CDs use 16-bit signed integers to represent each sample. So each sample is an integer between -2^15 and 2^15-1. A typical consumer CD player has two RCA outputs coming out the back of it, one for the right channel, one for the left, and no volume control. If the right channel on a CD track is simply a (discretized) 1 kHz sine wave with an amplitude of 2^14 "counts", what will be the amplitude (in volts) of the signal on the RCA output? (Just to be clear, when I say "amplitude", I mean one half the peak-to-peak amplitude. So the maximum of my digital sine wave is 2^14 counts, and the minimum is -2^14 counts.)

  • \$\begingroup\$ So, to clarify... You're asking for the peak output voltage of the audio signal on the RCA line-output from your home stereo CD player? \$\endgroup\$ Commented Feb 29, 2016 at 0:37
  • \$\begingroup\$ Yes, basically. If you told me the peak possible output voltage on the RCA line out, that would tell the answer to my question. I.e. if the peak possible output voltage was (2^15-1)/2^15*Vscale, the answer to my question would be Vscale/2. \$\endgroup\$ Commented Feb 29, 2016 at 17:39

3 Answers 3


There isn't a strict standard, but "nominal" line level for consumer-grade gear is typically -10 dBu, which is the voltage that would dissipate 0.1 mW in a 600 Ω load. This works out to about 0.245 Vrms, or 0.346 Vpeak.

Now, the level you are asking about is relative to the full-scale range of the DAC. The key question is: What is the relationship between dBFS and dBu? This depends on how much "headroom" the designer of the CD player decided to build into the DAC. A typical relationship is that full-scale is 10 dB above the nominal level, which means that the clipping level of the DAC (full scale) is 1.095 Vpeak (0.775 Vrms, or 0 dBu).

Therefore, a -6 dBFS signal (half the voltage) would be 0.549 Vpeak, or 0.388 Vrms. YMMV

  • \$\begingroup\$ OK, thanks. And not to quibble, but I really want to be 100% clear on this: that Wikipedia article also says that the line level for consumer devices is -10 dBV (rather than -10 dBu). Which is correct? \$\endgroup\$ Commented Feb 29, 2016 at 17:47
  • \$\begingroup\$ The difference is only 2.2 dB, which is hardly worth worrying about. I deal mostly with professional gear, and work mainly with dBu. But it's a tiny adjustment of the input pad either way, and falls within the scope of YMMV. \$\endgroup\$
    – Dave Tweed
    Commented Feb 29, 2016 at 18:05

I made a .wav file with a full-range 1 kHz sine wave on it, burned it to an audio CD, and played it out a Yamaha CDC-565 CD player, and measured the amplitude of the resulting signal on the RCA output (this CD player has no volume control). The amplitude was 3 V, almost exactly (i.e. 6 V peak-to-peak):

Yamaha analog output

I took the same sound file and played it on my Sonos system, and recorded the analog output from a Sonos bridge (that's a box for hooking your Sonos system up to a traditional component stereo system). Sonos lets you set the Bridge output to be "Fixed" or "Variable". The "fixed" setting is designed to be line level, with the intent that you control the volume on your receiver (or preamp). The "variable" setting allows you to control the volume via the Sonos software, by modulating the amplitude of the signal on the RCAs coming out the back of the Sonos. With the Bridge output set to fixed, the amplitude of my test file on the RCA was again 3 V:

Sonos Bridge fixed output

With the Bridge output set to variable, and the volume on the Sonos software set to the maximum, the amplitude on the RCA was again 3 V. In this case, by turning the volume down in the Sonos software, I could reduce the amplitude on the RCA.

So two different devices for converting digitally-stored audio to a line-level analog signal, of very different vintages, both produce a 3 V sine wave when the 16-bit digital signal is of amplitude 2^15 counts. This leads me to wonder if this isn't a common practice.

The question asks for the peak amplitude of a signal half this large, which would be 1.5 V.

Several online sources (see below) indicate that a typical consumer CD player, when playing a full-range sine wave, produces an analog output with RMS amplitude equal to approximately 2 V. This implies a peak amplitude of sqrt(2)*2 V ~= 2.828 V. The RMS amplitude of the 3 V sine waves produced by the equipment I tested would be 3/sqrt(2) V ~= 2.121 V, which could be construed as "approximately" 2 V, I suppose.

So it seems like the final answer to the question is that the peak voltage of a half-full-range digital signal will likely vary across makes and models, but will probably be around 1.5 V.



"The processor under test is driven by the digital code representing a full-scale or 0dBFS 1kHz sinewave. [...] The standard CD-player output voltage is 2V RMS, with units varying between 1.74V on the low side (the Audio Research DAC1) and a whopping 7.2V on the high side (the Theta DS Pro Basic). Most CD players and processors put out between 2.2V and 3.5V."


"Yes. Digital full scale = 2Vrms. An unwritten standard."


"Before digital sources, nominal audio levels were specified and the quality of the product determined just how high a signal could be accommodated before clipping would occur. With the advent of the CD player, suddenly a source was available that had a maximum output level that could not be exceeded. That level was specified to be 2 Vrms."


According to the Wikipedia Article:
Line outputs usually present a source impedance of from 100 to 600 ohms. The voltage can reach 2 volts peak-to-peak with levels referenced to −10 dBV (300 mV) at 10 kΩ. The frequency response of most modern equipment is advertised as at least 20 Hz to 20 kHz, which corresponds to the range of human hearing. Line outputs are intended to drive a load impedance of 10,000 ohms; with only a few volts, this requires only minimal current.

So, if you're wanting rectified voltage/Vpk, that should be <=1V.


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