An inverting loop made so by the mosfet inverting to positive input, your are BEGGING for oscillation, no matter the mosfet C!!!!!!!! ....SERIOSULY!!!!!!!!!!!!!!

Wow, the 100k will be dissipating 6.4Watts when the voltage is set to 0V output!

No over temp (high risk in this), no over current or short circuit protection. You might find what useful what a KZfaq channel "FesZ Electronics" did. :)

Omg you are a madman

Well, it is now 3 years later... sadly there is no more video about your work on the amp. So i save my time and do not have a look at the 2nd part.

I made a similar circuit, used Fallstad simulator, i adjusted the circuit to deliver 1 amp @ 1around 9 V.p.p. swing from a 2 x 15 V transformer, so 2 x 19.6 V. Using BD137, BD140, it follows the input voltage perfectly, pcb made, on it's way. wonder how it sounds, also included 2 pole Sallen Key filter. How much watts would that be would your recon? P = U x I, states like 19 ,6 watts, right or wrong?

thank you very much

Magnifico, grandioso !! ⭐⭐⭐⭐⭐

This is excellent! Where is the rest of the videos?

Op amp drive huh🙃

just found this, its v good. any follow up?

Very well executed and thoroughly tested with great effort, I am very impressed! I am just wondering why the op-amps are functioning correctly, as they are controlled by two potentials: 15V from the power supply and at the sensor input from the output potential 0-150V. I did not know that this was possible at all. Or is it a special function of this type of op-amp? I assume that both potentials are referenced to ground, is that correct?

Hi.a question what is difference between video and audio amplifier pls answer me if you know

I wanna have your bench 😢 goddamn with your set up. At my pace , it would take me prob 10years to have those goodies. Thank you for the vid, I will copy it and test a few things. I can't believe I understand everything, I'm amazed at myself.

The output capacitor should be at least 10 times larger than the input capacitor because the current through the output stage will be at least 10 times greater. This is the most basic LAW of designing a circuit.

What a load of RUBBISH. The output of the PNP emitter-follower can never have a swing of 10v because it simply follows the capability of the collector of the first transistor ( plus and minus the developed voltage across the 100n). It is no wonder Indian students don't have a clue about designing circuits when the teacher does not have a clue. For a start, you don't use 100n on the input and 100n on the output because the output has a much lower impedance than the input and the capacitor has to reflect this. The output capacitor should be 10u. The minimum output voltage will be the voltage across RE1, plus 0.3v plus 0.7v for transistor 2, minus the developed voltage across the output 100n capacitor. You can't work anything out until you have the value of the output load resistor. See my website as I have been teaching electronics to 25,000,000 visitors for 50 years at talkingelectronics.com

Where do you get 2k headphones from? Have you been living in a cave?

Very nice. Thanks.

I have a question when it comes to the inductance value of the load. What if you're designing a general purpose audio amplifier rather than pairing it to a specific driver of known parameters? Y'know, what a lot of car audio amplifiers are, what a lot of unpowered bookshelf speaker amplifiers are, what a lot of the DIY enclosures use, etc. Is there a good inductance value to use for this? Do you just pick a generalized range for inductance, such as if I'm looking at an amp for car audio subwoofers where the driver Le (the notation for inductance in T/S parameters) is commonly around 6.5mH at the high end with a handful of outliers being within the 8.5+mH range, just from glancing at some databases, would it be fair to just use Zl=Rl*1e-2 and just have the extra overhead for versatility? In the example of anything but car audio subwoofers, I know drivers are typically <<1mH, again from glancing at the same database, so would a general purpose amp just use Zl=Rl*1e-3 for the typical upper end with ample overhead?

Great job!

Out of my own curiosity, regarding Ideal #61-484, when you are in diode mode, and there is a short, does it make a beep noise like Fluke 87V? I am talking about when it is in diode mode. NOT in Continuity Mode with noise. Thanks.

If you make the ratio of the collector resistor to the emitter resistor 4 to 1 you will have a gain of 4. If you make the collector resistor 10k you will get a current of 1 mA. But. The emitter resistor will have to be 2k5 and the collector current will be less than 1mA. So you have to reduce the collector resistance.

Here is what the instructor should say. We have to know how powerful the input signal is. Suppose it is 2 volts with a current capability of one tenth of a milliamp.

Why does the base-bias bleed have to be much higher than the base current? All you are doing is attenuating the input signal. None of this is described or mentioned and the students are just sitting in a daze.

None of the students have a clue

Why would you add an emitter follower stage to increase the current from 1 mA to 2mA ??? I would climb out of this class as fast as possible.

