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I'm working with a chip that outputs audio signals to headphones that are referenced to a virtual ground. Since I'm connecting the audio signal to a speaker instead of headphones, I'd like to implement a basic audio amplifier on the same board (sharing the same power source and ground).

I'm thinking of using an LM386 since they're a jellybean part.

However, I'm not sure how I would go about doing this. I've read the data sheet of the LM386 but the application notes don't really go into handling a virtual ground. The virtual ground is roughly 1.2V when referenced to actual ground and can source/sink about 150mA before it goes into short circuit protection mode.

All the circuits I've come across just ground both the inverting input of the amplifier and the input signal ground (why?). Obviously, if I did this, I would short the virtual ground to actual ground.

The schematic I am specifically referencing is the one below (source):

enter image description here

Is it just a simple matter of connecting the virtual signal ground (instead of actual ground) to the inverting input of the amplifier or would I have to look into building in a isolated power supply to power the LM386?

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Connect your "virtual ground" (or as I would call it an AC ground) via a capacitor to real ground. This prevents the DC levels on your virtual ground getting upset by a conducting connection. Make sure the capacitor value is reasonably high; in your circuit you are using 1 kohm input resistors so, at 20 Hz, the capacitors impedance should not be higher than 1 kohm. Sounds like 10 uF should do the job. I'd use a non-polarized one although it will probably be OK with an elecrolytic.

You may need to put a 10 uF capacitor in series with each 1 kohm resistor too.

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  • \$\begingroup\$ Thanks, I appreciate the answer but I'm wondering if you could provide a short explanation on how/why this works? Essentially, we're AC-coupling the AC-ground as you like to call it, to the real ground but how does that affect the signal going to the non-inverting input of the amplifier? \$\endgroup\$ – tangrs Feb 24 '17 at 12:26
  • \$\begingroup\$ AC coupling simply re-sets the DC level of the signal. It will not affect audio signals as audio signals are AC and pass straight through the capacitor (and only get a DC voltage shift in between). \$\endgroup\$ – Bimpelrekkie Feb 24 '17 at 12:32
  • \$\begingroup\$ I'm suggesting caps on all three wires feeding the input and real ground. \$\endgroup\$ – Andy aka Feb 24 '17 at 12:32
  • \$\begingroup\$ @Andyaka Ahh I see. So we're removing the DC components from the audio signals. In that case, I think it's even possible that the virtual ground doesn't need to be connected at all since the input signal has the DC component (of the signal referenced to real ground) removed already. However, I'd have to check with the data sheet to make sure. \$\endgroup\$ – tangrs Feb 24 '17 at 12:39
  • \$\begingroup\$ It will likely be a bit noisy if you don't connect grounds capacitively. \$\endgroup\$ – Andy aka Feb 24 '17 at 12:43
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The LM386 does not use a virtual ground as it has ground referenced inputs. It does have a DC biasing voltage at the output around which the output swings. The circuit sets this DC level by itself.

You're confusing matters a little as you talk about virtual ground as that is often used at the input side of amplifiers. But here you want to use it with the output load. Which is possible though.

I do not like how that circuit is drawn as it suggests that you can "just" connect an audio signal to the AUDIO inputs. That is not the case, look in the datasheet of the LM386, the DC level of input pin 3 needs to be ground, that is not explicitly provided in this schematic as the AC coupling caps are missing.

Also a resistor to ground is often added but not really needed as the chip has a 50 k resistor from pin 3 to ground already.

The output of the LM386 will be at a certain DC level and it will not be the 1.2 V you have as virtual ground, it will probably be around half the supply voltage. This means you must use an AC coupling capacitor like the 1000uF cap in the schematic. This 1000 uF is for an 8 ohm speaker. Check the impedance of your headphones and scale that value accordingly, so if the headphones are 30 ohms, use 220 uF or so.

You can connect the ground side of the headphones to the virtual ground of 1.2 V provided it is decoupled well enough. I'd use at least the same value as used for the cap on the signal side of the headphones. So for 30 ohms headphones, 220 uF should do the job.

Edit:

After unconfusion it appears that the 1.2 V DC level is at the input. Then simple AC couple the audio signals into the LM386, like this:

enter image description here

Note the 10 uF capacitor on the left and the 10 kohm potmeter. Instead of a potmeter you could use 2 resistors to have a fixed ratio.

Connect LEFT_AUDIO and RIGHT_AUDIO each with a resistor (1 kohm for example) to the left side of the 10 uF capacitor.

The 10 uF capacitor will block all DC so the DC level at the right side of the capacitor (which is the input of the LM386) will be ground level.

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  • \$\begingroup\$ I'm actually talking about the input side of the amplifier. The output of the audio chip is not AC coupled. The signal from LEFT_AUDIO and RIGHT_AUDIO might be between 0 to 2.4V volts referenced to ground but be between -1.2V to 1.2V when referenced to the virtual ground output of the chip. \$\endgroup\$ – tangrs Feb 24 '17 at 12:22
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    \$\begingroup\$ OK, so the input signal has a DC level on it, then you must AC couple it because as I wrote, the input must be at DC ground level. I would AC couple the audio signal into a potmeter (for volume adjustment) and then follow the application circuit as in the datasheet. \$\endgroup\$ – Bimpelrekkie Feb 24 '17 at 12:29
  • \$\begingroup\$ Thanks, as per the comment on the above answer, I think there's probably no need to even connect the virtual ground to anything if it's already AC-coupled though the high-pass filtering cap at Vin. \$\endgroup\$ – tangrs Feb 24 '17 at 12:42

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