RC High Pass Filter Explained

[@albertargilagaclaramunt3693]

Amazing video, thanks!!! I saw 20 videos of audio people talking about crossovers and hi-low pass filters, yours is the only one that really explains it well.

[@srivatsaa.r.9936]

sir, yr video on pASSIVE HIGH PASS FILTER AND AN EXAMPLE OF DESIGNING A CIRCUIT IS VERY GOOD AND CLEAR. GOOD EXPLANATION. THANKS S.VATSA, BANGALORE

[@tusharjamwal]

I was thinking about getting into some basic analog signal processing for radio signals as well as things like designing cross overs for custom speakers. This helped a lot. Thank you 🙏

[@arsalansyed4709]

bro you are actually a legend at teaching

[@kimjong-un3870]

Your intro sounds like your making makeup tutorials :D

[@electricalknpwledge3511]

well said Mr. president

[@srivatsaa.r.9936]

sir yr video on BAND PASS FILTER AND YR EXPLANATION IS EXCELLENTLY BROUGHT OUT. IT HELPED ME TO UNDERSTAND BETTER. THANKS S.VATSA. BANGALORE

[@deanmichaelk]

Thanks for the video. What did you mean when you mentioned the next stage will get loaded if you choose the value of R in Ohms? I just don't understand the concept!!

[@circuitsanalytica4348]

Hello Dean If value of R is very small, when a load is connected, effective value of input impedance of the next stage will be very small and effective value of output voltage falls or it leads to loading....

[@tanmoydutta5846]

Loading effect is what occurs when you directly cascade the output port of one network with the input port of the next (without providing any buffer or active components in between). Basically, the input voltage at the second network is not what we would normally obtain as the output of the first network (here, filter) assuming infinite impedance. Here, the second filter would draw some current from the first filter and thereby, further reducing the gain of the first filter. So, in order to reduce this loading effect, the impedance of the second filter must be as high as possible.

[@indianlad23]

Thanks.. Would be helpful if you could also solve some questions..

[@ALLABOUTELECTRONICS]

Already covered questions related to it. Please check the second channel related to quiz. You will find there. In case, if you are not able to get it, then let me know here.

[@ganeshdilli2699]

I'm glad to see this video,it solve my doubts about hpf thank u.........

[@circuitsanalytica4348]

Yes Ganesh, really nice video....

[@SomaSaiKarthikDuvvuri]

Sir why didn't we place the Buffer in Second Order Low Pass Filter as we have placed for the Second order High Pass Filter

[@abhijithanilkumar4959]

Oh sir so in Rx phase shift oscillator we are using a 3 stage cascaded RC high pass filter in feedback path right So is it like they pass all frequency components of the noise above this cut off frequency of 1/2πRC but only this 1/2πRC frequency makes transfer function of feedback network Beta=1/29 and with A=29 we can satisfy Barkhausen criteria and they sustain And for all the ones above this frequency that has passed this feedback filter network do not satisfy Barkhausen criteria thus won't be sustained??? Sorry for such a long question

[@circuitsanalytica4348]

Dear Abhijith, in RC Coupled amplifier, for sustained oscillations barkhausens criteria must be satisfied. Total phase shift is 360 degree only for those frequency to which the oscillator is tuned to.

[@sledge2742]

all about electronics is nothing but life saver

[@circuitsanalytica4348]

Yes Sledge, you are right.....

[@armyofteachers]

Superb video keep up the good work ✌

[@sumanapain6894]

Very nicely described... Thank you so much

[@deanmichaelk]

Also at 11:39 of this video, you said, " So, now suppose, in your design if you want that at 1 KHz..." I'm not quite sure what we want at 1 KHz. "that" refers to what? Thanks!!

[@pratosh666]

He meant if you don't want that magnitude of noise at 1KHz then higher order filters are needed to reduce the magnitude of that said 1KHz noise even further. By now you could have figured it out. This is for those who have the same doubt and opened your comment. Correct me if I am wrong.

[@ethiofunnyvideos8251]

nice explanation keep it up👍👍

[@alimar1897]

Perfect explanation, thanks

[@heerbrahmbhatt6917]

Very well explained

[@lynns4122]

This was insanely helpful. Thanks!

[@on1ytheb3st]

So would just the positive terminal coming off an amplifier go into this circuit and the resistor gets its own ground or does the positive and negative terminals from the amplifier go into this circuit?

[@aryamehta007007]

why you sometime use sometime Xc = 1/(jwC) sometimes Xc = 1/(wC) ? i am confused.

[@ALLABOUTELECTRONICS]

I have used Xc= 1 /wC, when I am talking about only amplitude of Xc, or I would say it is |Xc|.

[@sundaranarasimhan58]

ALL ABOUT ELECTRONICS I did not get..

[@ALLABOUTELECTRONICS]

When phase information of Xc is not required and only amplitude of the reactance is required at that time I have used |Xc|= 1/wC. I hope now it will get clear to you.

[@circuitsanalytica4348]

Hello Arya, the term "j" make it a complex number. Both are correct, first one is a vector (with j) and second one is scalar....

