos. It’s been a long time since I was doing electronics stuff but getting back to it. I did notice a few things that I think may be incorrect. Like 5 minutes into the video a term is written 4A squared. But it should be 2A squared. Further on the correct squared value 4 is still used so it didn’t influence the result, it’s more a notational slip I guess. Then after about 13 minutes you opened a new sheet to calculate the switching loss. But if I’m correct here you left out the current so in my opinion the loss should be double of what was calculated there.
Nevertheless it was really instructive and helping me to get up to speed again.
Hi Peter, the (2A)^2 = 4A^2. The square is referring to the units and not the value as an A^2*Ohm = W. I probably should have been more clear on that notation. Actually the loss in the inductor is slightly greater as there a loss associated with the ripple current. That is there is an AC RMS voltage associated with the ripple current that I ignored. On the switching loss calculations you referenced, the calculations are correct. However, I should have included the units when switching to the next page. The switching loss is (1/2)(15V)(2A)(ton+toff)(fs). This would be 15W*(ton+toff)*fs. I did the calculations on those first three terms when jumping to the next page and should have place the units in that equation. Sorry for the confusion. Thanks for watching! Best wishes on your designs. -Dr K
It is writen as 4A^2 rather (2A)^2, so 4A^2 is correct notation.
Best video in the internet
Thank you.
Beautifully Explained ! Thank you very much.
You are welcome . Thanks for watching. -Dr. K
I have seen your both video on buck converter it carries valuable information.🙏
Sanjay, thank you. Best wishes on your designs. -Dr. K
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
Thanks Sir for this great video. I hope you all the best 💯
You are welcome. All is well. -Dr. K
First of all thank you for sharing your knowledge, I have just one question why you didn't consider the capacitor??
Hi Mohit, hope the videos are useful. Please reference part 1 of this series for the capacitor sizing ruclips.net/video/IpoI6ERn5zM/видео.html, Best wishes on your design. -Dr. K
@@powerelectronicswithdr.k1017 Thank you for your time Sir, but i am asking "Why you didn't consider capacitor in efficiency calculation? "
@@mohitarora8754 Yes, there is loss in the capacitor and it's based on (Irms)^2(Resr). The Irms should be about Delta_I/(2*sqrt(3)). For this design, I recall talking about using MLCCs which have an ESR that is often a magnitude lower than electrolytic caps. Yes, there is some loss, but its smaller compared to the loss in the fly-back diode and the MOSFET, at least for this application, which is not always the case. If you compute this power loss for a 10mOhm ESR, you will see it is pretty small. Great questions! - Dr. K
Dr K i have a COB LED which says 10W ,12 to 14 volts , I am confused if constant current or voltage is required for the LED - Thank you so much in Advance.
Hi, the answer to your question is "it depends." You will need to look at the COB LED specifications. LED drivers have similar characteristics to buck converters in that they provide a lower output DC voltage, perhaps around 12-14 volts from a higher input voltage. When LEDs are connected in series, the nominal output voltage is approximated by the forward voltage drop times the number of LEDs connected in series. Once the LEDs are on, the intensity is controlled by the current flowing through the LEDs. Each LED has a slightly different forward voltage drop, and often times we use current to control the duty cycle of the converter. The control of the converter (voltage, current or both) is not something that I've covered in my videos. Keep me posted on your design. Lighting systems are a great application for power electronics. -Dr. K
@@powerelectronicswithdr.k1017 Thank you Dr K for the explanation , unfortunately i cannot find any details of this COB strip other than 10W 12 to 14 volts, i am still searching for the details , i will update you soon- Thank you again for your time !
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.
Big fan here.
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
Hi sir your video is very good
I have one doubt. Why inductor loss current calculate 4A we have 2A only sir
Hi Yaser, it is (2A)^2 which is 4A^2. Sorry for any confusion on that. Yes, the current is 2A. Also note that there is an RMS component to the loss to, but that is typically much lower. Best wishes on your design. -Dr. K
Great video.
Really thank you so much 💛
You’re welcome 😊
Very Well Explained. Sir how to convert Bidirectional DC i.e +311V and -311V to 72volts?
Hi, it depends. Are those DC bus rail values are those from and AC source? - Dr. K
@@powerelectronicswithdr.k1017 yes
@@m.hassanbaig6885 Some of it depends on where your neutral/reference point is for the DC side and if you need galvanic coupling between your AC source and the DC load.
you are the best a grate thanks
Sir,Do you have video talking about diode loss and reverse recovery loss
Sorry, but I do not have one yet. This is an important topic and there is very little information on RUclips covering the reverse recovery loss and reverse recovery time of a diode. Thank you for the suggestion. I'll see if I can pull something together. -Dr. K
@@powerelectronicswithdr.k1017 thanks
Sir, when you calculate the switching loss the turn on and off current is not Io right? if the ripple is very huge I think can't not just use Io times Vds right?
That's correct. Using Io for the current is an average. If you know the ripple, then the switching loss for turn-off can be estimated at Vds*(Io+Iripple/2)*t_off*fs the switching loss at turn_on can be estimated as Vds*(Io-Iripplie/2)*t_on*fs. If t_on and toff are the approximately the same then a good estimate for the total switching loss is Vds*Io*(t_on+t_off)*fs. This all assumes continuous conduction mode (CCM). Hope this helps and best wishes with your design. -Dr. K
@@powerelectronicswithdr.k1017 Thank because most text book usually use average value so they don't tell why so I'm confused, and I have one question the 1/2 value is because the triangle, the vds and Ids the area alway's triangle? I see some text book use 1/6 but I don't know why , If I want to more accurate how should I do
@@rj8528 Hi, the 1/6 term comes only when the load being switched is purely resistive. For a resistive load, the instantaneous power p(t) = v(t)i(t) and v(t) = -(Vin/t_on)*t + Vin and i(t) = (Io/t_on)*t during the turn-on interval. Therefore p(t) = -(Vin*Io/t_on^2)*t^2 + (Vin*Io/t_on))*t from t=0 to t=t_on. The curve of the instantaneous power for a resistive load is an upside-down parabola with a max height of Vin*In/4. The arear (energy-dissipated) is the integral of p(t) and this equals (1/6)Vin*Io*t_on. When the load is inductive, the voltage Vin gets clamped to a constant value because the fly-back diode is still conducting while the MOSFET is turning on. This diode clamps the source node at near 0-volts relative to ground until the diode becomes reverse biased again. In this case p(t) = (Vin*Io/t_on))*t and integrating results in (1/2)Vin*Io*t_on. One of my videos illustrates the turn-on and turn-off voltages/currents for a MOSFET. Hope this helps. -Dr. K
@@powerelectronicswithdr.k1017 if the load is resistor the area I need use 1/6. Normally use 1/2 because the real situation doesn't have pure resistor load ?
@@rj8528 Yes, that is correct and for the Buck converter, the load is inductive. Therefore, use 1/2. -Dr. K
thx
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