What's All This Input Impedance Stuff, Anyhow?

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Um, CJ, there's also stuff like phase angle involved, as well as resistance and reactance. It can be a pretty deep subject. Ask an RF circuit designer! I own a book that's entirely about the subject of impedance. Logically enough, its title is Impedance :grin:

But granted, most of the time in audio, we're thinking of it in simple resistive terms.
 
DC theory
AC theory
L C R circuits

must think of them all at once
to view things in isolation can lead to wrong assumptions

eventhough things can get complex it is cool to chase the dominant factor through a given circuit but never forget that there can be a hidden special case ... like resonance ... these things can set a circuit off and not necessarily in the range that you are trying to work in


:roll: RF ... such a black art
 
I've done a modest amount of RF design and yes, it does feel like a "black art" sometimes. It's a much different frame of mind. Layout is critical. Trying to create a complete design on paper or using a simulator is laughable when you get into the MHz region and above. Well, at least in my opinion.

Show of hands: how many of you know how to use a Smith chart? :wink:

smithchart-thumb.png
 
Pease's fable answers the false theoretical objection to the one opamp difference amplifier. Adding identical impedances to both inputs causes only gain error, not CMRR error.

It does not change the fact that in the real world, where balanced lines are never exactly balanced, it is not half as good as the two opamp differential amplifier.

This isn't new news (even though it keeps getting rediscovered). The old DBX and Valley People boxes used good 2-amp inputs.
 
I have a question about the third opamp in the diagram, A3. I sort of understand it, but nowhere near to the point of being able to explain it-

Figure_01.jpg


with R1+R2 in series to ground after the positive input, this reflects a load the amplifier must drive, I get this.

On the negative side though, with R2 in the feedback loop, even though this is feedback, why does this not present some kind of load in series with R1 as seen by the output of A1?

This is very basic, but it reflects my sophistication with theory and application. Is there an easy way of explaining this?

thanks

dave
 
[quote author="soundguy"] Is there an easy way of explaining this?
[/quote]
I don't know if I can offer you an easy way - but it might help to remember one of Horowitz and Hill's 'golden rules' of op-amps when trying to figure out what A3 is doing - and what PRR is intimating.

  • Rule 1: The output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
So, the output terminal swings to get the feedback network (the upper R2) to bring the input differential to zero (give or take offset errors!). And the formula that would apply to A3 in this configuration, assuming that input voltages V1 and V2 were applied to the ends of both R1's, would be
  • Vout = R2/R1 (V2 - V1)
Now, I don't know if that's any more help, but sometimes a little extra information, and the odd key point can aid one's understanding.
 
This one was mentioned in PRR's original preamp design thread: http://www.thatcorp.com/1200desc.html
This is the same thing that all you smart people (and I sincerely mean that, as on this front I'm a blithering idiot) are speaking of, is it not? It looks like the description of the Valley People/dbx solutions, but I could be wrong.

Jeez, you never know who to trust in these web articles.

"Now I don't believe in nothing no more. I'm going to law school."
-Jimbo Jones

Bear
 
Oh yeah,
Isn't that Smith chart showing what Stephen Hawking was talking about with the point of infinate gravity in the black holes? :grin:

Joel
 
> This one was mentioned in PRR's original preamp design thread: http://www.thatcorp.com/1200desc.html

I didn't mention that.

It is however a good implementation of the circuit Pease is discussing (plus a frill due to Whitlock, not mentioned in Pease).

If you take off the input buffers and CM driver, just one opamp and four resistors, this is THE most common "balanced input" in today's world.

> It looks like the description of the Valley People/dbx solutions, but I could be wrong.

No. One design used widely, including my DBX 160 and several Valley boxes, is two opamps. One is a simple inverter. The other is a simple 2-input summing amp. The summer gets one input direct from the jack, and the other input is the inverted signal from the other side of the jack. Impedances are balanced every which way (not so for the 1-amp "diff" input). Both forms have the objection of large resistors and hence significant noise, but generally not a problem at line level. The resistors do protect things from abuse, always an important if non-theoretical consideration in the real world.

No, I'm objecting to Pease's apparent vindication of the one-opamp diff-input. He writes third-person, so maybe he is trying to stimulate our thoughts instead of simply being Pease's Word. The fable's conclusion is correct: IF you buffer both inputs, the one-opamp diff-input (total three opamps) is essentially perfect. But as DBX et al show, there is a two-opamp circuit just as good, some ways better. (Interesting that Pease works for a chip-maker....) And in any box under $999, what you usually get is the one-opamp form, without buffers. It is easy to show "ideal" results... if you drive it with ideal test signals. But in many real situations (as simple as just taking an unbalanced output into the "balanced" input) the CMMR really goes away (though not as bad as a plain unbalanced input).

The Smith Chart is fun, but I couldn't find my butt on it with a trail of breadcrumbs, and I can't think of any audio reason I would want to.
 
Hi!

Isn't this 3-opamp design referred to as the 'instrumentation' amp? As implemented on Burr-Brown chips such as the INA134?

****EDIT****

I just realised the INA134/137 chips are just differential line receivers and not in a 3 amp configuration.

The Whitlock modification is the bootstrapping to increase common-mode input impedance among other things....its covered in this article.

http://www.jensentransformers.com/an/ingenaes.pdf

Tom
 
How difficult is it to make a variable-impedance input? Can it be done with pots or would you have to do it hard-wired steps or relays? And what things would you need to change... for both transformer and electronic type inputs?
 
[quote author="PRR"]The Smith Chart is fun, but I couldn't find my butt on it with a trail of breadcrumbs, and I can't think of any audio reason I would want to.[/quote]

That was kinda my point. RF involves the use of some design approaches and tools that are completely alien to AF designers.

The best, most useful book I've seen on the subject is the remarkably slim but info-packed RF Circuit Design by Chris Bowick.
 
Dave,

Short answer is, the inverting input looks like ground as far as the input signal is concerned. It doesn't see the 100K feedback resistor.

The long answer has already been covered...
 
[quote author="cjenrick"]To me, impedance is just a shortcut way to say volts divided by current. With frequency thrown in sometimes for good measure.[/quote] I'm picking on CJ here (sorry dude, I totally respect your trafo work :thumb:), but impedance always involves frequency...and phase. I love it when people tell me that they measured the impedance of a speaker with a VOM...No, you didn't! Not without a signal generator, anyway! :grin:

Funny that Pease brings this subject to light just now. Our Sr. Engineer has been schooling me of late on this very subject of the difference in the impedance of the + and - inputs. I hope I am taking enough notes. I am fortunate enough to have this forum to generate topics that I can discuss with my own "personal PRR" so to speak. And I hope I encourage others when I say that I already have a BSEE and 10 yrs experience in audio R&D...but I still don't know everything there is to know!

[quote author="TomWaterman"]Isn't this 3-opamp design referred to as the 'instrumentation' amp?[/quote] Not exactly, see figure 6 of the article you reference for the "typical" IA config. Though they both may work the same in practice.

[quote author="soundguy"]no, Im asking why the 100K feedback resistor doesnt translate as a load.[/quote] Maybe an easier way to think of this is that the o/p and the - input are both low impedance points??

In regards to the Smith Chart...Dude, nice Spirograph! :razz:

I think this calls for head down
Peace!
Charlie
 
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