[@Infinitesap]

@@ALLABOUTELECTRONICS More information is needed here.

[@mayurshah9131]

Very well Narration

[@sais.here.5869]

sir , i was unable to understand the part where you assumed the value of Resistance as 10 Kohm.......what is the next level loaded you are reffering to?....plz do reply

[@ALLABOUTELECTRONICS]

when you connect the circuit next to the filter, the connected circuit will act as a load to the filter. Let's say the input impedance of the circuit which we want to connect to the filter is Zin, then the effective load or R-value for the high pass filter is R || Zin. I hope it will clear your doubt.

[@sais.here.5869]

@@ALLABOUTELECTRONICS oo so net impedance will be the parallel combination of both ....now I get it Thanku very much sir

[@neelofersideeqi]

@@ALLABOUTELECTRONICS but as far i know, when impedances are in parallel, net impedance gets reduced. so how does it increase the loading? please explain

[@israayusuf3712]

What does it mean that the circuit might get loaded by the value of R at 8:34?

[@ALLABOUTELECTRONICS]

When you connect the load to this filter (Let's say speaker), then the impedance of that load will be in parallel with resistor R. And in many cases, the (like in case of a speaker) the value of load is in ohms. So, that will change the effective value of the resistor R and hence the cut-off frequency also. Let's say in your filter design R is 1Kilo ohm and the load is 50 ohm then effective resistance R will be less than 50 ohms and your designed cut-off frequency will get changed. So, that is the loading effect on the filter. So, in your design value of R should be appropriate. If it's too small then it will draw too much current and if it is too high then value of capacitor will not small (and might be comparable to parasitic capacitance of the circuit) The best way to avoid loading effect is to isolate the load from the filter circuit, or in general, for any circuit it is true, and it can be easily done by using OP-AMP as a buffer. To know more about it you can check my video on the active low pass and high pass filter.

[@israayusuf3712]

That's a very good and clear explanation. I really appreciate it :')

[@circuitsanalytica4348]

Hello, when a load is connected to the filter, effective value of R will change, leading to a shift in cutoff frequency, possibilty of loading. So to avoid that, it will be better if a buffer amplifier is used in between....

[@bhavanihm7122]

Excellent

[@mdshaonmia7215]

Excellent ❤

[@sanveersookdawe]

What does it mean for a stage to get "loaded"?

[@ALLABOUTELECTRONICS]

I have explained it in the Butterworth Filter video. Here is the link: kzbin.info/www/bejne/opSZgodui8-kjMU If you still have any doubt after watching it, let me know here.

[@Infinitesap]

Could you elaborate a bit about the j thing?

[@jefejulioelgringojudeo6020]

I love the accent. Is that Hindi?

[@sundaranarasimhan58]

Jefe Julio el Gringo Judeo yes it's hindi

[@dhanurbanpanim]

Punjabi

[@russellborrego1689]

I can cut an existing rca cable in half, solder the resistor + capacitor into the two wires, put it all back together... And that will work fine, right?

[@circuitsanalytica4348]

Hello Mr.Russell, It will work for frequencies greater than cutoff frequency....

[@leyvarecio3699]

What about the phase change in higher order filters ?? 7:29

[@raghavendrasaicherukuri5832]

At 2:50 why the cutoff frequency is 1/√2 of input voltage??

[@ALLABOUTELECTRONICS]

Because at cut-off frequency, the power reduced by 3 dB or the power becomes half. So in terms of voltage, it becomes 1/√2.

[@raghavendrasaicherukuri5832]

@@ALLABOUTELECTRONICS why the power reduce to 3dB

[@noweare1]

when finding phase of transfer function how does tan (angle) = 1/(1-j/wrc) simplify to tan(angle) = 1/wrc I have looked into by trig books but can not find how this is solved. Thanks

[@ALLABOUTELECTRONICS]

If you have complex number, (A + jB) / (C + jD), then phase angle,= tan -1( B/A) - tan-1 (D/C) Comparing it with, 1/(1- j/wRC) Here in the numerator B=0, A= 1, C= 1 and D = -1/wRC If you put all these in the above expression then you will get it. I hope it will clear your doubt.

[@kevinha8256]

could you please inform me where to look for the formula (A + jB) / (C + jD)??? and also when you w=0 then angle =90!! but wouldn't that be tan-1(1/0*R*C)???? you cant divide by zero right???

[@SamerAbreek]

How did R/(R+ 1/jwc) become 1/(1-j/wcR) at 6:03 ?

[@rupeshkarar]

First divide both numerator and denominator by R and then put (-j) in place of (1/j) as they are same.

[@ZambiblasianOgre]

@@rupeshkarar You're a legend, thank you so much for explaining this. I have been stuck because of this for a week.

[@saboorsaboor704]

what am i doing here? I am a taxi driver.

[@U2BER2012]

Did Uber drive you here?

[@Eplemos4Life]

saboor saboor xDdddd

[@Eplemos4Life]

Quarantine day: 8

[@ladytalksalot4097]

No profession is barred from learning, silly goose. Now go wreak havoc with your new knowledge!

