Impedance matching resistors for digital output

Question

My circuit has to emit an S/PDIF digital audio stream over an RCA connector. According to the specs (from Wikipedia) the signal should have an output impedance of 75Ω and a peak-peak voltage of 0.5V.

I've found a schematic for a TTL to COAX converter:

but it's meant to work with 5V digital output, while my device (MAX II CPLD) works at 3.3V. So I'm trying to adapt this schematic for 3.3V by changing the resistor values.

As far as I understand the base circuit, including the input device, should look like this:

^{simulate this circuit – Schematic created using CircuitLab}

I have to compute Rx and Ry so that I can properly set up a voltage divider from 3.3 to 0.5V, and make a current of (0.5V / 75Ω =) 6.67mA flow from my device to the load. But I'm struggling in setting up the system of equations.

So far I came up with this one for the voltage dividing: $${1 \over {{1 \over R_y} + {1 \over 75}}} = 0.18R_x$$

And this one for the currents: $${3.3 \over {R_x + {75R_y \over R_y + 75}}} = {{0.5 \over R_y} + 6.67mA}$$

But when "developing" them they become really long. So I'm not sure if the approach is actually right or if I just did some steps wrong.

I'd also like to know the total current, to check if it's below the maximum source current of the CPLD.

Answer 1

The 390Ω || 91Ω will give you 73.8Ω looking into the TTL gate output from the load end which should be terminated in 75Ω. So this part had been set-up for matching the source to the tx-line and the 75Ω load. All you need to do is simultaneously ensure that the values of Rx || Ry = 75Ω, and that the voltage division generated by your Rx-Ry voltage divider is 0.5V into a 75Ω load to give you the whatever current you're after. The final Rx and Ry values should satisfy both conditions.

One equation will resolve to (Ry∙Rl)/(Ry∙Rl+Rx(Ry+Rl))∙V=0.5 the other will resolve to (Rx∙Ry)/(Rx+Ry)=Rl

Rl is your load impedance and V is your source voltage. You'll need to solve the simultaneous equations for Rx and Ry.

Answer 2

There is a very simple way to calculate the resistors.

The starting point is what you know already, unloaded signal is 1V, loaded signal is 0.5V, output and input impedances are 75 ohms.

If you short circuit the output, you get I=U/R=1V/75R=13.3mA short circuit current, so you need to get same short circuit current with 3.3V supply, so recalculate the resistance.

Then when you have that resistance, you need the unloaded voltage. Since you have 3.3V-1.0V=2.3V over the resistance you just calculated, you know the unloaded current.

Then you use that unloaded current and 1V to calculate the other resistor.

And then you can check if you end up with similar values used on other 3.3V SPDIF transmitters.

Answer 3

If it’s 5.5 V logic @ 3.3V consider the output already as 75 ohms +/50% then scale voltage with a 75 Ohm pad design. But 50 Ohms +/-50% at 5V.

verify yourself using Vol/Iol=Zol then load it to expect 50%.

103R Shunt and 220R Series divider ratio R/(R+Rs)=30% but source is ≈ 50R so series R reduces by 50 Solution Rs= 170, Rsh=103 Voc=1V, @ 75 Ohm Open cct.

I though it was 1Vpp no load and 0.5Vpp matched load. (?)

Thus 1V=30% of 3.3 with source = ——— Also If it was 3.6V logic (74ALCxx) @ 3.3V consider it as 25 Ohms and do the same.