The whole idea of an amplifier is to increase the current and the impedance of the circuit decreases as you move from input to output. It is total ignorance to specify. 100n at the input and 100n at the output.. You are not teaching your students anything. Just a very poor instructor who needs paperwork to back his presentation.

What he should say is this: The output impedance of the headphones is say 16 ohms and the gain of the emitter follower circuit is say 100. This mean the first stage sees the headphones as 1600 ohms.

I don’t like the way he says the first stage see a much reduced load due to the emitter follower stage.

Where did you dream up the impedance of any type of modern headphone is 2k ?????? It might be from 8R to 32R

Your dome is pushed-in. Not serious about the business.

Are you ever going to finish the 25W audio amp design? I has been 4 years since this was started and 3 years since you said you would restart/complete this series. Really appreciate this series but just looks like we'll never see it completed... 😒

which sensor you have used for feedback ?

nice!!!!

EXCELENTE VIDEO

I wish electronics was taught like this in college. Info is easier to absorb because of an objective which is to design.

no part 3? 😢

It's beautiful

Man this is unbelievable, thanks for the video

Hİ, which are you using books ?

Cool idea! Did you try it with only local feedback for the op amp?

1:40 This circuit has quite a few design flaws. The input stage has a very low bias current of abt. 1.7mA, leading to low transconductance and current starving of the voltage amplifier stage at frequencies above 1kHz, which in turn causes poor slew rate and distortion (mostly 2nd harmonic). It's also hugely disbalanced, because one transistor draws abt. 0.46mA (0.7V/1.5k), while the other one draws 1.7-0.46 = 1.24mA. The whole point of a differential pair is to be perfectly balanced. Deviations in transistor currents over 1% cause a significant increase in distortion (2nd and 3rd harmonic). The best solution here would be to use a current mirror - the increase in complexity and cost is negligible compared to the great improvement it makes. It's also beneficial to increase the bias current to abt. 6mA and add emitter degeneration resistors of abt. 33 Ohm to the differential pair. The dominant pole capacitor (220p) is too big. The whole point of the dominant pole compensation is to bring the open-loop gain down to unity just before the phase shift at the output approaches 180 degrees, thus turning an amplifier into an oscillator. The value of Cdom is determined by the current sourcing/sinking capabilities of the transconductance stage and the VAS collector impedance. The open-loop gain here is not that high to begin with, due to the aforementioned low transconductance and current starving of VAS. All of this leads to a poor slew rate and great increase in distortion at higher frequencies. The VAS collector is loaded by a current sink, which is correct, but there is never a need to adjust the exact current flowing through it, so the 10k potentiometer is useless. The 2.24k pot is placed between the Vbe multiplier's base and collector, which is a timebomb, because if it fails, the bias voltage will increase to a maximum, likely causing the output transistors to blow up. It's always a better idea to put it between the base and the emitter. The 3K resistor between the driver transistors' emitters is too big. Reduce it by a factor of ten and put a 0.1-1uF capacitor in parallel. Its purpose is to provide a small reverse-bias voltage to the output transistors, making them turn off faster. Doesn't hugely matter in this relatively low-power circuit. It's always a good idea to couple the lower leg resistor of the feedback circuit (10K) to ground through a large (470uF+) capacitor, thus bringing the DC gain down to unity. The input pair is never going to be perfectly matched, and you don't need to amplify the DC offset voltage. The 220K will then have to be reduced to the same value as the input bias resistor (10K in this case) to match the impedances. It's always better to keep the value of these resistors as low as practically possible to reduce the Johnson noise (the exact values will ultimately depend of the signal source's output impedance). It's also necessary to put a small (4.7-33pF) capacitor in parallel with the upper leg resistor of the feedback circuit. This circuit is likely to work fine without it because its dominant pole frequency and slew rate are very low and therefore the Nyquist stability is not compromised - but the noise and distortion figures most definitely are. That's all. :)

MAn i have been lookin at this .Symbolic amplifier design..There is only one book i have not been abel to get >and they use mathematica ..You solved all my needs

WHAT U THINK "CASECODE+DUAL POLE COMPENSATION 🤔 I THINK ITS POSSIBLE TO DO THAT, IS IT?🤔

Outdated croc of shitt😮

Sir please explain the circuit , make a detail video on this circuit

High order harming is way too high, their bad can be described as n^2/n, not good performance on such design.

Great video thank you 👍👍🇩🇿🇩🇿

Very creative use of power rails, lol. I’m surprised it even works as well as it did - there must be all sorts of non-linearities between the signal and power line circuits in an op amp. Kudos for creativity.

Alvrg soy mexican. En estados unidos están a otro nivel. CHido el amplificador.