RC Band Pass Filters - How To Design The Circuit
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- Опубликовано: 23 окт 2024
- This electronics video tutorial provides a basic introduction into RC band pass filters. It explains how to calculate the two cut-off frequencies, the resonant frequency, and the bandwidth. It also explains how to design the circuit by calculating the capacitance values need to construct a circuit given two cutoff frequencies and given the center frequency and bandwidth.
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I hope you're making a 6 figure salary teaching at a college! This is one of the best educational videos I've found on RUclips in the line of Electronics Engineering. Some of the algebra is irrelevant in my opinion but it can be used to find values in different ways than mine, so it is what it is. Thank you for this video.
Another outstanding lecture! I heard the term band pass filter although I had no idea what it meant. You're really doing a great service to all scientists and engineers with these videos!
Thank you!!!! Saved my life. Was reading the script of our Professor for hours but didnt understand until this video
From your video, now I understand the meaning of corner frequecy. I have been heard this term for long time and on various occasions, now I got it!! thank you so much!
I've just understood that you are not an Organic Chemistry Tutor, you are a tutor formed from organic chemistry.
🤣
😂
Makes sense
What can I say but 'Thank you' for an insight I took for granted but actually lacked. Keep up the great job👍👍👍
After a long staring competition with my assignment question which was getting me no were, I turned to RUclips and found your video. And Thank-Goodness I did!
Hope you keep the videos coming, Im going to need them.
you're a blessing to our generation
most reasonable expalnation I have ever heared. thanks for sharing your knowledge.
Dude, your videos are killer. Straight to thew point and concise.
Thanks for providing all of this wonderful free content! It really resonates with my interests...
You basically are knowledgable about everything I ever knew to exist!
Good explanation and examples. I used to call your f res as geometric average, and f average as arithmetic average :)
❤❤❤ from Chennai, Tamilnadu India
Priceless video. Thanks!
Thank you so much sir
Respect from 🇮🇳❤️❤️🙏
If I'm not mistaken, matching our resistors value to the reactance resulting from the defined cornered frequency of the capacitor will improve filter stability and Q?
I know it's ultimately also a voltage divider.
You’re a life saver!!!!
Please sir more of these thank you
great job bro thx from Turkey
You are a gem
Thanks alot my guy
Thanks this helped so much!
thank u man you are the best
Thank you, finally i understand it.
Love this channel
I really love this video. Cheers for the share!
Terimakasih banyak Mr.
U have saved my life.
Hi, Why didn't you solved problem 3 with the method that used to solve problem 4(by placing the resonance frequency in the mid and calculating Fc1 and Fc2?
And if I used the method in problem 4 in problem 3, Fc1 and Fc2 are 1.5KHz and 10.5KHz respectively.
But as per the video, they are 3KHz and 12KHz, if it is 3 and 12 KHz, then Fr should be 7.5KHz right?
you can find capacitors like that if you combine series and parallel capacitors untill you arrive in the deserved value
first of all thanks for reading my comment ..i have to design a passive band pass filter which range between freq of 174MHz - 230Mhz .. so my question is can i use this RC circuit for my design or should i need the RLC circuit instead ?
THANKS!
I wish my exam questions were this simple 😅
شكراً 🙏
Amazing
Ill make a tone control circuit for my amplifier
Thank you!
15:12 Why not complete the square by (9/2)^2 to give the expanded form of (f1^2) + 9f1 + 20.25 = 36 + 20.25
to give (f1 + 4.5)^2 = 56.25 then square-root both sides to get
f1 + 4.5 = 7.5
f1 = 3
Thankyou!!
can you make a tutorial of an RL band pass filter please
ขอบคุณ
This is quite problematic... When you charge a first filter with a second one, the first filter transfer function is changed, as it s now more an open circuit... In your case you should up the second filter equivalent impedance by using R2=R1*10 in the second filter and dividing C2 /10... Or should I have listened to the whole vid ? doesnt seem to be mentionned.
Thats what happens when Org. Chem. Tutor does electronics ?
It isn't 'problematic'. It is _wrong._ The issue you raise is not mentioned. I sat through the whole thing and waited. He doesn't seem to be aware that the voltage-divider equation assumes zero source impedance and infinite load impedance, and that the presence of another RC section before or after violates one or other of these conditions. His results are invalid in every case.
Plss sir Notice my problem.
An 18.5hp overhead crane is used to lift at a rate of 3.5m/minutes. How much load can it raise at this lifting speed.
Plss help me to answer this problem sir.
Thank you and Godbless.
You need to convert horse power to Watts
And because power in watts is energy over time you can calculate the energy witch is Power by unit of time.
Then with this potential energy you can calculate the weight. (E = m.g.h)
I hope this will help you
What about the H function?
for Mhz frequency can I use RC filter? Why dont you use inductor for the filter? why why why? @The Organic Chemistry tutor
Making a BP filter out of only RC elements give you a raw model of the BP filter. Using coils will give better results but sometimes you are in a position where you can't use or have coils, so in that case an RC BP circuit comes in handy
@@Tom-jc9lo Hhlmhlh, Today my last day for submission of capstone project and i was writing about the filter part. And, i was thinking, the answer of this question. IT'S UNBELIAVEBLE. 🤣🤣🤣🤣
It sure is unbelievable. Unbelievably wrong. Hope you didn't use it, or you would have failed.
@@EJP286CRSKW no i didnt use inductors. But i am still curious about the reason.
Inductors don't come in precise values; have significant series resistance; and below RF are rather large. People avoid them.
damn good video
I LOVE YOU
u do not calculate the load yet. do u have another video which calculates the load?
I know for cutting of High frequency there are used coils (inductors) not resistance ...
There are quite enough low-pass RC circuits out there to prove you wrong. You can use LC, or LR for that matter, or RLC, but it isn't compulsory.
@@EJP286CRSKW this works on line crossover before amplification ? Because in audio we don't use nanofarad but micro. And resistance of ohms or 0.1 ohm not kilo-ohm
Rubbish. In audio I have used every capacitance between 3.3pF and at least 10,000uF, and every resistance between 0R1 and 10M.
Please be aware this vid is NOT reliable ... Always check the poster credential and comments below...
How _not_ to design a bandpass circuit. You can't just stack up passive filters like this. None of these circuits will work to the specifications given. The two filter sections load each other, and the transfer functions are therefore not as stated. Totally misleading video.
I'm sorry for all the apparent EE students below who have failed because of taking this on trust.
1
I agree
R2C2. These aren't the filters you are looking for...............
15:18 This is incredibly unrealistic. The only reason you can solve this by factoring is because it's a custom-crafted artificial test problem. In the real world, if you needed to figure out something like this, the answer would almost certainly not be a convenient integer, and solving with factoring would be a huge pain. Therefore, it's really kinda stupid to teach people to do it this way. You should just be using the quadratic formula instead.
Sar aap samjhane se pahle mujhe bata do kya aapane ismein kaun sa pregnancy use frequency ne koi example nahin hai Kali calculate karke kya karne wale hain show the physically example and frequency calculator
What can I say but 'Thank you' for an insight I took for granted but actually lacked. Keep up the great job👍👍👍