Question about X and Y differential ILDA signals

[HankLloydRight]

Hello folks.. I've got a question about how to handle the X/Y differential signals.

From everything I (think I) understand, the X-/Y- signals are just the X+/Y+ signals inverted. And X+/Y+ signals are 0v to +5v and X-/Y- are 0v to -5v.

Say I'm building an ADC for ILDA signals (input), I know I can run the X+/Y+signals into two ADC channels, and then invert the X-/Y- signals and run them into tow more ADC channels -- but that uses up four ADC channels. But is that necessary?

Consider these two other options instead:

1. Invert the incoming X-/Y- signals with op-amps, and then average them with the X+/Y+ signals (in theory they should be the same, right?) -- and then feed that average X/Y signal into the ADC using two ADC channels

2. Only sample the X+ and Y+ lines (using two ADC channels) and just ignore the X-/Y- signals, and just mimic the X-/Y- signals in software by inverting the X+/Y+ data stream before writing out the digital data stream.

Any flaws in these approaches? Thanks.

[HankLloydRight]

I guess I'm confused because looking at several ILDA DAC op-amp circuits -- the output of the DAC chips are 0v to 4.096v (or 5v, depending on the DAC).

That DAC output signal is fed into an inverting op-amp to produce the X-/Y- signals (so -5v to 0v) and then that signal is inverted again for the X+/Y+ signal (0v to +5v).

So I don't see how X+/Y+ could be "-5v to +5v" as described here: https://pangolin.com/blogs/education/ilda-laser-pinout

There's clearly something obvious I'm missing, I just don't know what. Thanks

[james]

If you could measure voltage from either + or - to ground, you would find both positive and negative voltage on both sides. The signals each have the full swing of the display area, they are just 180 degrees out of phase. So when they come into the scanner amps, they are subtracted from each other. A positive minus a negative is twice the positive. Does that make sense? You can drive just the positive input and ground the negative. If you use twice the gain on just one input, you get the same difference.

Differential is used in long cable runs and usually with very small signals like microphones. Any noise picked up in the cable would be the same in both the + and - wires and since the signals are subtracted from each other you get noise minus noise equals no noise.

To answer your question more precisely, you only need to send the positive signal and ground into an ADC.

All of the information is there.

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[HankLloydRight]

If you could measure voltage from either + or - to ground, you would find both positive and negative voltage on both sides. The signals each have the full swing of the display area, they are just 180 degrees out of phase.
.

This is what I've been reading elsewhere, but on my own DAC, X+ is 0 to +5v (and it does not go below 0v), and X- is the inverse of that 0 to -5v. But it works..so shrug?

So when they come into the scanner amps, they are subtracted from each other. A positive minus a negative is twice the positive. Does that make sense?

THAT makes total sense -- and I had not realized how those signals were used inside the projector/scanner amps. Thank you! So that's why they say that the outputs are 20volts peak-to-peak. It all makes sense now.

Differential is used in long cable runs and usually with very small signals like microphones. Any noise picked up in the cable would be the same in both the + and - wires and since the signals are subtracted from each other you get noise minus noise equals no noise.
To answer your question more precisely, you only need to send the positive signal and ground into an ADC.
All of the information is there.
.

Great. I had thought that the differential signals were there to reduce noise, but I didn't know how that worked. Now I do.

Thanks for the help.

[james]

This is what I've been reading elsewhere, but on my own DAC, X+ is 0 to +5v (and it does not go below 0v), and X- is the inverse of that 0 to -5v. But it works..so shrug?

Do you have a scope?

If you make a few frames and play each one as a steady display, you would be able to more accurately measure your voltages.

A solid line across each edge of the display square (one edge in each frame) would do it.

You would be able to read a steady voltage at all four extremes (with respect to ground).

-X +X -Y +Y

Then you'll know if there is some dc offset on any of your signals (even if it does null out between the phases).

[HankLloydRight]

Do you have a scope?

If you make a few frames and play each one as a steady display, you would be able to more accurately measure your voltages.

A solid line across each edge of the display square (one edge in each frame) would do it.

You would be able to read a steady voltage at all four extremes (with respect to ground).

-X +X -Y +Y

Then you'll know if there is some dc offset on any of your signals (even if it does null out between the phases).

Yes, I have a scope, and I've done exactly what you describe using either a square wave generator or the Quadrature test pattern. I even have those test points on my PCB for scoping. The problem is I don't remember the exact results!

Unfortunately I'm leaving for a week long trip tomorrow, but as soon as I get back, I'll set up the test again and post the results.

[james]

Dean knows a lot about this stuff.

[HankLloydRight]

I realized in my dreams that my X and Y offset pots might be bringing my X+ output below zero, and the gain pots up to +/-5v. So I'll have to confirm this when I get back next week.

[james]

I'm pretty sure, if you use op amps and everything is DC capable all the way through (no caps in series with signal) the only way to get negative voltage with respect to ground is if you have a bi-polar power supply; like ( -10v : ground : +10v ).

I'm sure there is a lot about this that I don't know. But I did find some information from Texas Instruments about op amps and different configurations for differential with and without a bi-polar power supply.

If it's not complicated, I can understand it when I see it. I just can't come up with it out of thin air, like Dean can.


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[HankLloydRight]

I do have a bipolar power supply.