The timestamps for the different topics covered in the video is given below: 0:33 Frequency Response of the Op-Amp 1:25 Role of Internal Compensation Capacitor in the Frequency Response of the Op-amp 2:58 Gain Bandwidth Product of Op-Amp 5:40 Gain Bandwidth Product of Non-Inverting and Inverting Op-Amp
Hi from Vermont! I like this style of presentation where it's practical, graphical with good narration. Also you finished thoughts at about the right point for me. I found it helpful to pause periodically and write it down for myself and contemplate a bit. Thanks!
BW is decreases when we use multiple op amp cascaded and you are saying that we can use high Band width product or multiple op amp to increase BW. It is decreased to 64KHz from 100KHz
01:31 Op-Amp has a low cutoff frequency and internally compensated with composition capacitors. 02:44 Op-amps are internally compensated to ensure stability at high frequencies and have a low open-loop bandwidth. 03:56 The gain bandwidth product of an op-amp determines the cut-off frequency in a closed-loop configuration. 05:06 The gain bandwidth product of the op-amp is 10^6 Hz. 06:13 The cutoff frequency of an op-amp depends on its gain. When the gain is high, the cutoff frequency is equal to the unity gain frequency. But when the gain is low, the cutoff frequency differs between inverting and non-inverting configurations. 07:31 The cutoff frequency of an op-amp configuration is equal to the unity gain frequency divided by the closed loop gain. 08:42 Using multiple stages of identical op-amps increases the bandwidth while maintaining the same gain.
Congrats for your work! During this video presentation I think you can differentiate Gain (notated with A[dB]) from Amplification (Vout/Vin). I think you mix them using the same "A" notation for both of them, as depicted at 4:24 (for example). Good luck!
Ur explanation is like brahmin reading slokas!! U gave a breathless speech ! It's awesome and knowledgeable but y such hurry!! Please kindly Give us some time to understand the lecture.
I was a bit confused at first too. The aim is to get a gain of 100. Since the GBP is 1 MHz this results in a bandwidth of 10 kHz if one stage is used. However if you use two stages each with a gain of 10 you achieve the gain of 100 and a BW of 64 kHz according to the formula.
An amplifier is required to have a voltage gain of +70 dB over the frequency range 0 to 8 kHz. Design a suitable circuit employing operational amplifiers (op amps) that have an individual gain-bandwidth product (GBP) of 106. Draw the circuit diagram and indicate suitable values for the resistors used please help!!
Let's say you are using a single op-amp which has a gain-bandwidth product of 1 MHz. And you want to attain the gain of 100. In that case, the maximum frequency of the signal should be less than 10 KHz. (1MHz/100). So, effectively you can not amplify the signal which has frequency more than 10 KHz. On the other hand, if you amplify the signal in two stages, using two op-amps then you can achieve the same gain of 100 through 2 stages. At the same time you can also increase the effective bandwidth of the two stages. (i.e 64 KHz)
Is gain bandwidth product is only applicable for op-amp consisting capacitors either in feedback or input ckt because in simple ckt there is no one components whose parameters depends upon frequency.
I do not understand how putting two op amps in series will increase the bandwidth The value of the fc is 100kHz and putting them together makes it 64kHz.. i do not understand how it increased? Sorry am new to this Referring to the last part
Let's say you are using a single op-amp which has a gain-bandwidth product of 1 MHz. And you want to attain the gain of 100. In that case, the maximum frequency of the signal should be less than 10 KHz. (1MHz/100). So, effectively you can not amplify the signal which has frequency more than 10 KHz. On the other hand, if you amplify the signal in two stages, using two op-amps then you can achieve the same gain of 100 through 2 stages. At the same time you can also increase the effective bandwidth of the two stages. (i.e 64 KHz) I hope it will clear your doubt. If you still have any doubt then do let me know here.
I am not educated with electronic . I am looking for low gain amplifier for my 5.1 decoder. For soft sound . I hear tda or TPA series but don't like them due to sharp sound . So I looking for bi- amp development my self. According I am wish 50 watts for tweeter and 80 watts for woofer. How can I achieve my goal ??? Any alternative teqnic to solve my problem.
