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building an active balanced ground?

post #1 of 128
12/26/10 at 12:41pm
Thread Starter 

I've built one Bottlehead/Speedball Crack amp and have been thinking of trying out a solid state project (maybe an M3, or a beta22 if I'm crazy, though I should probably start with a mini^3 for sanity sake). I've been reading up on AMB and Tangent, etc, and I think I'm at the point where I know just enough to be foolish/dangerous.

 

I was also reading up on Meier's concept of active balanced ground (http://www.meier-audio.homepage.t-online.de/concerto.htm , also on the symphony page) and thought to myself: "hey, I think I can do that"... famous last words.

 

From what I can tell, I would apply a -(L+R)/4 to the left, right, and ground channel. So basically, working on a three board setup like the beta22, my new inputs (denoted with *):

L* = L - (L+R)/4

R* = R - (L+R)/4

G* = - (L+R)/4

 

Does this look about right?

 

That all said, I'm not actually sure *how* to implement it. I know I can make a simple summing circuit by using a couple resistors...

http://www.tkk.fi/Misc/Electronics/circuits/linesum.html

http://www.rane.com/note109.html

 

... then taking a quarter of that output (umm, run the output off 4 equal resistors in series and tap one resistor?) to get my (L+R)/4 value. Then use another couple summing circuits to get your L* and R*. Except I don't know how to get the negatives, and I think simply flipping the +/- would be a bad idea, unless maybe I just ran resistors off all 4 wires

 

 

 

But, I know there are issues with attenuation of signal when you start splitting and combining things willy nilly. A summing circuit loses 6 dB? That's sort of a quarter right? er...

 

 

So assuming anyone was able to follow that mess, am I completely off my rocker?


Edited by Armaegis - 12/26/10 at 2:18pm
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post #2 of 128
12/26/10 at 2:09pm

You lost me after "I've built one Bottlehead..."

 

The meier link doesn't work.

 

 

It seems to me like an awful lot of work to avoid reterminating the headphones, or using AMB style zero volts active ground.

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post #3 of 128
12/26/10 at 2:28pm
Thread Starter 

Oops, I accidentally got a comma stuck in the link. It should work now.

 

Well the whole "active balanced ground" thing is sort of a halfway step between active ground and fully balanced operation. I just figure, if the architecture of the active ground in your amp is basically the same as your L/R channels, then the whole L*/R*/G* circuit can be simply be inserted in between the volume pot and the amp channels (or before, if you use a 3 channel pot?).

 

As to why? Well why not? It's expensive to go fully balanced with all your gear.

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post #4 of 128
12/26/10 at 5:49pm

It would work. It appears to me that the potential difference, as seen by the drivers, would be the same.

 

All you gotta do now is figure out how to arrive at that perfect sum of L and R without distorting or polluting either.

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post #5 of 128
12/26/10 at 6:07pm


 

Quote:
Originally Posted by Armaegis View Post

From what I can tell, I would apply a -(L+R)/4 to the left, right, and ground channel. So basically, working on a three board setup like the beta22, my new inputs (denoted with *):

L* = L - (L+R)/4

R* = R - (L+R)/4

G* = - (L+R)/4

 

Does this look about right?

 

 

Almost right.

 

G = -2 * (L+R) / 4

 

This is easy to obtain with opamps...and we know Jan Meier loves opamps.  There's nothing wrong with them, I like the sound of some of his stuff.

 

I think this will give you -(L+R)/2.  Then feed that into a voltage divider and then into the input of the opamps driving the gain stage.  Your gain stage will likely be an inverting gain stage.  The amp would be inverting, and not non-inverting.  Just a quick guess, I haven't looked at this in detail, but the theory, to me, seems interesting and somewhat logical.

 

Resistor values may not be ideal.

 

sum_div_by_2.png

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post #6 of 128
12/27/10 at 3:45pm
Thread Starter 

 

Quote:

 

Almost right.

 

G = -2 * (L+R) / 4

 

 

No need to double the G since there are two ground wires (in theory).

 

 

The first link in my OP says there is an attenuation of 6 db. Is this due to the stereo->mono conversion, or due to the resistors in line? If they are small enough resistors, would there be no (appreciable) loss in signal? Granted, I understand there still needs to be some resistance in order to not create a short between L and R.

 

I would like to stay away from opamps for now as that adds a level of complexity that I'm not really ready to tackle just yet.


 

Quote:
Originally Posted by holland View Post


 

Quote:
Originally Posted by Armaegis View Post

From what I can tell, I would apply a -(L+R)/4 to the left, right, and ground channel. So basically, working on a three board setup like the beta22, my new inputs (denoted with *):

L* = L - (L+R)/4

R* = R - (L+R)/4

G* = - (L+R)/4

 

Does this look about right?

 


This is easy to obtain with opamps...and we know Jan Meier loves opamps.  There's nothing wrong with them, I like the sound of some of his stuff.

 

I think this will give you -(L+R)/2.  Then feed that into a voltage divider and then into the input of the opamps driving the gain stage.  Your gain stage will likely be an inverting gain stage.  The amp would be inverting, and not non-inverting.  Just a quick guess, I haven't looked at this in detail, but the theory, to me, seems interesting and somewhat logical.

 

Resistor values may not be ideal.

 

sum_div_by_2.png


I stumbled upon this link... http://www.ecircuitcenter.com/Circuits/opsum/opsum.htm

 

So it looks like I can get my quarter value if my output resistor is 4 times the input resistors. Hmm, maybe it's easier with the opamps...

