[MatthewK]

The AD8397 has a buffer built-in to it. So what would happen if a BUF634P was added?

Theories?

 

[berat23]

you can get very low output impedance with this combo

 

[amb]

It'd be a waste of parts. AD8397 already has plenty of output current capability (just as much as BUF634) and its output impedance is even lower than BUF634.

 

[MatthewK]

The Denon D2000 is 25 ohms with an input rating of 1,800mW, so that's why I ask.

Is there a single-channel version of the AD8397?? Unfortunately I don't see one.

 

[majkel]

AD8397 is a dual op-amp 310mA peak of output, and it's cheap. BUF634 is a single 250mA buffer and is expensive. With the AD8397 you save money and space. With the BUF634 you save the battery because of it's low supply current. BTW, you can wire the AD8397 to use it for one channel to double its output current.

 

[MatthewK]

Oh, that sounds interesting. So I could get two AD8397's and wire them to be used as single-channel? How would I do something like that? I'm not very experienced with soldering, but I think I could do a simple Brown Dog adapter SOIC-to-8DIP. Beyond that I'm not sure..

Thanks for the help BTW.

 

[majkel]

One AD8397 = 2x BUF634 roughly. When you wire one AD8397 like this - you use one per channel with 620mA peak.

To get 310mA per channel - one AD8397 is enough for both.

 

[Heady]

I don't want to be a spoiler but the AD8397 is a real stinker of an opamp to work with and I don't think using it this way will work. In fact, I doubt it will work in a browndog SOIC-DIP adaptor either.

Read about the AD8397 in the PINT and Mini3 threads.

 

[NelsonVandal]

I've used it on BrownDogs without problems. There was no sign of oscillation and the sound was just the same as in Mini3. It's easy to solder decoupling caps on the BrownDog. If you don't use the extra resistor inside the feedback loop (R5 in the Mini3 schematic), you can solder an 805 SMD resistor directly on the opamps pins. If that's too dary you can solder a 1206 or 805 on the BrownDog. Then you'll have short feedback loops and caps in very close proximity to the opamps pins. Why shouldn't it be stable?

Why on earth use dual AD8397? 30 mA of Current is enough for most situations, why > 600 mA.

 

[majkel]

Never a single problem with the AD8397 but I agree, it's nothing spectacular in terms of sound.

 

[jcx]

its not practical to parallel dual op amps in a way that could simply be "rolled" into just any other single op amp feedback circuit

the AD8397 isn't very suitable to the A47 style paralleling (which is shown incorrectly above anyway)

The AD8397 doesn't have "rail-to-rail" input common mode range so the follower op amp won't work well for output V swing near the supply rails - you would be wasting the main feature of the AD8397 its ability to swing output V "rail-to-rail" with high output current

I sketched out a sim of a possible way to parallel AD8397:
http://www.head-fi.org/forums/f6/ad8...ml#post2301447

but no one has built it so it really can't be recommended to the inexperienced

 

[majkel]

It's shown very correctly, don't treat the A47 like encyclopedia. I wouldn't worry about the proximity to the rails either. Who needs to be so close to this?
The final word is that almost no modern op-amp sounds better when biased in class A. The only exception I know is AD825 but it's a slight change, not necessarily better sounding, just different, and maybe even further from my preferences. The explanation of this fact is simple - most good op-amps have got a diamond buffer on the output pin.

 

[jcx]

majkel,

your sketch is missing the "current sense" resistor so the unity gain , no resistor, follower will hog all of the current and no V will be developed across the 10 Ohm R so it won't add any current to the output until the follower hits its output current limitation - very nonlinear/high distortion operation

 

[majkel]

What "current sensing" resistor? The 10 ohm resistor is only to protect the output from high offset-equalizing currents between the two op-amp outputs. The offset "appears" on that resistor. I need no current sensing in this circuit, it works very well, and the follower is necessary only to double the current efficiency, nothing more.

EDIT: I know what you mean - I agree that the current draw might be unequal, so the resistor should be kept as low as possible. I was using the dual resistor version some time ago but this one just sounds better.

 

[jcx]

I must admit my 2nd comment about the nonlinear operation of the added current was based on a misreading of your schematic, but maybe I caught the intent

on closer inspection I now believe your sketch simply can't work to add together the op amps output currents

I think you have misdrawn your intended circuit - even allowing for it not being a "A47" style current sharing scheme:

look at the upper op amp’s noninverting input connection – as shown the output (from the 10 Ohms) and the + input are shorted with a wire, if the circuit behaves as anything recognizable it would be a floating current source of ~ v_os/10 Ohms


I think theoretical analysis would show that op amp output stage distortion with a linear, mostly resistive load like dynamic headphones would be minimized by equal current sharing - as long as the current sharing resistors can be at least several times lower resistance than the op amp OL Zout and the Load