Dear Robert: Very nice video! Do you say that gain in filter is : 20 * log (vin/vout) , but how we are using second order filter gain would be : 2* [20 * log (vin/vout) ], right ? . And why do you replace "vin/vout" for " fo/fsw" ? Do you have any reference material where i can read the details of this transfer function?
Hi anonymous, the input filter is a second order and will have a 40dB/decade. Once we now how much attention we need from 20 log (Ice_maximum/ IFF_from_iprimary), the we can subtiture to find at what resonce frequency you need to make the input filter. You can also use log-log paper and draw -40db slope and where it intercept the 0db is the frequency. Try it
Please can you send me the video link for all your practical flyback design and the transformer winding procedure or the the link to access all your videos. Thanks
Nice Explanation. We are calculating L and C at particular resonance frequency so we are suppose to get oscillation at that frequency as well which we don't want. Shouldn't we concentrate on fundamental frequency? Can you please explain in brief why we are using capacitance with resistance in series to avoid resonance peak again.
SAMEER GARG Hi Sameer. The capacitance with a resistor is for damping the resonant frequency. At resonant frequency, the output impedance of the filter is at its maximum impedance. This can create a problem if you connect a power supply to the filter. The phenomena called the Middlebrook criteria. Basically it states that the output impedance of the input filter must be much lower than the input impedance of the power supply. If you go to my channel, look for "Input filter effects on a power supply" or try the link below for further explanation. th-cam.com/video/8VzUt0zS36s/w-d-xo.html
Robert Bolanos Thanks. Attached link cleared my all doubts. Is it mandatory to use input filter with all power supply or its depend on harmonics associated with fundamental switching frequency?
Sir I am also in designing field from last 15 years and currently working on an SMPS for EMC compliance. You have such a deep knowledge in this field that I feel that maybe I have only 1/10th of the knowledge compared to your knowledge. I can understand how much effort you put in this field to get such a remarkable knowledge. I believe that this type of videos have limited viewers as it is basically for technical people but I feel that it is million times better than any other viral videos. I wish I could have a mentor like you. Good wishes from my side. Regards: Sandeep Kumar
Hi Mr Roberto. In the min 18:00 you explain that is second order trans funtion ( 40dB*log [fo/fsw] ), but is it always that formula (one theorem) or that funtion is depend of Input filter design ?
A second order low pass filter means that there are two poles and each pole contribute a -20db/dec. Since both poles are on the same frequency then the transfer function simplifies to the equation I gave in the lecture.
Do you say that gain in filter is : 20 * log (vin/vout) , but how we are using second order filter gain would be : 2* [20 * log (vin/vout) ], right ? . And why do you replace "vin/vout" for " fo/fsw" ? . Best Regards.
Is standard showing only differential current limit? Or it shows the total conduction emission limit? If the case is later, how to segregate the differential and common mode current limits?
Hi Ashok, The MIL-STD 461 charts shows is combined. However, the common mode ussually occur at higher frequencies because common mode noise is generated by the parasitic capacitance from the mosfet, transformer and any stray capacitor associated in the primary. So a good rule is to pick the geometric mean of the high frequency range, (2MHz to 50MHz). (2MHz*50MHz)^0.5=10MHz. You then assume an atteniation of 80dB which means that the common mode resomance whould be 10KHz. The common mode filtering using the Ycaps to filter out the common mode noise and should be know. In DC to DC, this capacitance may have to be limited for safety purposes. Then you calculate the common mode inductance by,m assuming Ycap = 10nF, Lcommom=1/((2*pie*10KHz)^2*Ycap). I hope this makes sense.
![fellow fellow - พรุ่งนี้ไม่มีใครรู้ feat. INK WARUNTORN [OFFICIAL MV]](http://i.ytimg.com/vi/QP6JyjYx_W0/mqdefault.jpg)
Spot on!
Thank you dear Robert. It was so beneficial
You are very welcome
I must say ,very good video. I learn a lot. Thank you.
Very good video, thank Robert a lot
Great video sir
Dear Robert: Very nice video! Do you say that gain in filter is : 20 * log (vin/vout) , but how we are using second order filter gain would be : 2* [20 * log (vin/vout) ], right ? . And why do you replace "vin/vout" for " fo/fsw" ? Do you have any reference material where i can read the details of this transfer function?