I am not educated with electronic . I am looking for low gain amplifier for my 5.1 decoder. For soft sound . I hear tda or TPA series but don't like them due to sharp sound . So I looking for bi- amp development my self. According I am wish 50 watts for tweeter and 80 watts for woofer. How can I achieve my goal ??? Issue how to collaborate two different amp with diffrant gain? It's possible to develop amps as per requirement with both gain will be same below 20 dB is this possible ????
Sir at 4:24 we draw the straight horizontal line with gain 40 drawn in the "open loop FR" to get the frequency upto which it gives constant gain of 40 in closed loop configuration right??? Thanks
Sir, understood all the topics well,but this video i m trying to get,but don't know why i m not getting it well. Sir my 1st question is ,why do the gain of an op-amp circuit only involving resistances too depend on frequency...and how exactly this depend on frequency?is it only because of some internal structures as you said?? Please help sir🙏
Yes, that's because of the op-amp internal structure. As I mentioned in the video, even without any external resistor, in the open-loop condition also the op-amp has a certain frequency response (The exact response depends on the poles and zeros of the op-amp structure). At very low frequency, it provides very high open-loop gain, and the frequency increases, the gain reduces. With external resistors,the closed-loop gain reduces. For more info, please check this article: www.allaboutelectronics.org/gain-bandwidth-product-of-the-op-amp/
If we try to get the gain of 100 using single op-amp, then cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, overall cut-off frequency of the overall circuit ( two pair of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thank you sir, i got it In that timestamp frame, there is a typo mistake as 100kHz instead of 10kHz Then this will justify 64>10kHz
in the end of the lecture the cut off frequency is reduced from 100khz to 64 khz .So actually the bandwidth is reduced.But u said that it will increase. How?
If we try to get the gain of 100 using a single op-amp, then the cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, the overall cut-off frequency of the overall circuit ( two pairs of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
Sir what is meant by constant gain. i mean you have said that till now we were using constant gain and gain is constant till a certain band of frequency and after that frequency it becomes 0. Can you please elaborate it sir?
The gain of the op-amp is a function of frequency and the frequency response is very similar to a low pass filter. So, up to certain frequency, the gain is constant but after that it reduces and at one frequency it will become 0dB (unity gain). For ideal op-amp, the gain should be constant and bandwidth of the op-amp should also be infinite. So, if you consider ideal op-amp, then it has a very gain for all frequencies in the open loop condition. But for actual op-amp in open loop condition, the gain is constant only up to few Hz, let's 100 Hz or so. But when it is used in closed loop condition then that frequency will increase (depending upon the closed loop gain, as the gain-bandwidth product is constant) I hope it will clear your doubt.
That Op-Amp itself has a frequency response. Ideally an Op-amp should have infinite bandwidth, but this is not the case with practical ones, so an Op-Amp is not an ALL Pass Filter, but a wide band LPF. Therefore, it has a frequency response mainly due to the BJTs used to manufacture that IC.
Here, the gain is in decibel. So, if you convert it into the normal gain it will be equal to 10. A simple formula to represent gain dB is 20 log (Gain) . So, if your gain is 10, then in dB it will be equal to 20*1. If the gain is 100, then in dB it will equal to 20*2=40, and so on. For more information, you check my video on decibels. ruclips.net/video/ta1sUTiJNkY/видео.html
@@ALLABOUTELECTRONICS at 9:25 by cascading two op amps the bandwidth is reduced from 100kHz to 64 kHz whereas you said that we cascade to increase the bandwidth. Please explain ??
Yes, got it. I have already put some notes on the website for some opamp topics so that one can refer it if find any difficulty while watching the video. You will find the link on the channel page.
Its because of the internal compensation capacitor. It adds one pole in the transfer function. Because of that, it reduces at -20 dB / dec. Its similar to the first order low pass filter, where in the frequency response, the gain starts reducing at the rate of -20 dB/ dec after -3dB frequency. I hope it will clear your doubt.
