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The datasheet for the AD667 DAC shows how to connect the output to internal feedback resistors in different ways, in order to get different output voltage ranges.

  • For output range of 0 to +5 V, it says to connect the output to pin 2, a 5k feedback resistor.
  • For output range of 0 to +10 V, it says to connect the output to both pins 1 and 2.

Pin 1 is also a 5k feedback resistor, which is in series with the pin 2 feedback resistor. So it seems that the feedback current would all go through pin 2, because the pin 1 feedback resistor is in parallel with a short.

It would make more sense to me if it said to connect pin 9 to ONLY pin 1 to get an output range of 0 to +10 V, that way the feedback resistance is 10k, twice as much as 5k, twice as much gain, so 10 V range instead of 5 V range.

Can anyone explain why they connect pin 9 to both pins 1 and 2?

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  • \$\begingroup\$ Connecting just pin 1 to pin 9 gives you a 20V range, not 10V. You should include more of that table instead of cropping off the fourth column. \$\endgroup\$
    – Finbarr
    Commented Jun 12 at 0:17

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The difference is in the input codes that you failed to highlight. Offset binary is bipolar so it needs less gain to swing from the null point. Straight binary needs twice the gain (for a given bit) to operate in this unipolar mode of operation.

Electrically, what you are saying is correct. The reason they want you to "short" upper 5k is simply to keep the input from floating.

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  • \$\begingroup\$ but the input is connectoed to 5k, not floating \$\endgroup\$
    – Jasen Слава Україні
    Commented Jun 11 at 23:50
  • \$\begingroup\$ @JasenСлаваУкраїні It's floating if you don't connect pin 1 to pin 2. Sure, internally Pin 1 is connected to a resistor. But where are you going to connect that pin with respect to "outside the chip"? Remember, that pin 1 resistor goes into the high impedance input of the opamp so it is very sensitive - normally, you can tie unused inputs to either, ground, V+, V-, or VCC/2 depending on the IC. But here it's part of the amp FB so it has to "ride" with pin 2 and pin 9 to stay equipotential during operation. \$\endgroup\$
    – MOSFET
    Commented Jun 12 at 0:08
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    \$\begingroup\$ @JasenСлаваУкраїні Also, you don't want pin 1 to be an antenna while left open. Interference will couple into the FB path. Another reason is if there is any stray leakage currents on the board. This will manifest as an offset in output and compromise performance in your precious "true 12-bit DAC". \$\endgroup\$
    – MOSFET
    Commented Jun 12 at 0:13
  • \$\begingroup\$ I understood your explanation for why we don't want to leave an input pin floating, but can you explain in more detail why the unipolar output range doubles when connecting pin 9 to both pin 1 and pin 2? \$\endgroup\$
    – charles
    Commented Jun 12 at 0:34
  • \$\begingroup\$ @charles see my comment on the question, you're missing the bigger picture \$\endgroup\$
    – Finbarr
    Commented Jun 12 at 0:38

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