Electronic – Using differential line drivers for digital signals

I would approach this as two problems, generating a sine wave, and making a balanced line driver. Other answers have covered the sine wave generator, and it's an easy thing to research, and I have nothing to add there. However, I'll say some things about the differential line driver.

As some others have said, the canonical way to do this is with a transformer. Transformers work great, but are big and expensive. In audio applications you will need an even more expensive transformer to avoid introducing unacceptable distortion. However, if you want to look exactly like a dynamic microphone, this is your best option, since a transformer simulates more of the properties of the windings on a dynamic microphone than any other method.

However, any balanced audio signal you get from any modern device that is powered probably won't have a transformer these days, because of cost. Powered (condenser) microphones may fall into this category; mixing boards and preamps almost certainly do. I highly recommend you read Design of High-Performance Balanced Audio Interfaces for a survey of common techniques and a detailed explanation of the relevant concerns. Also see Balanced Transmitter and Receiver II from the same site.

There's one part of that later article in particular I'll summarize here: What's important is that the impedance of both lines is the same, so that noise will result in the same voltage, so that it can be rejected as common-mode. Having an opposite signal on the negative side doesn't matter at all. In that article, there's a schematic, under the section Hey! That's cheating:

See the article for detailed discussion, but you can plainly see that pin 3, the negative side of the signal, is just a connection to ground through a resistor. As it turns out, if you disassemble a lot of professional audio equipment, this is precisely the type of line driver they use. It's because it has quite a few advantages:

• Simple
• Easy to balance
• If pin 3 is connected to ground on an unbalanced input, nothing bad happens

The only critical part here is making sure R2 and R3 are exactly equal. Use 1% or better resistors, or balance them with a Wheatstone bridge for best common-mode rejection.

"Ground" as a concept needs some clarification. If you have just one signal line and one ground, then yes, it's hard to tell the difference. But if there's anything else going on, it matters.

All AC signalling involves current flow, even if it's being measured as a voltage on the receiver. At a minimum, you have to charge/discharge the parasitic capacitance of the receiver, and the signal wire will have a capacitance to ground as well. Note that the common mode behavior only holds if the two wires have the same length and are a constant distance from each other.

So if you have a signal wire and a ground wire, then the current in the signal wire must be matched by a corresponding current in the other direction in the ground wire. If you have lots of signals, the ground wire will contain a mashed-together copy of all the signals. Therefore it's advantageous for each signal to have its own ground. If you look at VGA, you'll notice that each signal gets its own ground because of this. If you look at 80 pin IDE, each pair of signals has a ground wire between it in the ribbon cable. Those are to prevent the signal wires inducing currents in each other ("crosstalk").

Once you've accepted that each signal must have its own matched ground, it's more natural to embrace the two as a matched pair, disconnect one from ground and connect the two of them together via a termination resistor network, and drive / read them as a differential signal.