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I recently bought an USB sound card to do some measurements/experiments with. What I want to do first is to build a simple amplifier for the headphone output.

Therefore I took the sound card apart and traced out the in- and outputs. Interestingly the ground of the female audio jack is not connected to the USB ground, it is DC biased with about 1.67 volts with respect to USB GND (half the AVDD supply voltage).

The problem I now face is that I don't know what to do with the audio jack's ground on my amplifier board. The amplifier board uses the USB power supply and obviously I can't connect the audio jack's ground directly to the USB ground...

Can I just pull the audio jack's ground to USB ground via a resistor?

PS: The amplifier chip I intend to use is a MC34119 (Datasheet) and the USB sound card uses the VT1620A chipset (Datasheet)

Here is a simple schematic of the sound card (the male audio jack should be female actually):

My schematic

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I would assume it's applying that DC-bias because it is cheaper to use an amplifier to generate a mid-rail then use proper series-capacitors to decouple the line out signals.

Basically, the line outputs need to be able to swing positive and negative. The general way this is done is with a series capacitor in each line-out connection, which blocks the DC. However, to get low enough frequency response with headphones, which can be as low as 8Ω, you need a very large capacitor, which is expensive/physically-large.

I would guess they have a additional output driver in the IC that simply buffers a resistor divider, and by using that, they remove the need for the series capacitor.

If you need to reference the outputs to a different circuit, you should use series capacitors to block the DC. Ideally, you should have a high-impedance connection to the USB ground, for any DC, and then the local ground of your circuit should actually be coupled to the HP_COM with a series capacitor.


They actually mention this indirectly in the datasheet:

enter image description here

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  • \$\begingroup\$ so my amplifier circuit and the sound card don't have any direct commong ground when I couple the audio jack's ground with a capacitor to my local ground? \$\endgroup\$
    – sled
    Mar 25 '14 at 17:21
  • \$\begingroup\$ @sled - Somewhat. Basically, you don't want huge voltage differences, but you do want to allow for some ground differences between the local grounds for each USB cable (e.g. voltage drop along the ground wires in each USB cable, etc...). Therefore, you tie your device's local ground to the computer ground with a big resistor (say, ~10K or something like that), and then couple the soundcard "ground" to your local ground with a capacitor. That way, if for some reason, the soundcard drives the ground or something, it will still be present in the signal on your board. \$\endgroup\$ Mar 25 '14 at 23:18
  • \$\begingroup\$ Basically, I am considering it's possible (though very unlikely) that the soundcard device could modulate the "ground" output for some reason or other. Therefore, you want to use that signal as your local ground reference, so you see the entire output signal magnitude. By allowing AC from the pseudo-ground to couple into the device's local ground, the entire rails for your device will be modulated by any ground variation, which indirectly presents said data on the input lines. \$\endgroup\$ Mar 25 '14 at 23:21
  • \$\begingroup\$ If your device has an external ground, the ability to have ground modulation would not really be necessary as much, though having a good AC ground for the audio signal is a good idea anyways. \$\endgroup\$ Mar 25 '14 at 23:21
  • \$\begingroup\$ TL;DR - I'm probably overthinking it. \$\endgroup\$ Mar 25 '14 at 23:22

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