Hi Ben, it is a lovely presentation and I am going to go through the remaining episodes too. I am also looking for some knowledge on designing high power RF PCB designs and what additional measures need to be taken for high Power PCBs (PAs rates for 40Watts x 4 Tx chains)...are you planning to cover that too?
Fantastic, sir! I've been out of the industry for quite some time and looking to find my way back. This is an excellent review! Thanks for the easy to follow language.
When discussing the charges on the conductors at 17:00 did you say this backwards? First you showed the Positive in the middle with the electrons at the edge attracted to negative. Wouldn't it be the opposite case? Electrons are attracted to positive and repelled from negative? You said it in the case below as well for the next phase of the sine wave. Just seems wrong in my head, but maybe I'm missing something.
Hi Ben, Thank you so much for these great and simplified video series. I just have some problems with the calculated wavelengths in most of the videos. For example: Video 1 (time 5:12) => Landazero = 300/2400 = 12.5 not 16.67 Video 2 (time 25:40) => Landazero = 300/(2400*(4.1^0.5)) = 15.4mm (606 mil) not 736 mil Video 3 (time 25:39) => L = 16.2mm not 19.061 and so on. Am I correct or I am doing something wrong. Thank you again for your time in advance. Regards,
I think its like this: Video 1 must be in vacuum/air. Video 2 must be in FR4 PCB, the epsilon of FR4 is about 4, and the speed of light, or electromagnetic energy, is reduced by a factor of the square root of the material's epsilon (dialectric constant). I havent seen video3 yet....
@@MrSemperfidelis225 yah, this was pointed out in another comment shortly after video was published. That's what you get for not actually doing the calculation live, and the pressure of standing in front of a camera. We're not perfect, but I think the value overall of the video is high - and this maybe confirmed by the number of views...
is it possible for radio frequency be used for internet connection? because current internet signal used today from smartphones are only good outside your house unless you use router and antenna which is very costly..
He is referring to a limiting case where the trace length appears to be nearly invisible to a signal with a certain wavelength, which then means you can ignore the impedance of the trace. This is when the wavelength is very long compared to the trace length. So when the wavelength is longer, the allowed upper limit on the mismatched trace length is also longer. This concept is often referred to as a critical length, but the calculation of a critical length is a bit complex as it requires calculation from the input impedance; this is why so many guidelines just state that the critical length is 10% of the signal wavelength. I have explained it in much more detail here: ruclips.net/video/NYAoqxlKW-U/видео.html
Hi, I have few questions. Let me know if I am wrong, does the length of RF anteena onboard must be multiple greater or multiple smaller of wavelength we calculated from formula? Is that right? But it cannot lies in half of multplie. Like 6.1cm then 12.2cm and 18.3cm will be okay, but If my anteena length is b/w 10cm it would ve wrong.
Hi Altium, Hi Ben. I would like to suggest such practical courses, please prepare a certificate for participants. Later on, the attendances can present it as a skill certificate for a job.
I really appreciate the effort Altium is putting in making high quality contents like these. Thank you Altium
Glad you like them!
I have been looking for a series like this one for a while!
Thank you very much for your effort!
Glad you enjoy it! Ben is currently working on the next two episodes so please stay tuned!
You really encourage me to get Deep-Dive into the subject that make our core every day consumed technologies wich we barely know the fundamentals !
Sir, I absolutely love you. I’ve been so dumbfounded on this topic, but you’ve opened my eyes! ❤️🙏🏾
Glad it was helpful!
Anyone else stuck on his 3/24 calculation?
3/24 = 1/8 = .125 m = 12.5 cm
Why is it 16.67 cm?
Yes! It's 12.5cm for sure
Hi Ben, it is a lovely presentation and I am going to go through the remaining episodes too. I am also looking for some knowledge on designing high power RF PCB designs and what additional measures need to be taken for high Power PCBs (PAs rates for 40Watts x 4 Tx chains)...are you planning to cover that too?
Fantastic, sir! I've been out of the industry for quite some time and looking to find my way back. This is an excellent review! Thanks for the easy to follow language.
Glad it was helpful!
GREAT PLAYLIST! I wonder if you have done some EMC video? i have no clue where to start, i'm watching the Rohde & Schwarz Playlist.
