I was looking at this schematic for a two-channel audio power amplifier.

amplifier schematic

The amp supports a bridged mode by means of an XLR connector and a switch. When the switch is in the XLR position, pin 2 (Signal+) is sent to the left channel and pin 3 (Signal-) is sent to the right channel and the speaker is connected between the + outputs of the amplifier.

However, as far as I can see, there is no differential stage before the power amplification stage to actually filter any noise from the balanced input. Therefore, any noise that gets onto the input wire will be amplified and sent to the speaker.

Is this correct, or am I missing something?

Here is the amplifier in question:

amp panel

  • \$\begingroup\$ well, I'd agree, but then again, this is mostly a block diagram with electronic schematic "elements" in it; so, for simplicity, it might be omitting input electronics. \$\endgroup\$ Jan 18, 2019 at 10:47
  • \$\begingroup\$ why don't you actually try? \$\endgroup\$ Jan 18, 2019 at 10:47
  • \$\begingroup\$ Shouldn't be needed. XLR is balanced - common mode noise should cancel out. \$\endgroup\$
    – JRE
    Jan 18, 2019 at 10:48
  • \$\begingroup\$ @MarcusMüller I don't actually have this amp so I can't play with it. \$\endgroup\$
    – friedo
    Jan 18, 2019 at 10:54

2 Answers 2


What you're actually (technically) asking is: Does this amplifier have common mode rejection?

Common mode is the noise/disturbance that is the same on both wires of a differential signal. Read more here.

Assuming the schematic is accurate, then the answer is No, this amplifier does not have common mode rejection, but read on!

Indeed there are circuits that can reject (we circuit designers don't call it filtering) common mode signals. This circuit does not have such common mode rejection.

Fortunately, you will still get the common mode rejection that you want!

How does that work?

The actual cancellation of the commonmode signal will happen at the speaker.

Suppose there's a common mode disturbance on the input. As it is common mode it will be present on both signal lines (of the XLR connector). As the amplifiers are equal this signal will also be amplified equally and appear equally at both output terminals with the same amplitude and phase etc.

Then at the speaker this signal will also be common mode, so present at both speaker terminals when measured with reference to ground. But the speaker itself is "floating" between both outputs, it has no ground connection so the speaker does not "see" that common mode disturbance! That means the common mode signal is ignored at the speaker.

  • \$\begingroup\$ Thank you! I suspected it was something like this, but couldn't quite figure it out. Edit: It's perfectly obvious now: the difference between the + and - on the speaker terminals will be zero where there is noise. Cool. \$\endgroup\$
    – friedo
    Jan 18, 2019 at 11:06

The circuit is either weird or incomplete.

Usually the mono signal is fed directly to one power amplifier and an inverted version to the other power amplifier. There is no sign of an inverting stage in this unit.

There is nothing in the diagram to support your idea that the XLR input is balanced (although XLR connectors are standard in balanced audio circuits). Having a balanced input would solve the problem of the lack of inverting stage.


simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple unbalanced to balanced with ground reference.

The balanced input signal would need to be ground referenced rather than floating. This could be achieved using a transformer or by electronic output.

What I'm curious about is whether the inverted signal is ever actually used to filter noise, or just to drive the bridged speaker.

That's inherent in the design. Any common-mode noise on the input will come out as common-mode noise on the output affecting both speaker terminals to the same extent. Since the noise doesn't create a differential signal across the speaker the noise will be inaudible.

  • \$\begingroup\$ Well, I think it's clear that the designer is expecting the XLR input to be balanced since an inverted signal is required for bridging anyway. What I'm curious about is whether the inverted signal is ever actually used to filter noise, or just to drive the bridged speaker. \$\endgroup\$
    – friedo
    Jan 18, 2019 at 10:56
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Jan 18, 2019 at 11:08

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