Voltage gain in dB = 20 log (Av) So, here the gain dB is 40 dB. That means log (Av) = 2, or Av = 10^2 = 100. For more information of dB (decibels) please check this video: ruclips.net/video/ta1sUTiJNkY/видео.html
It's great. However, a little bit too fast. Electronics is something that I actually don't like and have to study. So your preciseness is working like a charm to me... however a little more patience would have been better
Because according to graph, when fc - 10 Hz, the gain is 100 dB. If you convert dB to gain, then it is 10^5. Gain in (dB) = 20 log (Gain) I hope it will clear your doubt.
Gain bandwidth for the op-amp is defined for open loop gain only (In datasheets). In fact, as discussed in the video, it is the product of open loop gain and the unity gain frequency. So, in open loop configuration if you are operating at DC or low frequency then the gain of the op-amp will be equal to open loop gain (As defined in the datasheet). And as the operating frequency increases, the open loop gain reduces. (As shown in the graph in the video). I hope it will clear your doubt. If you still have any question then do let me know here.
It is fast, and I have to watch the video multiple times, but he is very concise. Everything you need for a very detailed understanding is in his videos. Sometimes you have to go back and watch previous videos. However a Glossary of all the terms and their meanings would be really helpful.
with the designed or required gain, when we need to find the maximum allowable operating frequency then this equation is useful. Gain bandwidth product of the opamp is already given in the data sheet of the opamp. Using that, and the required closed loop gain, it is possible to find fcl.
If try to get the gain of 100 using single op-amp, then cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, overall cut-off frequency of the overall circuit ( two pair of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
For the solved examples, there is a seperate channel. ALL ABOUT ELECTRONICS-QUIZ. On the channel page, you will find the link. There is a separate playlist for the op-amp. In case, if you are not able to find it, let me know here.
The timestamps for the different topics covered in the video is given below:
0:33 Frequency Response of the Op-Amp
1:25 Role of Internal Compensation Capacitor in the Frequency Response of the Op-amp
2:58 Gain Bandwidth Product of Op-Amp
5:40 Gain Bandwidth Product of Non-Inverting and Inverting Op-Amp
ALL ABOUT ELECTRONICS
You rocks.🤘
Great job.
5:55 sir did u provide a dervaition of bandwidth of inverting and non-inverting amplifier?
i finally found this indian tutorial, thank god, how was I supposed to learn in any other way
Hi from Vermont! I like this style of presentation where it's practical, graphical with good narration. Also you finished thoughts at about the right point for me. I found it helpful to pause periodically and write it down for myself and contemplate a bit. Thanks!
BW is decreases when we use multiple op amp cascaded and you are saying that we can use high Band width product or multiple op amp to increase BW.
It is decreased to 64KHz from 100KHz
yup that didnt make any sense at all.the first point of choosing a opamp with high gbp was clear but on cascading it the bandwidth reduced
@@sushanttiwari3078 Look at @8:04 and then at @9:20. Using the same Opamp BW increased from 10kHz to 64kHz. Hope this helps.
100KHz was for open loop unity gain bw, not for the closed loop. For closed loop it was 10KHz and was improved to 64KHz.
01:31 Op-Amp has a low cutoff frequency and internally compensated with composition capacitors.
02:44 Op-amps are internally compensated to ensure stability at high frequencies and have a low open-loop bandwidth.
03:56 The gain bandwidth product of an op-amp determines the cut-off frequency in a closed-loop configuration.
05:06 The gain bandwidth product of the op-amp is 10^6 Hz.
06:13 The cutoff frequency of an op-amp depends on its gain. When the gain is high, the cutoff frequency is equal to the unity gain frequency. But when the gain is low, the cutoff frequency differs between inverting and non-inverting configurations.
07:31 The cutoff frequency of an op-amp configuration is equal to the unity gain frequency divided by the closed loop gain.
08:42 Using multiple stages of identical op-amps increases the bandwidth while maintaining the same gain.
No waste of time "concise,precise and sharp to the point " excellent
At 6:12 you said that "I will provide a separate note for the derivative in the gain band width " plz provide it sir .
Is there any update on this one, @ALLABOUTELECTRONICS?
