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I have seen a few different ways of adding DC bias to an audio signal. I have simulated them and they all give me similar results, but I can't figure out why choose A over B or C. My audio source will be a Line Level Audio -2V to +2V AC passed through a 220uF coupling cap and then a low pass filter(RC, 2 pole). The signal will be read by an ADC.

First way is using a Voltage divider: Simple Biasing Circuit

enter image description here

This is pretty self-explanatory and I understand how it works. I have also seen this same design using a diode but could not find an example.

Next example: How to read an audio signal using ATMega328? - picture is from endolith's answer.

enter image description here

Another one that I have seen is: I don't quite understand this FET-BJT preamp circuit

And the schematic is for a pre-amp, and there are 2 versions and both add a bias.

enter image description here

My question is what is the best practice for adding the bias to an audio signal? What are some of the other ways to add a DC bias to the signal?

Edit/Update: Looking at the answers - using the second one looks like it will work best for my application, using something like this. Are there any other improvements I can make? Other then Stable Vref/power rails.

enter image description here

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    \$\begingroup\$ Small note, you say you have a 220 uF decoupling cap. I think you might be referring to a coupling cap. \$\endgroup\$ – Kellenjb May 18 '11 at 12:16
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    \$\begingroup\$ The signal is going to be read by an ADC, for a BPM counter(beats per min) And Kellenjb you are correct coupling cap and not decoupling - fixed \$\endgroup\$ – jsolarski May 18 '11 at 12:31
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    \$\begingroup\$ As a note: I use the first solution with no problems but my circuit is definitely not a high fidelity one as it is on a model aircraft with lots of other noise (motors, servo, wind etc.) \$\endgroup\$ – Thomas O May 18 '11 at 16:11
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    \$\begingroup\$ You're intentionally trying to isolate low frequencies with those RC filters, right? \$\endgroup\$ – endolith May 19 '11 at 14:17
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    \$\begingroup\$ @endolith Yes I am intentionally trying to isolate low frequencies, under 3KHz \$\endgroup\$ – jsolarski May 23 '11 at 6:08
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Don't use the first circuit. Any noise or spikes on the power supply will be mixed with your signal. Because the bias point is connected directly to the signal, you can't filter out power supply noise without also filtering out the signal.

Do use the second circuit. It produces a mid-point voltage that is tightly coupled to ground, so the DC component is half the supply, but the AC component (noise and spikes) is filtered out by the capacitor. That's not a complete circuit, though, you still need to connect it to your signal.

This is what you're trying to do:

simple DC biasing

The output is the same as the input, just shifted upward by 2.5 V. The resistor on the input ensures that the input side of the capacitor is at 0 VDC, to prevent pops during connection. The resistor on the output side of the AC coupling cap biases that side to the DC bias voltage. If your circuit already has a clean, low impedance DC bias voltage source, connect to that. Otherwise, you can use circuit #2 to generate the bias, like this:

Circuit showing DC biasing of an AC signal

(The simulation takes a loong time to reach the DC bias value, though. Hit the "Find DC operating point" menu entry to settle it.)

The DC bias voltage is produced by a voltage divider and capacitor to filter out power supply noise. Note that if you use the same Vbias point for multiple signals, they can crosstalk through this point. Larger bias cap reduces crosstalk. Larger coupling capacitor improves low frequency response. But make them too large and they'll take a long time to charge when you flip the power switch.

The 3rd diagram is not a biasing circuit; it's a microphone preamplifier.

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  • \$\begingroup\$ As for the pre-amp would that circuit bias signal if I were using a un-amplified(phono) signal or microphne? or should it just go through a pre-amp that doesnt bias and bias the signal near the filter? \$\endgroup\$ – jsolarski May 19 '11 at 7:47
  • \$\begingroup\$ @jsolarski: I don't understand your question about the preamp. That circuit is a high-gain mic element preamplifier. Are you just looking to buffer the signal before it goes into the ADC? \$\endgroup\$ – endolith May 19 '11 at 14:19
  • \$\begingroup\$ as to my earlier comment, I will have to post another question when I get to that point, my concern is I have multiple signals, some are line level and some will be very low level mic inpputs or phono inputs. I just need a way to keep the levels in the same range, when plugging in different sources. \$\endgroup\$ – jsolarski May 23 '11 at 6:25
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    \$\begingroup\$ @jsolarski: It'll be like any audio mixer input, then. You'll need some variable gain control for the unpredictable low-level inputs, and maybe separate jacks for the line-level, depending on the connectors you want to use. For a BPM counter, low-noise probably isn't that important, so you could just use a single op-amp gain stage with a wide gain range. Alternately, a narrower gain range and a 20 dB pad switch. \$\endgroup\$ – endolith May 23 '11 at 14:06
  • \$\begingroup\$ You say that the OP's first circuit will have the power supply noise amplified but your first circuit has the power supply directly connected to the output. Ins't a capacitor missing there? How is the first circuit filtering the power supply AC (noise)? \$\endgroup\$ – SpaceDog Apr 10 '18 at 2:29
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The simplest method is the first image you linked to. It will do the job, but has a big draw back for your application. If your supply lines have any noise on them, the noise will be added to the signal you are trying to measure.

The second method is almost identical to the first method. It's big advantage over the first method is that noise on the supply lines wont have as big of an effect on the signal itself.

The third method is over kill for what you are wanting to do. It is designed to give higher power outputs, but since you are just reading it with an ADC there is no reason you need it.

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    \$\begingroup\$ If the voltage from which the ADC derives its reference feeds the resistor divider, I don't see that having that voltage coupled into the audio feeding the ADC will be much of a problem. Indeed, if there's any wavering in the ADC reference, I would think having it coupled to the ADC input with 50% amplitude would be better than having the reference move and the input not, although the audio signal would be degraded in any case. If the ADC reference is an absolute voltage which is not available on an output pin, your bias should likewise be an absolute voltage (from a regulator or whatever). \$\endgroup\$ – supercat May 18 '11 at 16:42
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The first circuit, the simple resistor divider, is by far the easiest, quickest, and cheapest solution. It is also the solution that most audio circuits use. Unless you are wanting pro-audio levels of performance, this is the method that I would recommend.

The "correct" solution would be to have a separate power rail that is at the bias voltage. Run your audio signal through a DC blocking cap then have a resistor to the bias power rail. This approach has less noise and harmonic distortion than the simple resistor-divider-- although the difference in performance is only important for those in the pro-audio world and not worth bothering with for most.

One case where the "correct" solution is worth the trouble for the average circuit is when the ADC itself provides the bias voltage rail. Some ADC's will output that voltage and all you have to do is use it. This is nice because you can get better precision than any other solution. Sometimes I have had issues, however, where I had to take this output from the ADC and run it through a unity gain op-amp based buffer so it had the drive strength to work properly.

The other two solutions you mention would work, but I wouldn't bother. They are somewhat tweaky and don't offer any important benefits that the simple resistor-divider or the "correct" solutions provide.

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