 

You know, I'm sure this is way more complicated in implementation. There's probably all sorts of DC offsets and irregularities that we aren't accounting for.

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post #7 of 128
12/27/10 at 4:39pm

What would be wrong with having the two inputs of your opamp be left and right, the output being the difference.

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post #8 of 128
12/27/10 at 4:58pm
Thread Starter 
Quote:
Originally Posted by digger945 View Post

What would be wrong with having the two inputs of your opamp be left and right, the output being the difference.


I'm not sure I understand what you're saying here... confused_face%281%29.gif

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post #9 of 128
12/27/10 at 6:38pm

Instead of trying to combine the L and R signals before the opamp, let the opamp itself be the summing device(for the active ground channel alone). The opamp will output the signal needed for the blue "balanced ground" line on the Meier website. Actually,  the active ground signal is inverted from the sum of the signals, thus the "Balanced" label.

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post #10 of 128
12/27/10 at 7:27pm
Quote:
Originally Posted by digger945 View Post

Instead of trying to combine the L and R signals before the opamp, let the opamp itself be the summing device(for the active ground channel alone). The opamp will output the signal needed for the blue "balanced ground" line on the Meier website. Actually,  the active ground signal is inverted from the sum of the signals, thus the "Balanced" label.



the circuit shown (minus caps) is the G channel.  The opamp circuit shown is generating     - (L + R) / 2 .... (caps not put in place to block the necessary voltages).

 

feeding L and R directly into the inputs won't function as you think, as it will be a comparator and return V- or V+ depending on which of L or R is greater at any given point in time.

 

you may be able to replicate the G channel and instead of G into the +, you feed in L or R.  You would need to replace the 2K resistors with 4K.  I'm not really sure about the results there, but it may give the corresponding L or R channels.  The resistors would need to be finely matched though.

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post #11 of 128
12/27/10 at 7:36pm


 

Quote:
Originally Posted by Armaegis View Post

 

Quote:

 

Almost right.

 

G = -2 * (L+R) / 4

 

 

No need to double the G since there are two ground wires (in theory).

 

Huh?  I'm not sure that's correct.  V1 + V2 + V3 + V4 = 0.  Sum your equations.  I don't think it works out.

 

Quote:

Originally Posted by Armaegis View Post

 

You know, I'm sure this is way more complicated in implementation. There's probably all sorts of DC offsets and irregularities that we aren't accounting for.

 

 

You need blocking caps, of course, that's a given.

 

 

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post #12 of 128
12/27/10 at 7:47pm

nevermind that was dumb.  G is 2x the "gnd" components.  There are 2 "gnds", but on the wire it's summed up, so my original post, IMO, is correct.  The G channel is 2x the factor.


Edited by holland - 12/27/10 at 8:28pm
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post #13 of 128
12/27/10 at 8:48pm
Thread Starter 

Yeah I think we're agreeing on the same thing, just having a semantics issue.

 

Something I just thought of though, Meier's equations assume that there are 4 wires running through the headphone cable. For the single sided cables though, sometimes there's only one ground cable. That would make the G* value = -(L+G)/3

 

 

Quote:
You need blocking caps, of course, that's a given.

 

Man I don't even know what those are...

 

 

Bah maybe I should just try drawing something up with resistors and let you guys tell me if it looks like it might work or not.

 

 

edit: also, do we have any problems with attenuation of the signal due to resistors in the path? Or is this only a concern when outputting directly into the headphones?


Edited by Armaegis - 12/27/10 at 8:50pm
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post #14 of 128
12/27/10 at 9:34pm
Quote:
Originally Posted by Armaegis View Post

Yeah I think we're agreeing on the same thing, just having a semantics issue.

 

Something I just thought of though, Meier's equations assume that there are 4 wires running through the headphone cable. For the single sided cables though, sometimes there's only one ground cable. That would make the G* value = -(L+G)/3

 

I hate this editor, it totally blows.  Using Meier's equations, it does call for 4 components, so G would need to be 2/4 and the other components 1/4, to sum up to 1.  Using neg(L+R)/3 doesn't meet his primary equation of (L+G factor) + (G factor) + (R + G factor) + (G factor) = 0.  You are correct it is divided by 4, but the G channel itself needs to be summed of 2 G factors.

 

Quote:
You need blocking caps, of course, that's a given.

 

Man I don't even know what those are...

 

Just caps in series to block DC from passing through.

 

Bah maybe I should just try drawing something up with resistors and let you guys tell me if it looks like it might work or not.

 

 

edit: also, do we have any problems with attenuation of the signal due to resistors in the path? Or is this only a concern when outputting directly into the headphones?


There shouldn't be a problem, but there's no complete circuit.  I don't really know how Meier does it, but I think the hint lies in his crossfeed application.  I'd bet it's a similar topology, but instead used to generate the Gfactor.

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post #15 of 128
12/27/10 at 10:35pm
Thread Starter 

Why would you need to double the ground channel? You're tapping two sets of wires off it, so your voltage is the same in each except now you have two of them, so it all balances out. (I suspect we're still agreeing on the same thing but the communication is just misfiring)

 

The -(L+R)/3 was in the case where the headphone cable only has three wires running through the main length (Meier's calculations assume 4, with a separate ground wire for each channel). For the three wire case, you want:  (L+G) + (R+G) + G = 0

 

If you used -(L+G)/4 in the three wire, your audio is still fine, but now you still have an electric field.

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