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When I was looking for some information how to convert unbalanced audio signal to balanced, it was hard to find info other than "use transformer" or "buy X device to do this".

Then I found this circuit:Converter circuit

source: http://www.fivefish.net/diy/balanced/default.htm

but I'm not sure if this is correct solution. I mean that on this schematic you can see low pass filter which cut-off freq is about 140 kHz, which I think is way to high (changing 220p to 1n would help?), also where op amp in inverting configuration seems rather ok, but non-inverting shouldn't be done differently (according to https://www.electronics-tutorials.ws/filter/filter_5.html)?

Same thing with capacitor connected to pins 2 and 6 - I think it should rather be connected between output and those pins, not ground, and short between unbalanced signal + and gnd (or it's just compression of image that creates this illusion)

Anyway, I'm better in digital electronics than analog, so maybe someone with greater knowledge help me here with this.

EDIT 1:

Based on SSM2142, I created the following circuit that, I think (or rather I hope) should work correctly, making unbalanced audio signal balanced. Am I right or wrong?

schematic

simulate this circuit – Schematic created using CircuitLab

Instead of TL081 I was thinking about using LM358.

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  • \$\begingroup\$ This circuit just seems like such a waste... internally, an op-amp produces balanced outputs. It's just that most of them only bring one of the outputs through the final stage... but having two complete op-amps to do the job that one could do seems just weird. Look at an amplifier with differential outputs, e.g. THS4521. \$\endgroup\$
    – Jules
    Commented Jul 15, 2018 at 4:43
  • \$\begingroup\$ That circuit is pants. Do you want balanced (as in impedance balanced) or balanced and differential. Usually impedance balanced is quite sufficient. \$\endgroup\$
    – Andy aka
    Commented Jul 15, 2018 at 10:23
  • \$\begingroup\$ I wanted to do equivalent of this circuit, but using only "common" op-amp (like LM358 or similar). \$\endgroup\$
    – user193597
    Commented Jul 15, 2018 at 17:03
  • \$\begingroup\$ "I was thinking about using LM358." Please re-think. The 358 is terrible for real audio, and is only passable for something like an intercom. It has a relatively low gain-bandwidth product, but its real problems are crossover distortion and noise. \$\endgroup\$
    – AnalogKid
    Commented Jul 16, 2018 at 0:36

2 Answers 2

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You are correct about the feedback connections. The inverting and non-inverting inputs of both amps are reversed. Also, there is no reason for a full differential amplifier topology because the input is not balanced, so there is no "differential" other that the single-phase input and its GND.

For this most basic approach, start with two completely independent opamp circuits, one inverting and one non-inverting, both with the same gain magnitude. Note that the impedance of the inverting amp's feedback network directly sets that amp's input impedance, so keep it at around 10K or above.

A shortcoming of this approach is that if one side of the balanced output is disturbed, such as shorted to GND, the other side continues to make its full amplitude output. For a more complex balanced driver, look at app notes from THAT Corporation and Analog Devices. Both make chips for this function, and their internal schematics and explanations should be helpful.

http://www.analog.com/media/en/technical-documentation/data-sheets/SSM2142.pdf

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  • \$\begingroup\$ I found 3 IC that would do what I need, these are DRV134, THAT 1646 and SSM2142. Based on SSM2142, I created circuit which I added as edit to my original post. Will this circuit work as intended? \$\endgroup\$
    – user193597
    Commented Jul 15, 2018 at 22:34
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140kHz is actually reasonable, you want to be way above 20kHz to minimize in band phase shifts, and yes those opamps are drawn with the inverting and non inverting pins swapped.

However, there is a much simpler way to get most of the goodness, just make the source impedance to ground of the cold pin equal to that of the hot pin, but don't bother to drive the cold pin! A singled ended output stage plus 1 resistor and a cap gets you line balance that is as good as the complicated way, with 6dB lower maximum output as the tradeoff, but it still balances and you still get the interference rejection at the receiver.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the line balance depends on how closely the resistors (and at high frequency the caps) are matched, 1% = -40dB worst case, 0.1% = -60dB worst case. The caps should be taken directly to chassis as should pin 1 via the shortest possible route, and the earthy end of R2 should probably connect to circuit reference at the decoupling caps for the opamp. Circuit reference and chassis ground should be connected at ONE point.

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  • \$\begingroup\$ This could do with a schematic, Dan, to clarify what you mean by "hot" and "cold" as well as showing how a single-ended output could give a balanced output. \$\endgroup\$
    – Transistor
    Commented Jul 15, 2018 at 12:48
  • \$\begingroup\$ I think I understand what your proposed circuit is doing but I always understood "balanced" to mean a push-pull arrangement. What is your definition? Would you not set OA1's gain to +6 dB to counteract the loss in this design? \$\endgroup\$
    – Transistor
    Commented Jul 15, 2018 at 15:22
  • \$\begingroup\$ Nope, balanced refers strictly to matching the common mode impedance to ground, differential is the term you are thinking of. +6dB is fine if your output buffer has the required voltage headroom, but usually you are pretty close to banging off the rails at full shout anyway. \$\endgroup\$
    – Dan Mills
    Commented Jul 15, 2018 at 15:58

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