Excellent presentation. Thank you for the tip on using two stages.
It feels like this is guy is reading fastly from some book.
Whatever I got the concept that's what matters in the end. Thanks
Congrats for your work!
During this video presentation I think you can differentiate Gain (notated with A[dB]) from Amplification (Vout/Vin). I think you mix them using the same "A" notation for both of them, as depicted at 4:24 (for example). Good luck!
Ur explanation is like brahmin reading slokas!! U gave a breathless speech ! It's awesome and knowledgeable but y such hurry!! Please kindly Give us some time to understand the lecture.
Brahmins are meant read slokas 🥵 ! Typical North Indian caste proud girl 👧 shame on you
Watch at 0.5X speed🙂
Uske liye college professors hai Madam 🙂
Ryt
Please use 'Roll Off freq' .. Amplifiers don't just 'Cut Off' . The subscript of A-CL and f-CL is standard notation for 'Closed Loop'
by adding two idential opamps bandwidth is decreasing,with one opamp it is 100KhZ ,with 2 opamps its 64Khz..did i miss anything here?
I was a bit confused at first too. The aim is to get a gain of 100. Since the GBP is 1 MHz this results in a bandwidth of 10 kHz if one stage is used. However if you use two stages each with a gain of 10 you achieve the gain of 100 and a BW of 64 kHz according to the formula.
@@straydogg56 then the gbp changes? if not,then gain is not 100
Thank you for another flawless and perfect explanation on OP-AMP characteristics !
An amplifier is required to have a voltage gain of +70 dB over the frequency range 0 to
8 kHz. Design a suitable circuit employing operational amplifiers (op amps) that have an
individual gain-bandwidth product (GBP) of 106. Draw the circuit diagram and indicate
suitable values for the resistors used please help!!
@9:30 Bandwidth of op amp in reduced by cascading, it GBW product which is increased.
true
Let's say you are using a single op-amp which has a gain-bandwidth product of 1 MHz. And you want to attain the gain of 100. In that case, the maximum frequency of the signal should be less than 10 KHz. (1MHz/100). So, effectively you can not amplify the signal which has frequency more than 10 KHz.
On the other hand, if you amplify the signal in two stages, using two op-amps then you can achieve the same gain of 100 through 2 stages. At the same time you can also increase the effective bandwidth of the two stages. (i.e 64 KHz)
@@architdongre1372 its 6.4kHz and not 64kHz.
@@moiz6164 it's 64khz...go by the formula. Also, each stage now has the gain of 10, not 100 (gain for 1 stage)
isn't the bandwidth decreased when two stages of the opamp is used??
Is gain bandwidth product is only applicable for op-amp consisting capacitors either in feedback or input ckt because in simple ckt there is no one components whose parameters depends upon frequency.
I believe u made a small mistake at 9:13 .. Previously we found f_cl = 10 kHz bandwidth.. not 100 khz . Thank you for the amazing videos!
its maximum gain÷root 2 and then u measure the value its near come to 100khz
this video is way better explained than uni lecturer , good job
Fc= corner frequency , may be not cut off ?
(1:10)
Thank you for this wonderful explanation. You inspire us. I want to go buy some op-amps and build a circuit :). thanks dude.
Is there any relationship between the open loop and closed loop gain of an Op-Amp?
I do not understand how putting two op amps in series will increase the bandwidth
The value of the fc is 100kHz and putting them together makes it 64kHz.. i do not understand how it increased? Sorry am new to this
Referring to the last part
Let's say you are using a single op-amp which has a gain-bandwidth product of 1 MHz. And you want to attain the gain of 100. In that case, the maximum frequency of the signal should be less than 10 KHz. (1MHz/100). So, effectively you can not amplify the signal which has frequency more than 10 KHz.
On the other hand, if you amplify the signal in two stages, using two op-amps then you can achieve the same gain of 100 through 2 stages. At the same time you can also increase the effective bandwidth of the two stages. (i.e 64 KHz)
I hope it will clear your doubt. If you still have any doubt then do let me know here.
The cutoff frequency decreases, hence the bandwidth increases
Note that gain is increased from 10kHz to 64kHz.
