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I followed a tutorial on 8-bic DACs using an R-2R resistor ladder. I am outputting a sine-wave using data from 0 to 255 on pins 0-7. This goes through the pins into the R-2R DAC producing an analogue signal.

Now I would like to convert this signal into something that can be used by a piezo speaker. I understand (perhaps incorrectly) that the speaker cannot play this signal because it needs an AC signal, whereas the DAC output is 0 to +5V.

I've read that now I need to "bias" the DAC to electrically subtract 2.5v from the output to make it -2.5 to +2.5 volts. This signal should then be able to be fed into the speaker. Is this correct? How do I bias this signal, with a capacitor in series?

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  • \$\begingroup\$ Not detracting from Gerben's answer below, but if you were to ask this question in SE EE instead, you might get a more definitive resister / capacitor combination or even a different answer. \$\endgroup\$
    – Madivad
    Jun 29 '14 at 9:05
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You need to AC-Couple the signal. Just put a capacitor in series.

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  • \$\begingroup\$ Any sort of capacitor ? \$\endgroup\$
    – Kingsley
    Jun 25 '14 at 20:19
  • \$\begingroup\$ I'm not sure. I'd avoid polarized electrolytic caps. Just try some values, or do a google search. \$\endgroup\$
    – Gerben
    Jun 26 '14 at 7:54
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    \$\begingroup\$ It depends on how low of a frequency you want to be able to reproduce, note that the circuit above is a high-pass filter, so low frequencies will be attenuated (which is what you want - you are trying to attenuate DC and bring the zero signal level up to the half-way point). The R part depends on what values you use for your R2R ladder, since by design the R2R ladder has an output impedance of just R Ohms. The cutoff is 1/(2*piRC), so you can figure out how big the C needs to be for a given acceptable cutoff frequency. \$\endgroup\$
    – Zuofu
    Jun 29 '14 at 14:09

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