[@xXmayank.kumarXx]

10:29 do you mean 20kΩ "rheostat"?

[@shuvbhowmickbestin]

Didn't understand why you chose the resistance value as such.

[@circuitsanalytica4348]

Dear friend, you can refer my channel for more details.....

[@irocx8745]

Sir...what is meant by loading...which you mentioned in video...?

[@ALLABOUTELECTRONICS]

In the video of input and output impedance, I have explained the loading effect. You may check that video. And still if you have any doubt then do let me know here. Here is the link: kzbin.info/www/bejne/bZvaY5Kubcmsh5o

[@jairamrprabhu5696]

Unit of capacitance is Farad and not Faraday

[@circuitsanalytica4348]

Yes Jairam you are right....

[@hariprasadgowdakerenadka8450]

Iam not getting@ 9:23 How you select the frequency value for calculating the value of capacitor

[@ALLABOUTELECTRONICS]

Here high pass filter with a 10 kHz cut-off frequency is designed. So, f is already known. The value of R is selected as 10 kilo ohm. And with that value of R, the capacitor value has been found for 10 kHz cut-off frequency.

[@aritradutta6049]

what is the highest amount of freq. component the filter can pass?

[@circuitsanalytica4348]

Dear friend, highest frequency is infinite ideally....

[@saikiranreddyskr2743]

Sir what about LC filters...please make them if u can...and .thanks for these vedios....

[@ALLABOUTELECTRONICS]

Yes, in this analog filter series, I will also make a video on LC filters.

[@saikiranreddyskr2743]

ALL ABOUT ELECTRONICS Thank you sir

[@balakumar7262]

ALL ABOUT ELECTRONICS why we shouldn't use 10 mega ohm.. Explain detail

[@potlapallyakshith4669]

for second order high pass give the phase response

[@anassanass909]

Is this buttorworth high pass active filter...

[@ALLABOUTELECTRONICS]

No, it is not Butterworth high pass filter. The Q should be equal to 0.707 for the Butterworth filter. Please check my video on Butterworth filter for more information about the Butterworth filter. At the latter part of the video, I have explained how to design the Butterworth high pass filter. Here is the link: kzbin.info/www/bejne/opSZgodui8-kjMU

[@ALLABOUTELECTRONICS]

What here I was referring was for the higher order filter. (second order or more)

[@UMMULKAINATT]

when we take magnitude we square the denominator but not numerator ..why?

[@ahmadjabaly]

It comes from taking the magnitude of the values and the magnitude of a complex number |a+bi|=sqrt(a^2+b^2), all other values in the equation are real numbers therefor they don't change.

[@siddharthjindal6060]

what is the loading effect of which you are talking about in this video.

[@ALLABOUTELECTRONICS]

I have a separate video for that. Please check this video, you will get it. kzbin.info/www/bejne/bZvaY5Kubcmsh5o

[@circuitsanalytica4348]

Dear Siddharth, you can refer my channel to know about loading...

[@smrutishah7839]

nice information

[@circuitsanalytica4348]

Yeah Smruti really nice videos....

[@fahmidaakterdina875]

was expecting more mathematical problems

[@himanshuful]

HPF leads LPF by 90 degrees.

[@kaursingh637]

REASON FOR PHASE -- WHY MINUS TAN THETA ?

[@hadleymanmusic]

Oh my I found it

[@pracheerdeka6737]

I DONT UNDERSTAND ... HOW A LOW FREQUENCY IS CUT OFF?

[@circuitsanalytica4348]

Hello friend, can you be more specific, it didn't get you?

[@pracheerdeka6737]

@@circuitsanalytica4348 yes i dont get it. A capasitors will just increase the volts and it will produce bass and treble at high db.

[@robitops1547]

is a diode is called bufer, why?

[@circuitsanalytica4348]

A diode is not a buffer, an emitter follower is called as a buffer....

[@dakshveersinghchauhan6698]

what is Omega(c)

[@typicalidealist9138]

omega is (2 x pi x frequency)

[@jeffsam5495]

if its 20db per decade, it should be 7.07/20= 0.535v isn it???

[@ALLABOUTELECTRONICS]

It's a logarithmic scale. Please check my video on Decibels. Your doubt will be get cleared. kzbin.info/www/bejne/qpKUpIiKnq-Bobs

[@circuitsanalytica4348]

Dear Jeff, 20db per decade means change in gain is 20db in every decade. One decade, means multiples of 10s. ...

[@HR-ke1hv]

Vout=(R/sqrt(R^2+Xc^2))*Vin........ye Hona chahiye bhai....tune galat likha hai

[@cgi7543]

Bro didn't understand lot of things u did it by just explaining the derivation

[@douglawson8937]

you pronounce that word, "a ten you ate" attenuate...

[@bishalghoshb3412]

Thank🙏💕🙏💕🙏💕🙏💕🙏💕🙏💕

[@rajkishan1799]

Great...

[@ramc5988]

,👌

[@h.r.pmotivation9350]

hol