Hi anonymous, the input filter is a second order and will have a 40dB/decade. Once we now how much attention we need from 20 log (Ice_maximum/ IFF_from_iprimary), the we can subtiture to find at what resonce frequency you need to make the input filter. You can also use log-log paper and draw -40db slope and where it intercept the 0db is the frequency. Try it
Please can you send me the video link for all your practical flyback design and the transformer winding procedure or the the link to access all your videos. Thanks
excellent video ! thank you sir, for the valuable video !
Thank you Sanjeevi for the comment! I am glad that you found it helpfull
Nice Explanation. We are calculating L and C at particular resonance frequency so we are suppose to get oscillation at that frequency as well which we don't want. Shouldn't we concentrate on fundamental frequency? Can you please explain in brief why we are using capacitance with resistance in series to avoid resonance peak again.
SAMEER GARG Hi Sameer. The capacitance with a resistor is for damping the resonant frequency. At resonant frequency, the output impedance of the filter is at its maximum impedance. This can create a problem if you connect a power supply to the filter. The phenomena called the Middlebrook criteria. Basically it states that the output impedance of the input filter must be much lower than the input impedance of the power supply. If you go to my channel, look for "Input filter effects on a power supply" or try the link below for further explanation. th-cam.com/video/8VzUt0zS36s/w-d-xo.html
Robert Bolanos Thanks. Attached link cleared my all doubts. Is it mandatory to use input filter with all power supply or its depend on harmonics associated with fundamental switching frequency?
great video!
Thank you anaconnaught!
Awesome video, thanks
Thank you Kumar. Comments like your is what keep me making videos like this one. Thank you again for you comment. Robert Bolanos
Robert Bolanos your reply made my day, a huge respect for you and your knowledge.
Sir I am also in designing field from last 15 years and currently working on an SMPS for EMC compliance. You have such a deep knowledge in this field that I feel that maybe I have only 1/10th of the knowledge compared to your knowledge. I can understand how much effort you put in this field to get such a remarkable knowledge. I believe that this type of videos have limited viewers as it is basically for technical people but I feel that it is million times better than any other viral videos. I wish I could have a mentor like you. Good wishes from my side. Regards: Sandeep Kumar
Hi Mr Roberto. In the min 18:00 you explain that is second order trans funtion ( 40dB*log [fo/fsw] ), but is it always that formula (one theorem) or that funtion is depend of Input filter design ?
A second order low pass filter means that there are two poles and each pole contribute a -20db/dec. Since both poles are on the same frequency then the transfer function simplifies to the equation I gave in the lecture.
Do you say that gain in filter is : 20 * log (vin/vout) , but how we are using second order filter gain would be : 2* [20 * log (vin/vout) ], right ? . And why do you replace "vin/vout" for " fo/fsw" ? . Best Regards.
Hi robert. In the min 18:00 how did you get that transfer funtion in second order ?
Is standard showing only differential current limit?
Or it shows the total conduction emission limit?
If the case is later, how to segregate the differential and common mode current limits?
Hi Ashok, The MIL-STD 461 charts shows is combined. However, the common mode ussually occur at higher frequencies because common mode noise is generated by the parasitic capacitance from the mosfet, transformer and any stray capacitor associated in the primary. So a good rule is to pick the geometric mean of the high frequency range, (2MHz to 50MHz). (2MHz*50MHz)^0.5=10MHz. You then assume an atteniation of 80dB which means that the common mode resomance whould be 10KHz. The common mode filtering using the Ycaps to filter out the common mode noise and should be know. In DC to DC, this capacitance may have to be limited for safety purposes. Then you calculate the common mode inductance by,m assuming Ycap = 10nF, Lcommom=1/((2*pie*10KHz)^2*Ycap). I hope this makes sense.
@@RobertBolanos Thank you so much sir, for the clarification.
@@RobertBolanossir, how did you decide the high-frequency range? Will it be same for all the standards?
Amp(fsw)=40log(fo/fsw) instead of Amp(fo)=40log(fo/fsw) ? is it right at 22:13