Continue doing what you're doing! You doing it great! and thx
c = 299792458 m/s
Wifi : min 2.4000GHz and max 2.4835GHz
λ(2.4GHz) = c/F = 124.91 mm
λ(2.4835GHz) = 120.71 mm
When discussing the charges on the conductors at 17:00 did you say this backwards? First you showed the Positive in the middle with the electrons at the edge attracted to negative. Wouldn't it be the opposite case? Electrons are attracted to positive and repelled from negative? You said it in the case below as well for the next phase of the sine wave. Just seems wrong in my head, but maybe I'm missing something.
Hi Ben,
Thank you so much for these great and simplified video series.
I just have some problems with the calculated wavelengths in most of the videos. For example:
Video 1 (time 5:12) => Landazero = 300/2400 = 12.5 not 16.67
Video 2 (time 25:40) => Landazero = 300/(2400*(4.1^0.5)) = 15.4mm (606 mil) not 736 mil
Video 3 (time 25:39) => L = 16.2mm not 19.061
and so on.
Am I correct or I am doing something wrong.
Thank you again for your time in advance.
Regards,
I'm not Ben ... but you are right with the calculation... 2.4GHz is ~12.5cm and so on ... 16.67cm is for 1798MHz
I think its like this: Video 1 must be in vacuum/air. Video 2 must be in FR4 PCB, the epsilon of FR4 is about 4, and the speed of light, or electromagnetic energy, is reduced by a factor of the square root of the material's epsilon (dialectric constant). I havent seen video3 yet....
@@MrSemperfidelis225 yah, this was pointed out in another comment shortly after video was published. That's what you get for not actually doing the calculation live, and the pressure of standing in front of a camera. We're not perfect, but I think the value overall of the video is high - and this maybe confirmed by the number of views...
He did say approximately. 😅 Also, at 5:17 his fingers are only 12.5 cm apart, and he should get partial credit for showing his work.
You are amazing! Thank you very much for these videos
Very good content, very good explaination. Thumbs up!!
Glad you liked it!
Very nice explanation.
super thanks, Fantastic video. love you man, luv u Altium....😊
Glad you liked it
Beautiful lecture
is it possible for radio frequency be used for internet connection? because current internet signal used today from smartphones are only good outside your house unless you use router and antenna which is very costly..
I am newbie in High speed and microwave design but i have experience with Altium low speed design, Please guide me where to start.
What books you recommend if we need to dig into antenna technology ?
balinis
Great video, thanks!
Muchísimas Gracias, me ha ayudado mucho. / Thank you very much, it has helped me a lot.
Thanks altium!
This is so helpful! Informational without going over my head or being too simple. From a recent EE/Photonics undergrad.
12:17 the length of the traces should be larger for the larger wave lengths right? it sounds opposite what he is saying. I didn't understand.
He is referring to a limiting case where the trace length appears to be nearly invisible to a signal with a certain wavelength, which then means you can ignore the impedance of the trace. This is when the wavelength is very long compared to the trace length. So when the wavelength is longer, the allowed upper limit on the mismatched trace length is also longer. This concept is often referred to as a critical length, but the calculation of a critical length is a bit complex as it requires calculation from the input impedance; this is why so many guidelines just state that the critical length is 10% of the signal wavelength. I have explained it in much more detail here: ruclips.net/video/NYAoqxlKW-U/видео.html
Thanks for Nice knowledge.
Hi , perfect video
Excellent
thank you so much. this is very helpful.
Loved it. Thanks.
Hi, I have few questions. Let me know if I am wrong, does the length of RF anteena onboard must be multiple greater or multiple smaller of wavelength we calculated from formula? Is that right? But it cannot lies in half of multplie. Like 6.1cm then 12.2cm and 18.3cm will be okay, but If my anteena length is b/w 10cm it would ve wrong.
Quarter wavelength, normally.
thank u very much sir...
CAN YOU SUBMIT SOME TUTORIAL OR EXAMPLE OF DESIGN
Thank you
Welcome!
а так очень годный материал .спасибо!
18:36 "Fucking draw that upside down"
Hi Altium, Hi Ben. I would like to suggest such practical courses, please prepare a certificate for participants. Later on, the attendances can present it as a skill certificate for a job.
I want that shirt
viewed 6.11.2024
300/2400 = 1/8 = 0.125 = 12.5 см
3/24=1/8 =0.125)) плохо в школе учился))
Thanks, very informative
Glad it was helpful!