@@ALLABOUTELECTRONICS thank you so much sir thank you
@@sumitabhabanerjee1338 I don't think so can you justify your answer
Why non inverting is preferable when gain is low?? Please answer
At 4:44 you said below the cutoff frequency gain is constant but by seeing the graph it's varying. What is it?
the product of gain and frequency is constant. But gain will reduce
I am not educated with electronic .
I am looking for low gain amplifier for my 5.1 decoder.
For soft sound . I hear tda or TPA series but don't like them due to sharp sound .
So I looking for bi- amp development my self.
According I am wish
50 watts for tweeter and 80 watts for woofer.
How can I achieve my goal ???
Any alternative teqnic to solve my problem.
What if f(cl) of both op amps is different then how to calculate overall cut off frequency??
please give the complete note of the derivation....which were promised by you in video.
I am not educated with electronic .
I am looking for low gain amplifier for my 5.1 decoder.
For soft sound . I hear tda or TPA series but don't like them due to sharp sound .
So I looking for bi- amp development my self.
According I am wish
50 watts for tweeter and 80 watts for woofer.
How can I achieve my goal ???
Issue how to collaborate two different amp with diffrant gain?
It's possible to develop amps as per requirement with both gain will be same below 20 dB is this possible ????
Thanks for valuable video, which clear the concepts and save the time from surfing lots of websites
Sr...
If u can then pls solve some jam questions based on that topic
will cover it soon in the daily Quiz.
What if I use two opamps with different closed loop gain.how to find overall cut off frequency?
I will cover it in the upcoming examples on the second channel.
Sir at 4:24 we draw the straight horizontal line with gain 40 drawn in the "open loop FR" to get the frequency upto which it gives constant gain of 40 in closed loop configuration right???
Thanks
Yes, correct.
@@ALLABOUTELECTRONICS thanks
Does this lecture is enough to write in examination?
yup
Very very late reply..still I passed my exam..Thank u 🤗
Sir, understood all the topics well,but this video i m trying to get,but don't know why i m not getting it well.
Sir my 1st question is ,why do the gain of an op-amp circuit only involving resistances too depend on frequency...and how exactly this depend on frequency?is it only because of some internal structures as you said?? Please help sir🙏
Yes, that's because of the op-amp internal structure. As I mentioned in the video, even without any external resistor, in the open-loop condition also the op-amp has a certain frequency response (The exact response depends on the poles and zeros of the op-amp structure). At very low frequency, it provides very high open-loop gain, and the frequency increases, the gain reduces. With external resistors,the closed-loop gain reduces.
For more info, please check this article:
www.allaboutelectronics.org/gain-bandwidth-product-of-the-op-amp/
his 10 minute video is equal to my 1 hour lecture in uni.😅
what is meaning of overall cutoff frequency?,@8:57
Please make video on transistor...as soon as possible
Where are the notes that you are gonna to provide soon as per the lecture????
in this video i learnd about omething which i ree
ted
for the any op amp gain and b.w is different but when we use 2 stage or cascade op amp then why always gain. B.W product is constant ?????
1:15 cut off frequency is given where gain reduce by 3 dB but also, from fc gain reduces by 20 dB/dec. which is true?
3dB means 70.7% of maximum gain or 1/√2 times of maximum gain
9:34 how the bandwidth is increased as fcl is 100kHz and fcl' is 64kHz?
If we try to get the gain of 100 using single op-amp, then cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, overall cut-off frequency of the overall circuit ( two pair of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thank you sir, i got it
In that timestamp frame, there is a typo mistake as 100kHz instead of 10kHz
Then this will justify 64>10kHz
God bless you
Could have mentioned something about 3dB half power point.... Could have done it better
in the end of the lecture the cut off frequency is reduced from 100khz to 64 khz .So actually the bandwidth is reduced.But u said that it will increase. How?
