Still don't understand this

[SirNickity]

I posted a topic on this problem a year or two ago, but I still don't get it. I'm hoping recent research and a new perspective will help.

OK, consider two line-level circuits. One is powered with a 9v battery. The other is powered via mains and a transformer. Because audio is AC and will swing above and below ground, we need to split the voltage on the battery powered device into -4.5/virtual gnd/+4.5. The mains powered device is rectified AC from a center-tapped transformer and gives us +12/CT/-12.

I've read lots on how to couple two signals that might have DC offset, but so often, the ground is ignored. How the crap do you connect a device with +4.5vdc on the shield to a device that is referencing the physical center of the secondary windings and not blow something up? (Or at the least, crush half the waveform.)

I just don't get how this isn't a problem. Can someone please explain what happens to that offset?

I've been told "don't worry about it" but I can't do that. Why not? Because I really want to build stuff. I've stole-- eh.. incorporated some basic op-amp designs and chained them together into an active subwoofer input stage (L+R summing amp / input buffer -> SK variable lowpass -> volume control / output buffer) and have gone so far as to breadboard it and power it up with a dual-supply transformer. But I can't quite summon the courage to attach it to my receiver or plate amp until I know how this whole ground thing works. I may never use any single-supply sources, but there's obviously a proper way to engineer an input stage, and I'm not confident I've done it.

[Mooly]

Its not a problem because the "virtual" ground is floating and has no relationship to the mains powered supply. No current flows.

The batteries negative terminal is floating.

Draw it out and it will make sense.

[vacuphile]

Virtual ground is not very 'robust', so in this kind of battery setup, you would typically take the - of the battery for shielding. The output signal then has a DC offset of 4.5 volts, but no reason for worry. You can block this DC offset by using a coupling capacitor, and the same goes for the input. So, the input and output devices only see AC, and there is no DC potential between shield and signal.

Do you have schematics for the device you built?

vac

[Lazybutt]

SirNickity,

What you ask is at the same time quite simple to understand, and difficult since it is a bit abstract.....

I think what you have to understand is that there is no such thing as an "absolute" voltage. A voltage is always a difference of potential between two things.

In the case of your 9V battery, there is 9 volts of difference between the - pole and the + pole......

Where each pole of the battery stands compared to the potential of anything else is just a matter of how you connect them together... lets imagine you connect your battery - pole to a point that is +120 V DC compared to the ground, you will have the + pole at 129 V above the grounb ... and there's still 9V difference between your two poles.

Now the same applies to your secondary.. there is 12 Volts AC between each terminal and the center Tap, which means it oscillates between +16.9V and -16.9V (*)

If you connect the - pole of your battery to one of the secondary terminal, the + pole of the battery will oscillate between -7.9V and +25.9V versus the CT.... it's no problem as long as you don't attach the + pole to the CT

You still follow me..?

OK..... Where your secondary stands vs the primary potential is actually unknown because they are not electrically connected to each other.. the secondary creates voltage by ways of the magnetic field....
but in the end we don't care much where it stands because they are not connected together, and as Mooly said, no current flows (**), so the CT finds its potential vs earth by itself..... it might be close to zero, or close to 110 Volts, or maybe floating somewhere else, ....who knows.

You could think of voltages as altitudes or lengths.
You could imagine
- Your 9 V battery is a 9 m rod
- Your secondary is a 12m ladder

When the rod is on the floor, the altitude of the tip is 9cm
Now if you take the rod and climb halfway on the ladder, the altitude will be 15meter
You can even move up or down on the ladder, the length of the rod is still 9m, but it's absolute altitude is varying...

Now imagine you can even move the ladder on a mountain, etc...

I see what you are worrying about ... if you are trying to reference too many things versus each other, most likely you will create a short circuit.. for example if you attach simultaneously
- the rod in the middle of the ladder
- the ladder on top of the mountain
- the rod at the foot of the mountain
Something will happen which is not good to your components (like in previous example where the - pole is attached to one of the secondary terminal, and the + to the CT) .. but if you are not trying to reference the rod to the mountain, everything just finds its place.

In your case, what happens is your 9V battery will get elevated by a certain potential ..... as long as you connect your two line level circuits by only ONE point (the signal ground), there is no problem! What could be a problem is if you connected the two circuits at two points, or if you tried to attach the battery to the CT of the secondary ..... there you could have a short circuit.


(*) I'm assuming you understand where the 16.9V comes from..
(**) in reality they ARE connected together by one way or another, it might be through the earthing of your transformer, or through the materials in contact between the secondary and the primary (which resistance might be very very high, but not infinite... the secondary always finds its own voltage versus primary and vs earth) but the current flowing is so low that there is nothing to worry about...

PS: I realize explaining this is not as easy as I thought when I started the post, so please ask what part is still unclear, I think it will be easier to concentrate on the fuzzy parts

[SirNickity]

Wow guys, thank you for the thorough replies. I think I'm starting to understand but I need to let all of this sink in a bit. In the meantime:

Its not a problem because the "virtual" ground is floating and has no relationship to the mains powered supply. No current flows.

