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My DAC (AK4384VT) specifies 3.4Vpp output voltage. does it mean relative to 0V such that there is -1.7V to +1.7V excursion of the signal or does it mean relative to +1.7V such that the signal lies between 0 to 3.4V?

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  • \$\begingroup\$ Which DAC do you mean? \$\endgroup\$ – Andy aka Aug 1 '17 at 7:22
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    \$\begingroup\$ pp means peak to peak ... \$\endgroup\$ – Solar Mike Aug 1 '17 at 7:37
  • \$\begingroup\$ DAC is AK4384VT, yeah i know pp means peak to peak \$\endgroup\$ – ravikumar Aug 1 '17 at 7:38
  • \$\begingroup\$ @ravikumar: Please add the part number and link to datasheet into your question. You could fix the lazy capitalising of words in both title and post too to improve legibility and credibility. \$\endgroup\$ – Transistor Aug 1 '17 at 7:43
  • \$\begingroup\$ It could mean either; the datasheet will wake it clear which. \$\endgroup\$ – Brian Drummond Aug 1 '17 at 8:04
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https://www.akm.com/akm/en/file/datasheet/AK4384VT.pdf

Read Note 6 of datasheet:6. Full-scale voltage (0dB). Output voltage scales with the voltage of VREF, AOUT (typ.@0dB) = 3.4Vpp × VDD/5. If you supply 5V, you will get 3.4Vpp swing.

The analog output will swing at half of 5V, which is 2.5V. The peak voltage will then be 2.5V + 1.7V = 4.2V and 2.5V - 1.7V = 0.8V. The signal will always remain in positive DC region, it won't goes negative unless is a different kind of DAC that uses dual supply.

When you are connecting the DAC into another opamp circuit with a reference voltage of 6V. It is a must to add a DC blocking capacitor between them, with the + of the cap facing the higher DC voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ hi, does 2.5V mean the VCOM in section 2. Analog Outputs? But there is mention of 2Vrms from LPF, does this 2Vrms mean relative to 0V? \$\endgroup\$ – ravikumar Aug 1 '17 at 7:50
  • \$\begingroup\$ Page 20 refers to a dual supply opamp, so yes that one is refer to 0V. Single supply opamps will not have 0V reference. VCOM is the half VDD, derive from the internal circuit of the ADC. \$\endgroup\$ – Jason Han Aug 1 '17 at 8:03

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