If we try to get the gain of 100 using a single op-amp, then the cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, the overall cut-off frequency of the overall circuit ( two pairs of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS yes.Thank you sir
Why gain is 1? according to the curve the gain should be 0
the value of y in the x axis =0
But u r saying unit voltage gain at A=0
How
The gain on the y axis is in dB
When f =10^6
A=0
Den why unity gain? @9:57
Gain is not 1 according to ur curve
It should be 1,1
The curve should be strt from 1,1
Not from 0,0
Cz its a logarithmic curve
And log0= doesn't exist
Nice
Make a video on feedback amplifier, differential amplifier using bjt soon sir
It will be covered soon.
Sir what is meant by constant gain. i mean you have said that till now we were using constant gain and gain is constant till a certain band of frequency and after that frequency it becomes 0. Can you please elaborate it sir?
The gain of the op-amp is a function of frequency and the frequency response is very similar to a low pass filter.
So, up to certain frequency, the gain is constant but after that it reduces and at one frequency it will become 0dB (unity gain). For ideal op-amp, the gain should be constant and bandwidth of the op-amp should also be infinite. So, if you consider ideal op-amp, then it has a very gain for all frequencies in the open loop condition. But for actual op-amp in open loop condition, the gain is constant only up to few Hz, let's 100 Hz or so. But when it is used in closed loop condition then that frequency will increase (depending upon the closed loop gain, as the gain-bandwidth product is constant)
I hope it will clear your doubt.
Very well Narration
Nice video...but the Gain Vs Freq graph..is wrong
Deravation for non-inverting and inverting Op-amp ❓
why does an inverting opamp with a purely resistive arrangement have a frequency response?
That Op-Amp itself has a frequency response. Ideally an Op-amp should have infinite bandwidth, but this is not the case with practical ones, so an Op-Amp is not an ALL Pass Filter, but a wide band LPF. Therefore, it has a frequency response mainly due to the BJTs used to manufacture that IC.
At 3:48 in multiplying Gain and frequency how did we get 10×10^5 ? And how at 4:26 40db corresponds to 100?
Here, the gain is in decibel. So, if you convert it into the normal gain it will be equal to 10.
A simple formula to represent gain dB is 20 log (Gain) . So, if your gain is 10, then in dB it will be equal to 20*1.
If the gain is 100, then in dB it will equal to 20*2=40, and so on.
For more information, you check my video on decibels.
ruclips.net/video/ta1sUTiJNkY/видео.html
@@ALLABOUTELECTRONICS at 9:25 by cascading two op amps the bandwidth is reduced from 100kHz to 64 kHz whereas you said that we cascade to increase the bandwidth. Please explain ??
Where is the separate note for the derivative ?
A Superb video
Thank u sir for the video
Where is the derivation for inverting Op- Amp GBW?
awesome thank you!
Bro Can you please make video on everything on zigbee. Needed.
You are absolutely amazing
how did you get 10 to 5 power???
Antilog, 20logA=_ db
How come bandwidth is 10Khz or 64 Khz??It should be Fu-Fc. Right??
This cut-off frequency is found from the gain-bandwidth product. Because the gain -bandwidth for the op-amp is constant.
@@ALLABOUTELECTRONICS Thanks I got
Respected sir.Instead of speaking continuously,write something that you speak .. So that everyone will understand what u are speaking. Thank u ❤️
Yes, got it. I have already put some notes on the website for some opamp topics so that one can refer it if find any difficulty while watching the video.
You will find the link on the channel page.
why exactly at -20dB it will decrease or increase??
Its because of the internal compensation capacitor. It adds one pole in the transfer function. Because of that, it reduces at -20 dB / dec. Its similar to the first order low pass filter, where in the frequency response, the gain starts reducing at the rate of -20 dB/ dec after -3dB frequency.
I hope it will clear your doubt.
sir will you please explain in the example how does he get 10 power 5
100 = 20log10^5
I can tell this video would answer my questions if the writing was clearer and the accident was a lot less.
At 4:14 , I didnt understand how 40 dB correspond to 100. Can someone explain please.
Voltage gain in dB = 20 log (Av)
So, here the gain dB is 40 dB.
That means log (Av) = 2, or Av = 10^2 = 100.