I read this last night and it didn't quite seem obvious. After sleeping on it, I pieced some things together and I'm wondering if I'm on the right track. Let me explain my take on this: A circuit requires a return path, so tying grounds together doesn't make a complete circuit. The signal portion is already coupled via a capacitor, so we're not going to see any DC offset flowing via that path.

I think what you have to understand is that there is no such thing as an "absolute" voltage. A voltage is always a difference of potential between two things.

This idea I am familiar with, but it was as disheartening as comforting. Sure, there may not actually be a 4.5v offset between a battery powered device and a mains powere device. On the other hand, it could be hundreds, or thousands, even millions of volts for all I know. I've never measured the difference between the negative terminal on a battery and AC earth. I might try that... not sure what it would prove, though.

Where each pole of the battery stands compared to the potential of anything else is just a matter of how you connect them together... lets imagine you connect your battery - pole to a point that is +120 V DC compared to the ground, you will have the + pole at 129 V above the grounb ... and there's still 9V difference between your two poles.

Now this is beginning to illuminate things...

If you connect the - pole of your battery to one of the secondary terminal, the + pole of the battery will oscillate between -7.9V and +25.9V versus the CT.... it's no problem as long as you don't attach the + pole to the CT
You still follow me..?

The most comfortable I've ever felt with this idea, actually.

Where your secondary stands vs the primary potential is actually unknown because they are not electrically connected to each other.. the secondary creates voltage by ways of the magnetic field.

This had never really occurred to me. I knew they were isolated, but hadn't put it together.

I see what you are worrying about ... if you are trying to reference too many things versus each other, most likely you will create a short circuit.

Pretty much. Mostly that the two would try to equalize, at the demise of everything they were attached to.

I'm assuming you understand where the 16.9V comes from..

12 * 1.414, from rectification -- but not counting diode drop. Yep!

I realize explaining this is not as easy as I thought when I started the post, so please ask what part is still unclear

I will chew on this for a bit. What has been said makes sense, so it's a matter of accepting these principles and allowing my world view to adapt. :-)

Virtual ground is not very 'robust', so in this kind of battery setup, you would typically take the - of the battery for shielding.

Could you expand on this? What makes it problematic? If I used the - terminal for shielding -- meaning, 0v on the battery was equal to 0v on the input or output of the device with an audio signal -- what would happen when the AC waveform goes negative? Wouldn't the opamp(s) in the circuit in question have no rail low enough to produce that voltage?

Do you have schematics for the device you built?

I had thought I might post this on another forum for feedback anyhow. There are several points where I took textbook examples (the summing amp, input buffer, and Sallen-Key sections) and tweaked things to work together. I added the DC blocking cap, and I'm hoping I chose a reasonable value, and placed it properly. The final buffer is actually from ESP -- project 09 I believe. I kinda see how that part works, but I'm still making sense of the topology chosen, as I would have probably done things differently if I were trying to create it from scratch.

Thanks again everyone, I really appreciate your efforts. Hopefully, some lurker can benefit from this as well.

[sreten]

Hi,

Your 9V device does not have 4.5V DC on the shield, OV is always OV.

The OV is referenced from the 12-0-12 +/- mains connected device.

The 9V device needs to be fully AC coupled with capacitors. OV in
that device is the same as the other device. The output capacitor
will have 4.5V dc across it for a correctly biased arrangement.

You need to look at AC coupled single supply versions of typical
op-amps circuits, compared to symmetrical supply versions.
Lots of examples all over the place. Ground is ground.

rgds, sreten.

[SirNickity]

My question was more about theory than implementation. I've seen how it's typically handled, and I comprehend how that part works. I was having trouble reconciling the philosophy behind a single coupling capacitor on the signal connection while leaving the grounds to sort themselves out.

The part that really tripped me up was how you know "ground is ground" when they could be anything in relation to each other.

[vacuphile]

SirNickity,

Please have a look at your schematic, and let me point out some issues.

1) There is a decoupling cap on the input side, but not on the output. You need one there as well.

2) In this setup, you will need to create a virtual ground, and I suggest to make that part of the schematic if you take that route.

3) The reasons you need a virtual ground here are because the input buffer R-DC and output buffer volume control. To make it simpler, look for alternative schematics that do not need a virtual ground, e.g. use the input buffer opamp in non-inverting mode

[sreten]

Hi,

Yes, for 9V single supply to a dual supply, the single supply must be AC
coupled, the dual supply may or may not be AC coupled. The single supply
must be set up as +9V - +4.5V virtual ground - 0V actual ground, the actual
ground interconnection between the two is OV to the dual +XV - OV - -XV.

For the whole thing OV can be referenced to any point as its floating.

following are a dual supply and single supply versions of a circuit :





RX1 and RX2 (with CX1) create the virtual ground at the supply mid point.

rgds, sreten.

[DF96]

As others have said, there is no such thing as 'a voltage'. Voltage is always a difference between two points. The + side of a 9V battery is not at +9V, it is merely +9V higher than whatever the - side is. The battery establishes a potential difference of 9V; that is all.

Similarly, there is no such thing as 'ground' or '0V'. You can establish a reference conductor, and call it ground, and then refer all voltages back to it. Indeed you should do this. If you do this twice, your two 'grounds' will often not be at the same potential. Which one is 'true ground'? This is a meaningless question.