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diegomj1973 14th January 2013 01:23 PM

Opamp challenge
 
2 Attachment(s)
Taking Scheme A with its magnitude and phase versus frequency as shown in Figure B (blue curve), I make an invitation to the community to achieve similar frequency response (magnitude and phase), as shown in Figure B (red curve). To accomplish this, one has to use the same components shown in Scheme A (referring to quantity, type and value). Not allowed to remove or add components as those shown in Scheme A (you must use the same components, to achieve the objective of this challenge). The input signal must remain AC coupled. The high frequency response must remain similar (over 10 KHz). The voltage gain in the passband should remain similar (3.92). The input impedance in the passband should remain similar as well. The output is applied over the 10K resistor. This 10K resistor must continue to fulfill the function as output load in the new proposal. Another thing is that is not allowed in the attempt to generate offset problems at the same stage.

The answer to this challenge's available to me, since I am the mentor of this exciting discovery.

With this new interconnection between components, we can achieve these and many other improvements.

Everything is in the process of being patented. That is why the new scheme can not be shown yet. It can be applied to the field of audio, among other applications.

As soon as I get the title of the invention, I'm going to explain in detail.

regards

Scheme A:

http://www.diyaudio.com/forums/attac...1&d=1358169717

Figure B:

http://www.diyaudio.com/forums/attac...1&d=1358169717

jan.didden 14th January 2013 01:28 PM

Not sure I fully understand.
Are you asking for a different way to connect the same parts for the same results? If yes, what's the advantage of it, apart from the intellectual challenge? If the same parts give the same results, whats the improvements you mention?
I assume that there is some advantage that is the subject of the patent?

Why do you attenuate the input signal?

Did you apply already for the patent? This can take several years easily.

jan

diegomj1973 14th January 2013 01:47 PM

Quote:

Originally Posted by jan.didden (Post 3325323)
Are you asking for a different way to connect the same parts for the same results?jan

No. Different way to connect the same parts for better results.

Quote:

Originally Posted by jan.didden (Post 3325323)
I assume that there is some advantage that is the subject of the patent?jan

Yes. Improvement ratio between lower cutoff frequencies is proportional to the ratio of C2 to C1 and inversely proportional to the gain of the system (+1 in this case). It's what I can say for now.

I dimmed the entrance to the circuit simply because it is an example for the challenge, although it could have been a simple non-inverting configuration.

The patent is in process. Yes, it will take some time to emerge definitively.

I have concrete evidence that this is working very well. Even now I have it implemented in a pure class A amplifier.

This new connection may give characteristics next to the directly coupled systems, but with the advantages of AC coupled systems.

Regards

jan.didden 14th January 2013 03:56 PM

Quote:

Originally Posted by diegomj1973 (Post 3325341)
No. Different way to connect the same parts for better results.
[snip]Regards

Better in what way? You ask for the same (similar) passband and phase curve. If you want the same performance with the same parts it is not much of a challenge. If it really is a challenge, in what way must it be better?

jan

diegomj1973 14th January 2013 06:08 PM

Quote:

Originally Posted by jan.didden (Post 3325475)
Better in what way? You ask for the same (similar) passband and phase curve. If you want the same performance with the same parts it is not much of a challenge. If it really is a challenge, in what way must it be better?

jan

Let's see ... What you have to get are the red curves (the main objective, in the example given).

I give an example: if, in your attempt, you exchange C1 to C2, and vice versa, leaving all other components in the original position, obviously you can not get the red curve given. The lower cutoff frequency (-3 dB) of this change is around 1.26 KHz (far from around 6.67 Hz, corresponding to the red curve).

In a similar way to the example given I suggest trying to get to the red curves.

Am I clear?

PD: One advantage is a significant increase in bandwidth (to the side of low frequencies, using for this small capacitors). There are many other advantages.

Regards

sreten 14th January 2013 06:28 PM

Hi,

This is an exercise in tedium. There is nothing you can have "discovered"
about such a simple circuit that hasn't been done before and I expect
whatever you think you've "discovered" is fundamentally flawed.

I'm not excited. I'm bored. You can't do what you say. There is
no right answer because what you think it is, is simply wrong.

The challenge is to guess what you've got wrong.
My guess in no DC path for an input.

rgds, sreten.

jan.didden 14th January 2013 06:31 PM

Ahh now I think what you want.
I probably would want to use some form of bootstrapping.
Looking forward to see your idea!

jan

Mooly 14th January 2013 06:46 PM

Does the "new" version have to be stable if the input is left open circuit ?

diegomj1973 14th January 2013 06:46 PM

Quote:

Originally Posted by sreten (Post 3325683)
Hi,

This is an exercise in tedium. There is nothing you can have "discovered"
about such a simple circuit that hasn't been done before and I expect
whatever you think you've "discovered" is fundamentally flawed.

I'm not excited. I'm bored. You can't do what you say. There is
no right answer because what you think it is, is simply wrong.

The challenge is to guess what you've got wrong.

rgds, sreten.

This is not an exercise. In fact is laboratory tested. I have also implemented a pure class A amplifier, as I said.

And the proposal is not tricky!. Believe me that's how I say it.

This does not go against the laws of physics.

The improvement ratio of the lower cutoff frequency for this example is about 40.68 times.

Just as the circuit is so simple, I invite you to get similar results, without doubt your abilities.

regards

diegomj1973 14th January 2013 07:04 PM

GoatGuy:

Is everything okay with the analysis you've done, but this analysis corresponds to the example presented (which is deductible, for its simplicity)

Again, the question is what is the configuration that achieves the red curves?. Obviously, using all the components given (in another connection).

regards


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