Question about ground-relative voltages in an audio switching/mixing circuit - Page 1
Forum section: electronics / projects

author: Muny
author_rank: Contributor
timestamp: December 28, 2015, 08:32:53 pm
content: I'm designing a two-input two-output audio mixing/switching device which I presume will need to use some sort of analog switch or multiplexer to do the switching.I drew up this diagram to explain what it should be able to do:https://pcbx.us/muny/bvxz.jpg(embedding it here made the post very large)Each switch would need to have 3 channels (ideally the IC would be a dual 3-channel switch, one for left audio and one for right audio).I started hunting around for chips to do the job, and I came across this TI part:http://www.ti.com/lit/ds/symlink/ts5a3359.pdfThis has almost exactly what I think I need, but one specification worries me.As far as I'm aware, the audio signals I will be switching are AC, with a maximum peak voltage of 0.447v. That means that the switch will need to output a maximum of -0.447v at the bottom portion of a wave. That is almost outside the absolute maximum ratings of the chip, so I think I might be taking the wrong approach at this.Will I need to bias the audio signal so that the input to the switch does not reach below 0v? If so, would biasing it as soon as it enters the device suffice for the entire life of the signal through the circuit?

author: ThorTheNorseman
author_rank: Newbie
timestamp: December 29, 2015, 04:57:27 am
content: Yeah, normally what you do if you lack a dual supply is that you decouple the signal through a capacitor and bias it. The biasing part is done by passing the signal through the middle of a voltage divider across your supply rails. If you're going to use op amps or do filtering further down the chain though, you're going to need a low impedance virtual ground. One way of getting that is to buffer a voltage divider (outside of the audio path) through a voltage follower, and then use that as the virtual ground in your audio path.Supposing your original rails are GND (0V) and +5V, and you divide that in half with two identical resistors in the 1-10 kOhm range, then that's 2.5V into and out of your voltage follower. In your audio path, you mentally relabel your 2.5V output as VGND. Your original 5V rail is then +2.5V relative to VGND, and your original 0V rail is -2.5V relative to VGND. Op amps connect with V+/V-/GND to +2.5V/-2.5V/VGND. The ground of any RC filters would also connect to VGND. At the outputs of your circuit, you decouple with capacitors again, and use the normal GND.For line level signal inputs, you want a 10 kOhm impedance to the _cable_ ground, so it would make sense to use two 10 kOhm resistors for the input biasing circuit, and not fall for the temptation of simply biasing it with a single 10 kOhm resistor to VGND. Both approaches might work, since the idea of input impedance is basically to have a load for the signal to drive, but I think lacking a direct path to cable ground might have some negative implications for shielding and interference.And if you've got chips that aren't designed for dual rails, no problem: Voltages are relative. When chips have a minimum/maximum input/output voltage rating, that's relative to what your Vdd and GND is. If you have a dual supply of +/- 12V, and your need your chip to handle negative voltages, just ground it to the -12V rail and use GND as Vdd. Your chip will see a 12V supply and will think that -12V is 0V (GND) as far as it is concerned, because that's how you wired it. This is the beauty of voltage: It's a potential across two measured points in a circuit and not an absolute value. There is no true ground, there is only what you designate as ground.

author: Muny
author_rank: Contributor
timestamp: December 29, 2015, 05:48:06 am
content: Thank you very much; you have eliminated my confusion and then some! Well explained.