3
\$\begingroup\$

I have an 8-bit waveform description coming from an FPGA into the DAC, and I want to read the analog output on an oscilloscope. I figured I should implement this setup found in page 11 of the data sheet, but I am not sure how to do so.Basic Bipolar Output Operation. Found on page 11 of the data sheet

What I am uncertain about is the 10V between the two 10k resistors. Is it just that the node voltage will always be 10V? Or should a 10V source be connected there?

\$\endgroup\$
  • \$\begingroup\$ It seems like you have to supply the voltage . The output of DAC is basically current . The 10V is used so that the output can both source and sink current i.e you can get positive and negative swings across the output with reference to 10 V . For sanity check ,just check the output for all zeros case .Ve0 - Ve0bar =-9.9 -10 =-20 V i.e 10V below 10 V ref .The exact opp happens for all ones case \$\endgroup\$ – Kishore Saldanha Nov 15 '15 at 4:46
  • \$\begingroup\$ a measurement is usually indicated by an arrow, a circle is a connection. \$\endgroup\$ – Jasen Nov 15 '15 at 8:51
2
\$\begingroup\$

The DAC08 is a current-mode multiplying DAC (the output is a scaled version of the reference current). It can only sink current; it cannot source it. The schematic in the datasheet shows how the DAC works, it's basically just a current mirror with multiple outputs, each output scaled by the bit weight, with the outputs switched between the positive and negative DAC outputs. So you need to provide an external bias for the DAC to work - this is provided by the 10V supply fed through the 10k resistors. The DAC will then sink current through those resistors to produce the voltages listed in the table.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.