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What purpose of these resistors? Spdif level about 0.5Vpp. Do I need use feedback resistor if I will convert 3.3V SPDIF to 5V?

I want to make 4-way SPDIF Input/Output for WM8805 I2S<>SPDIF converters, and made a schematic. What should I do better?

Logic inverter feedback SPDIF Input SPDIF output

Yiffme
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2 Answers2

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They are intended to bias the inverter into operation as a linear amplifier, but this is likely to result in high frequency oscillation unless you use an unbuffered inverter such as the 74HCU04.

When it works, the input sits at about midway between the logic levels of Vih and Vil or about 1.4V in this case. Since the input is AC-coupled that doesn't matter much.

You can learn more from this web page. The author echoes my experience that a buffered gate as shown will often oscillate (he says 240MHz and getting very hot as a result). The reason is that buffered gates (the usual type these days) have two additional inverting gain stages internally and have inaccessible nodes that would be required to stabilize the circuit as an analog amplifier. In fact, with feedback, the circuit resembles a ring oscillator.

Spehro Pefhany
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  • actually, it's for dc biasing, see my answer – bobflux Aug 10 '19 at 17:19
  • The input is connected to GND via 100R-10nF-75R so if the layout has low enough inductance, it won't oscillate... also, for HCT, the input DC bias won't be exactly at the center between Vih and Vil (see my answer) – bobflux Aug 10 '19 at 21:13
  • I concur with Mr. Pefhany: an unbuffered inverter is what one uses in these circuits. – td127 Aug 04 '20 at 03:23
  • Also, the output impedance of the coax s/pdif should be 75 Ohms total: adjust your resistors so the parallel combination is 75. – td127 Aug 04 '20 at 03:29
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SPDIF signal is designed to have zero average voltage so it can be transformer or capacitor coupled.

However... a 74HC inverter doesn't take an input voltage centered around 0V, it wants an input voltage centered around its logic threshold, which is half supply voltage.

The resistor (along with the 10nF cap) simply acts as feedback to bias the input of the gate to half the supply voltage. This makes the circuit work.

Note there is a mistake in this circuit. The input is biased to the threshold of a 74HC gate (mid-supply) but the chip is 74HCT, which has a lower threshold voltage. It will kinda sorta work, but it will add a lot more jitter than using the correct threshold for a 74HCT chip, or using the same circuit with a 74HC chip.

Also, if you want to make a high quality DAC... just use a WM8805.

EDIT:

enter image description here

The fact 74HCT04 has an input threshold which is not at mid-supply has interesting consequences. If we feed an input signal with 50% duty cycle, with realistic rise and fall times, we'd expect the output signal to also be 50% duty cycle, but it is not! The duty cycle changes, because the input DC bias point is not exactly the midpoint between Vih and Vil, but it is skewed towards a bit higher voltage. HCT is a bad choice here.

bobflux
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  • Thank you. I want to make 4-way SPDIF Input/Output for WM8805 I2S<>SPDIF converters, and made a schematic. Check my first post again please, i uploaded two schematics. What should I do better? Is it important to place 75/110 Ohm input resistors after transformer? – Yiffme Aug 10 '19 at 18:10
  • @Yiffme WM8805 already has several inputs and an input selection circuit that works very well. Why not use it to select your input? It is the simplest solution... – bobflux Aug 10 '19 at 21:12
  • because I have two factory produced PCB – Yiffme Aug 11 '19 at 07:47
  • And i will not use inverter for Input, all inputs will be connect to WM8805 directly. Only 74AC04N for SPDIF output, with 0-3.3V input level (directly from WM8805). – Yiffme Aug 11 '19 at 07:54
  • @peufeu - it will correctly bias the device the switching threshold. It will not bias it at the 74HC mid point. – Kevin White Nov 15 '19 at 00:12