For more information of dB (decibels) please check this video: ruclips.net/video/ta1sUTiJNkY/видео.html
@@ALLABOUTELECTRONICS OK. Thanks for explaining.
Nice video sir
It's great. However, a little bit too fast. Electronics is something that I actually don't like and have to study. So your preciseness is working like a charm to me... however a little more patience would have been better
Yes, I have considered that suggestion and slowed down a little on the new videos. Probably you can watch these old videos at 0.75 x speed.
ARE bhai gain either 0 hoga ya kuch aur hoga unity kaise hoga
Frequency response of opamp is similar to response of LPF
How to reduce a bandwith? Any idea?
I didn't get your question. But if you increase the gain of the op-amp, then bandwidth of the op-amp will reduce.
Plz remove subtitle it is covering a lot of space
You can turn off the subtitles in the setting.
Notes please .
can u send link for notes
What?? U point to fc =10 and then say that gain =10^5... How?
Because according to graph, when fc - 10 Hz, the gain is 100 dB. If you convert dB to gain, then it is 10^5.
Gain in (dB) = 20 log (Gain)
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS thank you for your kind response. I get it now.
UR GOING TOO FAST....pls if u have time record this session again.tnx
what will be the next topic of opamp and when will be avaliable?
It will be on the slew rate of an op-amp. I will try to post it as soon as I can.
ok sir....
Unable to understand wt u r saying keep going slow
Watch the video at x0.75 speed bro.....
what about open loop gain sir?
Gain bandwidth for the op-amp is defined for open loop gain only (In datasheets). In fact, as discussed in the video, it is the product of open loop gain and the unity gain frequency. So, in open loop configuration if you are operating at DC or low frequency then the gain of the op-amp will be equal to open loop gain (As defined in the datasheet). And as the operating frequency increases, the open loop gain reduces. (As shown in the graph in the video).
I hope it will clear your doubt. If you still have any question then do let me know here.
ALL ABOUT ELECTRONICS yes sir now it is all clear to me.....thank you sir
U r too much fast & is very difficult to understand,,, please go slow
We r not as meritorious like u
i am runnin dis video in 1.5x speed!!!!! lol
@Ayon ghosh u r right
You r slow bro
It is fast, and I have to watch the video multiple times, but he is very concise. Everything you need for a very detailed understanding is in his videos. Sometimes you have to go back and watch previous videos. However a Glossary of all the terms and their meanings would be really helpful.
How to find bandwidth?
You mean the bandwidth of op-amp or the bandwidth in general ?
Bandwidth in general
I mean when we use that fcAcl=fu
with the designed or required gain, when we need to find the maximum allowable operating frequency then this equation is useful. Gain bandwidth product of the opamp is already given in the data sheet of the opamp. Using that, and the required closed loop gain, it is possible to find fcl.
Longest intro ever. Nice content btw
Nice one bro👍
100 kHz to 64 kHz how it is an increase
If try to get the gain of 100 using single op-amp, then cut-off frequency is 1Mhz/100 = 10 kHz. While using two stages with each having gain of 10, overall cut-off frequency of the overall circuit ( two pair of opamp) is 64 kHz. So, with this configuration, it can be operated till 64 kHz instead of 10 kHz. I hope it will clear your doubt.
Thank you very much
It is like you are reading from any where blindly....
Can you go slowly and give us some time to understand
very fast .nothing got cleared.
plz share your ppts
UR GOING TOO FAST
Watch the video at x0.75 speed bro..
neso academy jaisa nhi h
Define gain bandwidth in 2-3 lines simply anyone ?
PLEASE INCREASE NUMBER OF EXAMPLES SOLVED ONE
For the solved examples, there is a seperate channel. ALL ABOUT ELECTRONICS-QUIZ.
On the channel page, you will find the link.
There is a separate playlist for the op-amp.
In case, if you are not able to find it, let me know here.
i could not understand the concept
You may go through the notes which is provided on the website. It will help you.
www.allaboutelectronics.org/gain-bandwidth-product-of-the-op-amp/
3dB
Can you please.... Go slowly...
It is hard to catch your voices
ye lecture me kuch samajh ni aya
Content good but he speaks fast