I have a DAC (datasheet available here) that has a SLEEP mode (see page 14). When SLEEP mode is activated, my oscilloscope reads an output voltage of about 1.65V on each output, which is about half of the supply voltage if 3.3V.

I want to check that this output voltage makes sense, but the datasheet doesn't seem to provide the information. (It specifies that the output current is turned-off, but that's it.)

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    \$\begingroup\$ This is a current source DAC, so it does not control its output: you do, by providing a load resistor to ground. What value load resistor did you use? \$\endgroup\$ – Wouter van Ooijen Jul 7 '12 at 15:35
  • \$\begingroup\$ @WoutervanOoijen: The design has quite a few resistors. Which one exactly is the load resistor? (Please see the screenshot.) \$\endgroup\$ – Randomblue Jul 7 '12 at 15:52
  • \$\begingroup\$ A load resistor is one that you connect to an output (IOUTA and IOUTB), with one end of the resistor at the output and one end of the resistor to ground. It's not specified in the schematic. \$\endgroup\$ – Shamtam Jul 7 '12 at 15:59
  • \$\begingroup\$ R140 and R141 are your load resistors, they give you 1 V out per 20 mA. But there are other resistors which influence your load: R139 in series with R142||R237 is 720 ohm, in parallel with the 49.9 ohm gives 46.7 ohm. That's a 7 % error on a 0.1 % resolution DAC. \$\endgroup\$ – stevenvh Jul 7 '12 at 16:03
  • \$\begingroup\$ @stevenvh: Thanks a lot. Could you please detail your calculations? I'm a little rough on calculating resistance from the diagram. \$\endgroup\$ – Randomblue Jul 7 '12 at 16:37

The opamp output you are seeing is normal.

If we examine the circuit, starting with the opamp non-inverting input. We assume the input is infinite impedance so we first work out what voltage it's seeing:

We work out the bottom of the divider:

((R139 + R141) || R142)) = 1/(1/269.9 + 1/1000) = 212.54 ohms.

The top of the divider is R327 1k, so the voltage at the centre is:

3.3V * (212.54 / (1000 + 212.54)) = 0.578V

So the non-inverting input sees 578mV.

Now we look at the inverting side. We know that if the non-inverting input is 578mV, the inverting side must be at 578mV too. We also know that if the DAC outputs are off, then effectively the input is 0V, and the input resistance is R140 + R137.

So the input resistance is 269.9 ohms, and the feedback resistance (R326 || R136) is 1k || 1k = 500 ohms.
We assume the inverting input has infinite resistance, so we are left with a voltage divider.
With 0V at one side of the voltage divider, 578mV in the middle, the top side should be 0.578 * (269.9 / (500 + 269.9)) = 1.65V, which is what you are seeing.


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