Trying to design a simple balanced to unbalanced splitter...

[Mbira]

Hi all,
I'm sorry if this has been covered. I don't know what this device would be called, so I'm not sure what I'd search for! What I want to build would have balanced input that goes directly to a balanced output and also splits to unbalanced output. I have seen the jensen papers on using a traffo. The thing is I would like this to be cheap, easy, and generally transparent (no iron color). It'd be nice to be passive, but is that possible? No attenuation is needed other than the +4 to -10 step-down. I need this to be able to hook into two seperate systems-one balanced and one unbalanced obviously. Both outputs would be plugged in all the time, but only one system would be on at once.

Any ideas?

:guinness:
Joel

 

[Samuel Groner]

A quick-and-dirty-and-passive method is to run the + to the hot unbalanced input and the - trough a resistor to unbalanced ground. The resistor has to have the same value as the input impedance of your unbalanced input, which might be difficult to measure and could change when switching the device on and off.

As I cannot think of another passive and transformerless method, I would recommend a simple 4-resistor-one-opamp differential amplifier which will provide the unbalanced output. The balanced output is simply the input paralleled.

Samuel

 

[fum]

The +4 to -10 resistor network I've seen is to put 10K in series with the hot pin, and 2K post 10K across + and ground. That's giving you an attenuation ratio of 5:1, which will nock off 15db from the +4 signal (getting you in the ballpark of -10).

If the connecting line is going to be more than a few feet, you could alternately use a series resistor of 5K, and a 1K strapping resistor.

Maintaining the ratio maintains your attenuation.

Hope this helps.


Regards

ju

 

[Mbira]

Will there be a problem of loading down the signal if the balanced and unbalanced connection are both connected at the same time (even if one of the output systems is powered down)?

Joel

Samuel, do you have a schematic of such a thing?

Joel

 

[fum]

Will depend on the other source you are driving, but if the other source has an input impedance of 10K, your still driving a 5K load, which should still be ok.

Maybe someone with a bigger brain(than I) can chime in and say something too :wink:

ju

 

[Mbira]

I'm trying to fing a differential amp circuit using 5532's (or 5534's) ...
If no one beats me to one, I'll post what I find.

Joel

 

[Mbira]

So to be honest, when I said I was trying to find a circuit-I really meant that I am too dumb to figure this out on my own and I really hope someone posts a schematic that'll show me a differential amplifier using 5532's.



:guinness:

 

[Samuel Groner]

Differential Amplifier, first pic. Set the two input resistors to 20k and the feedback- and shunt-resistor (the one between opamp out and - in resp. the one going to ground) to 5k. Add a 50 ohm and a 220 uF cap in series with the output and you are done.

In my first post, I forgot that you want a 12 dB stepdown, so my passive method would need some elaboration. Maybe fum did this already, a bit in a hurry right now...

Samuel

 

[Mbira]

Something like this?

:guinness:

 

[daArry]

INA134 could be another option

 

[pstamler]

[quote author="Mbira"]Something like this?

:guinness:[/quote]

That'd work. If you hang a 100k resistor from the XLR pin 2 input to ground then both sides will see the same load. Match the 20k and 5k resistors as closely as you can. And if you hang a 10k resistor from the output of the 5532 to V- you'll bias up the output to single-ended Class-A and the thing will be remarkably transparent.

Peace,
Paul

 

[Samuel Groner]

Something like this?

Yep. Thinking about it, I would add a simple RFI filter. Split the 20k into a 19k (or what the closest standard value is) plus 1k. Then thrown in a 470 pF COG/NPO ceramic going to ground. So from the XLR we first meet the 1k in series, the cap and at the end the 19k. Hope I made myself clear!

Samuel

 

[Mbira]

A combination of both suggestions.

Like this?

 

[bcarso]

quote: "If you hang a 100k resistor from the XLR pin 2 input to ground then both sides will see the same load. Match the 20k and 5k resistors as closely as you can."

Common-mode signals already see the same load due to feedback, without the 100k.

 

[Samuel Groner]

Yes. I don't see the need for the 100k either.

Samuel

 

[Mbira]

Also, just to be sure, this will be fed line-level signals (after a preamp). I'm not interested in any form of amplification in this circuit. I just wanted to make sure that I was clear on that and that this will still work. I'll drop off the 100K and post another pic.

Joel

 

[Samuel Groner]

You can calculate the gain of this circuit as 5k/20k=0.25=-12 dB, that's right what you want for the +4 dBu to -10 dBV conversion.

Max. input level is well above +30 dBu, ample for your needs.

Samuel

 

[jvanslem]

[quote author="Samuel Groner"]You can calculate the gain of this circuit as 5k/20k=0.25=-12 dB[/quote]

Amatuer question:
How do we arrive at -12 dB from .25?

 

[Samuel Groner]

How do we arrive at -12 dB from .25?

-12=20*log(0.25)

Easy to remember are the 6 dB steps:

1=0 dB
2=6 dB
4=12 dB
8=18 dB
and so on...

and the other direction:

1=0 dB
0.5=-6 dB
0.25=-12 dB
0.125=-18 dB
and so on...

20 dB steps are to remember as well:
0.1=-20 dB
1=0 dB
10=20 dB
100=40 dB
and so on to both directions!

Samuel

 

[jvanslem]

[quote author="Samuel Groner"]

How do we arrive at -12 dB from .25?

-12=20*log(0.25)
[/quote]

Thanks for the explanation!

Follow up:
I know they represent resistor values in our circuit, but what exactly are we dividing in our (5k/20k) part of the equation?

If you have a reference link to explain it that's cool too.