Debalancer from unipolar supply

[conleycd]

Hi people,

I know of lots of ways either with specific ICs like the INA134, THATs, or a simple OPAMP to de-balance a signal.

However, is it possible to (cheaply) debalance using an Opamp with a unipolar supply.  I'm familiar with the opamp designs using unipolar supplies for guitar based effect- voltage divider 1/2 voltage into the opamp input (ground v-) with blocking on the output.

Is there a similar way to debalance a line signal with a single power supply (economically and space wise - the reason I'm trying to avoid bipolar supply and ac transformer to begin with)?

Much thanks in advance.

CC

 

[abbey road d enfer]

It's feasible, of course, and has been done in many pieces of gear. Basically, you have to reference the foot resistor of the + input to V/2 instead of ground. You have to have input and output caps also.

 

[conleycd]

Hey Abbey Road,

If I do the 1/2 voltage bias into the + of the opamp (like almost every guitar effect) - do I then just follow a traditional differential amplifier with the classic opamp and 4 - 20k resistors?

CC

 

[abbey road d enfer]

Yes exactly

 

[conleycd]

What do you think about this?  Do you think it will work?



I don't know if the image will show up right. 

But essentially, it would use a dual opamp - something like a TL072.  It is powered with 1/2 voltage into the non-inverting input and full supply to +V with -V grounded.  Signal is unbalanced by Opamp B and is buffered with Opamp A.
Thoughts?

CC

 

[Svart]

First off, you might not need any kind of voltage divider for offset.  Simply using series caps will allow the inputs to bias themselves to wherever they want to be.

Second off, that 100uf shunt before the second unity stage is going to short all of your AC signal to ground.  You just made a big low pass filter.

Here's some help if you do need to do a true single supply setup with biased inputs:

http://focus.ti.com/lit/an/sloa072/sloa072.pdf

http://instruct1.cit.cornell.edu/courses/bionb440/datasheets/SingleSupply.pdf  (see figure 8 )(I also suggest using figure 11)

 

[audiox]

Do you think it will work?

No it doesn't. You don't need the second op-amp stage at all. And the first stage is not correctly biased. You should connect the uppermost ground symbol in your schematic to 1/2 power supply voltage (resistive voltage divider and large capacitor to ground).

Remember that the debalancing stage is not necesserily required. If the cable is relatively short and amplification is not required, you can connect balanced output to unbalanced input directly.

 

[abbey road d enfer]

The first stage needs to be biased
The second opamp is not necessary

 

[conleycd]

Wow - thanks everyone.  I'm pretty green.

For some reason - I thought the Opamp "A" in a dual opamp required the 1/2 power supply - but that may have been my own self imposed delusion.  That was my reasoning for powering the buffered opamp and having the buffer at the end of it all.

I for some reason also thought that if I powered the Opamp receiving the balanced signal using the voltage divider then I would throw off the impedance of the non inverting opamp.  I understand that is not the case from the diagram provided.

Abbey Road - in terms of input impedance - is that determined by the 20k resistors or the 100k pull down resistors in your diagram (or both or a combination of other stuff)?

This circuit is actually so I can split a balanced signal - one to a balanced mixer input and one to an unbalanced source.  I know for short runs I could go balanced to unbalanced with just using "-" as the ground source.  

The unbalanced source is a mixer (pretty standard summing into the inverting input opamp) attenuates the signals before the mixer with a pot.  Because of that I think need an opamp first to unbalance before hitting the pot.  

If there was no attenuation I believe I could (correct me if I'm wrong) simply route "-" through a resistor that would be the same impedance as the input of the mixer to ground and then take my split directly at the input.  There was another thread on this some years back that I read.  The problem is the attenuation pots would create a change in impedance on the "+" signal.  Correct?

Thanks again.

CC

 

[Svart]

The 100k resistors set the load.  Technically the impedance would be set by the 100k + the 20k + the impedance of the opamp inputs, etc but impedance will change with frequency depending on layout, parts used, etc.  So just say that it's 100k and be done.

I'm not sure what you are asking about the attenuation pot..

The pot would be setup as a voltage divider, wiper would be your output to a unity stage.  This creates a known load on the output of the driving stage.  The input of the unity stage won't care about your change in divider value because it's impedance will be high enough.

 

[abbey road d enfer]

The actual input impedance is set by a combination of the 100k and the leg impedances; I use the plural because, as PRR mentioned some time ago, the leg impedance of the + input is different than the - input. The + in Z is 40k//100k = 28.5k, the -in Z is 20k//100k = 16.6k. The differential input Z is also 40k//200k = 33.3k.
If it is a worry for you, you can use 20k for the positive branch and 40k for the negative (see attached dgm), but then again, the differential Z will be different. I wouldn't worry at all about it.
You are worrying too much about impedances , but I reckon the diff amp will protect you against all sorts of nasties like fortuitious short-circuit or unbalancing.

