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Best/cheapest way to cheaply get a 5-bit DAC: R-2R, or R = 2^N

October 30Hits:28

I can either use an R-2R divider or a R = 2^N divider to get a 5-bit DAC.

  1. Which will be more accurate? R = 2^N I suspect, but;
  2. Which will be cheaper? I think R-2R because it can use standard values, but...
  3. I've heard R-2R have issues with monotonicity, when the output doesn't change for the same amount with each count. Would this be an isuse with 5-bit DAC - would any such monotonicity simply be overwhelmed by divider error?
  4. There's also the trade off of size, R = 2^N will be larger.
  5. Also there's the possibility of using an IC to do this. I'd be open to doing this as well, if it wasn't very big and could be interfaced over I2C.

The DAC is going to be used to feed a comparator. The comparator will use this reference to compare against an incoming video signal.


If it were me I would use an IC to do the DAC for me. The manufactures of the ICs take a lot of time and consideration for getting matched components and put them in the best configuration.

The TC1320EOA is probably a good solution for what you want. It has I2C and in an 8pin SOIC configuration. I bet you will be hard pressed to get a design this small if you take into account the traces from your micro as well as the resistors and possibly smoothing caps and such. It has 8 bits of resolution, which is more then you need, but I wouldn't think this would be an issue. I also think the cost of it will be fairly equal if you account for the more pins needed on your micro for a home brewed solution and the PCB space required.

If you have a spare PWM output on your micro, driving it into an RC filter will be the cheapest way to do this, and will take up very little space. This will be slower than the resistor arrays though.

Tags:resistors, dac

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