Accuracy of resistor ladders:Resistors used with the more significant bits must be proportionally more accurate than those used with the lower significant bits; for example, in the R-2R network shown above, inaccuracies in the Bit4 MSB resistors must be insignificant compared to R/32 (i.e., much better than 3%). Further, to avoid problems at the 10000 to 01111 transition, the sum of the inaccuracies in the lower bits must be significantly less than R/32. The required accuracy doubles with each additional bit--for 8 bits, the accuracy required will be better than 1/256 (0.4%). Within integrated circuits, high accuracy R-2R networks may be printed directly onto a single substrate using thin-film technology, ensuring the resistors share similar electrical characteristics. Even so, they must often be laser trimmed to achieve the required precision. Such on-chip resistor ladders for digital-to-analog converters achieving 14 bits accuracy have been demonstrated. On a printed circuit board, using discrete components, high precision resistors of 1% accuracy may be employed for a 5 bit circuit, however with bit counts beyond this the cost of ever increasing precision resistors becomes prohibitive. Often, even for a 5 bit circuit, the home constructor will use either a matched set (by measuring a batch of 100 5% resistors and selecting the best set of 15) or by trimming down individual resistors to a common value (by adding high value resistors in parallel).
I have sorted out the schematic, so it goes from D0 to D7, not up to D8 . After looking around on some component suppliers' websites, I have found the TLC7524CN from RS. at the minute, I am scrolling through the datasheet to see how it works and how to apply it for the analog circuit. Hopefully I will not get any noise from the IC, but I will still add in the capacitors. The only problem is that I do not know where to put them . As it is to filter noise from the input of the buffer, I would assume that it would go from the buffer input to 0 volts, however I do not know if it is right...Onions.
Full Scale Error ±2.5LSB ?
The shift register needs all the pins that have already been discussed if it is going to work.The latch pin is different as it does not need to be connected, although it can be useful ... By taking the latch pin low, you can shift out the data without the bits changing. This means that the output value will stay the same until all the bits have been shifted out, and the newvalue can be seen.
You are not following Figure 3 or 4 of the datasheet for the output to the op amp.Try it as suggested first to avoid damaging the chip.
It is possible to operate the current-multiplying DAC in these devices in a voltage mode. In the voltage mode, a fixed voltage is placed on the current output terminal. The analog output voltage is then available at the reference voltage terminal. Figure 1 is an example of a current-multiplying DAC, which is operated in voltage mode.
Yes - follow Figure 3 or 4.