 

[conleycd]

Hmmm.... Now I'm a little confused now.

Should I be using 40k into the non-inverting and 40k in the feedback loop with a parallel 47pf ish cap?  Then the 20k into the inverting input with 20k pull down down?

I understood the first schematic better or should I be using the new resistor values (in the second schematic) with the first?

Thanks.

CC

 

[abbey road d enfer]

Should I be using 40k into the non-inverting and 40k in the feedback loop with a parallel 47pf ish cap?  Then the 20k into the inverting input with 20k pull down down?

That's the contrary; the higher value is in the inverting path.

I understood the first schematic better or should I be using the new resistor values (in the second schematic) with the first?

The variation with different values has better balance of the impedance. But the practical benefits are almost non-existent. This last schematic was just to indicate the position of the resistors that should be modified. But you need to use the first schematic with the caps and pull-down input resistors.

 

[audiox]

Should I be using 40k into the non-inverting and 40k in the feedback loop with a parallel 47pf ish cap?

There is no difference in practice. And if you scale the resistor values, you must also scale the capacitor values because of CMRR.

If the impedances are your main concern, you should use a balanced input made of two inverting op-amp stages. As a bonus the gain is adjustable by chancing just one resistor. And you get rid of the low frequency CMRR decrease caused by the biasing network (you can use the non-inverting inputs for biasing only).

 

[conleycd]

Interesting... Sorry Abbey Road I meant to say '40k into the inverting'.

Well.. it might help if I explain better the application.  I am intending to split a balanced signal.  One to a balanced input of a mixer and one to an unbalanced input of another mixer.

So by using debalancing or unbalancing I can split the signal at the input to the balanced mixer and use the unbalanced signal to the unbalanced mixer.

The balanced mixer input will be 70' away and the other about 8'.  The balanced input's input impedance (into the microphone connector) is 2k and the other mixer will be whatever I set this debalancer input at.  So... I was thinking 20k. The output impedance of the device is from an M-Audio Delta 1010.  The specs don't list the output impedance but I'm guessing it's between 50 and 150.

From my calculations (correct me if I'm wrong) that presents a 150ohm output with an input impedance of about 1800.  This should be sufficient for the 1:10 rule - barely - if the output impedance of the 1010 is 150.

I am not really concerned about the input impedance of the de-balancer as long as it is over 20k based on this calculation.

Audiox - are you suggesting that I run into two opamps using their inverting inputs - at either unity or with some gain if needed?  How would I blend the two outputs?  Would it be that I then run the outputs it into into a differential amplifier (as described already)?

Hmmm, maybe a TL074 might suffice for all this - per channel.

Thanks again.

CC

 

[abbey road d enfer]

I am not really concerned about the input impedance of the de-balancer as long as it is over 20k based on this calculation.

Rightly so. The actual differential impedance is 40k when using 20k resistors. This should work very well. Don't fret.

 

[audiox]

The balanced input's input impedance (into the microphone connector) is 2k and the other mixer will be whatever I set this debalancer input at. 

But we are still talking about line level signal? If you are splitting mic level signal, you should use a transformer to avoid noise problems.
By the way: the mixers mic input impedance in "line mode" is usually more than 10 kohm because of the attenuator included.

Audiox - are you suggesting that I run into two opamps using their inverting inputs - at either unity or with some gain if needed?  How would I blend the two outputs?  Would it be that I then run the outputs it into into a differential amplifier (as described already)?

Check this. Just add 1/2 power supply voltage to the non-inverting op-amp inputs (use the voltage divider descrided earlier in this thread) and capacitors at the inputs and outputs to block DC. The gain equations are also included in the block diagram.

http://www.groupdiy.com/index.php?topic=29333.0

 

[conleycd]

Audiox - yup we are talking about line level.  The sound guys have been running the line outputs into the microphone inputs.  It actually hasn't worked too bad - no problems with clipping.  However, you are right with the pad the input impedance pops up to 10k without the risk of frying the sound card by accidentally hitting phantom power (unless there's some bipolar caps in the sound card - I doubt it).  There's maybe some sonic benefit too from not running through unnecessary electrolytic caps in the front of house board.

I like your schematic and boards.  Actually I've always admired your projects that you have going on here.

So I'll pop 1/2 voltage into the non-inverting input (just opamp "A" if I'm using a dual opamp right?) using a 10k resistor divider.  Should I use the blocking cap after the first opamp or second or both? 

Thanks again - really - everyone.

CC

 

[audiox]

So I'll pop 1/2 voltage into the non-inverting input (just opamp "A" if I'm using a dual opamp right?) using a 10k resistor divider.  Should I use the blocking cap after the first opamp or second or both? 

You should bias both op-amp sections and place DC blocking capacitors at input and output (as I said in the previous post). Interstage capacitor is not required.