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I am designing a driver circuit for S/PDIF, and in a couple different places, like this one, it specifies an amplitude of 500 mVtt. I'm not familiar with those units, and googling gives me nothing. However, I do see it specified as 500 mVpp in a few places, but I want to make sure one or the other isn't a mistranscription. Are they the same thing? Is mVtt a European term perhaps? What does the tt stand for?

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  • \$\begingroup\$ That simply looks like a typo. It refers to the same qty's and only occurs 2X. \$\endgroup\$ – placeholder May 17 '13 at 14:59
  • \$\begingroup\$ @rawbrawb No, it isn't a typographical error: the terms Vtt and mVtt are used for specifying signal levels in multiple independent references. \$\endgroup\$ – Anindo Ghosh May 17 '13 at 15:22
  • \$\begingroup\$ Karl, there is some discussion in the EE site chat around this question, that might be of interest. \$\endgroup\$ – Anindo Ghosh May 17 '13 at 15:40
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Vtt stands for Volts, Threshold to Threshold.

Vtt and mVtt units are used (apparently as differentiating terminology) for signaling systems where digital signals are carried over analog channels, and comparators are used for pulse stream regeneration at the receiving end. This allows a wide range of actual signal voltages to be used for transmission while keeping the specification fixed.

It is not equivalent to Voltage peak-to-peak, in that the actual signal over the medium by definition has a higher peak to peak amplitude than the designed threshold-to-threshold excursion.

S/PDIF is the one technology I can recall off-hand that uses this terminology, but yes, there are others.


Using a 500 mVpp signal for a 500 mVtt channel will cause a failure condition if the channel has the slightest attenuation. However, with S/PDIF this is typically not a concern, as the reception specification is for 200 mVtt, allowing both a high attenuation margin, and the ability to use a variety of waveforms (sawtooth or pulse for instance), not just sine waves.

Not sure why one would want to use a waveform with a higher upper harmonic requirement and thus a higher bandwidth, instead of a sine wave, though. I'm sure someone on the site could explain.

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