I have a differential audio signal which I want to use to drive a single-ended device (in this case, an amplified pc desktop speaker with line level inputs) with 10kohm impedance. I have wired up a TLV2462 op-amp as a differential amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

The opamp power is 5v relative to ground, which is not connected to the amp ground in the circuit above. The input common mode voltage is 0.75v, and the differential signal is about 0.5v p-p, so the inputs should be well within the rails.

If I put a scope between amp input and amp ground with the load disconnected, I see the output signal exactly as I want it (sorry for the bad cellphone pic of a cheap scope):

enter image description here

But if I connect the load/speaker, the output is clipped slightly below 0v:

enter image description here

I see exactly the same output if I just put a 10k resistor in the circuit instead of connecting the speaker.

Edit: In the answers and comments below, several people have commented on ground and amp ground not being connected. If I connect A to ground, then I see basically the same as the second scope image below, but the clipping is actually at ground, not a little below it. If I put a capacitor between amp ground and A (now ground) instead of connecting those directly, I get the output I want. But, I don't really understand why.

Is the ground connection and capacitor I describe above the "right" fix? Why does it work? If not, what should I be doing instead?

Since this is for music, I'd like to keep frequencies above 20 Hz, if that's practical. The impedance of the load is 10k.

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    You mention "not connected to ground in the circuit below" yet there is no circuit below and this confuses me. You also say the output is clipped slightly below 0V - an op-amp powered by +V and 0V cannot produce an output less than 0V. Also, where is 0V on the scope shots? – Andy aka Jul 8 '16 at 8:17
  • Whoops. I meant above. Edited. – archbishop Jul 8 '16 at 8:18
  • The scope ground is connected to the ground in the schematic, not the ground input of the op-amp (and the signal to Vout). 0V on the scope shot is where the yellow indicator on the left is; on the right side of the photo you can see the horizontal line with short marks along it. When I said the input common voltage is 0.75v, that is measured relative to the op-amp ground. – archbishop Jul 8 '16 at 8:24
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    Do you provide the op-amp with -5V supply so its output can go negative? Show the op-amp power supply connections in your circuit. – Dmitry Grigoryev Jul 8 '16 at 9:11
  • Voting to leave question OPEN for now. OP must show effort at understanding the problem and correcting it. – Sparky256 Jul 8 '16 at 21:54

Step 1:

PC speakers are crappy at best. Who knows what they call ground, how they relate it and where exactly it comes from.

Thus, Step 2:

You need to have ONE ground point in your system.

Which follows to Step 3:

You need AC coupling in your schematic, because you do not know what will be at the output, in terms of power-loops through whatever other connection, such as power-earth.

Gives Step 4:

The same ground loops may be causing your input signal to jump/drop to either power-rail and op-amps only really work with voltages between their power-rails. Many op-amps even only work with a (small) part in the middle of that region.

Making your schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Where the 5k resistors make a voltage divider to keep your incoming signals on average around 1/2 Vcc, as long as your signal source can cope with 5k/3 =~ 1.6k loading that should be fine. If your signal source is very weak, you may need a balancing resistor coming from a dedicated 1/2Vcc point, like so:

schematic

simulate this circuit

Whether your capacitors need to be in the range of single nF, hundreds of nF, or several μF, only someone knowing all the specs of the signal (and allowable star-up behaviour) can tell you.

  • I built the upper circuit. Wired up like this, the output is maybe 5mV dc. If I measure the two terminals of R7, I see the input sine wave, about .4v p-p, with about 2v offset at the bottom end, and 3v offset at the top. So, they are kinda near 1/2Vcc, but with the large DC offset between the two inputs, this doesn't really look like a differential signal at all? The specs for the chip driving the input signal say the allowed load is 16 ohm to OC (open circuit, I guess), so 1.6k should be fine. – archbishop Jul 9 '16 at 3:48

The opamp power is 5v relative to its ground, which is not connected to the ground in the circuit above.

Well here's one problem - the negative power rail being 0V does need to connect to the ground points. Not being connected to ground means the signal ground can float off relative to power ground and break rules on input common mode range.

Using an output capacitor will probably achieive what you want but you must have the power rails of your op-amp tied to signal ground: -

enter image description here

Choose C to avoid too much bass cut-off but without knowing what value Rload truly is or your lowest frequency I can't give a value.

  • I'm not sure how to use this schematic tool to indicate what I want, which is that the negative power rail connected to ground is not connected to the ground for the single-ended output signal path. – archbishop Jul 8 '16 at 8:26
  • So why did you state what I highlighted in my answer? That clearly says that your op-amp power supply is not referenced to ground in the circuit and the only ground in the circuit is at the bottom of R4 and the load. – Andy aka Jul 8 '16 at 8:29
  • Let me try again. I have a power supply. gnd and +5 are connected to gnd and Vdd+ on the op-amp. The same supply is powering the signal generator, which has the properties I gave above. the r4 and the "ground" input of the speaker are connected to each other, the ground input to the scope, and nothing else. I grant any of this may be wrong or mislabeled, but I believe it describes my circuit. – archbishop Jul 8 '16 at 8:36
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    You cannot have your power rails galvanically isolated from your signal 0V in regular op-amp circuits. Your expectations of your circuit exceed the reality. If you need an AC output that is symmetrical about ground use an output coupling capacitor or an extra negative rail. – Andy aka Jul 8 '16 at 8:46
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    Then the 3 dB cut-off point of a series 1uF feeding a 10k resistor is 15.9 Hz i.e. \$\dfrac{1}{2\pi RC}\$. You can make the capacitor bigger of course. – Andy aka Jul 9 '16 at 10:22

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