shayaan_musta said:
Hello experts!
Here is 3-bit digital to analogue converter.
I have a question related to this.
1)What does it mean if you have to say to perform operation on it with 101?
This mean you put 101 at the input according to the voltage I specified below. You should get +5V output.
2)And we can obviously conclude and see output voltage on oscilloscope. But by calculation how can you predict the answer(i.e. output voltage)? Is there any formula?
See below.
3)And What is the criteria if I want to add 1 more resistor at input(making 4-bit DAC) either it is last or may be it is the first resistor. And of course adding another input resistor must change the value of feedback resistor, then what it should be?
To make it work, you have to have precision voltage as input. Let say you use a +5V reference and "1" and 0V as "0". Your full scale is from 0 to +7V output. Each LSB is 1V.
The 10K input is the LSB, 5K is the middle, the 2.5K is the MSB in this case. The first stage of the circuit is nothing more than a summing circuit. With virtual ground on the +ve input and "1" is by putting a +5V on the resistor, when you put "1" at the 10K, you source 0.5mA to the summing junction. Feedback resistor of the first stage is 2K, so the output goes to -1V for the LSB. The middle resistor is 5K, so when you put a "1", you source 1mA into the junction, the outpur of the first stage will go to -2V. And the same for the MSB give -4V.
So I defined each LSB is 1V, then if I put 101 at the input, The current I will source to the junction is 0.5+0+2=2.5mA. The output of the first amp will go to -5V due to the 2K feedback resistor.
With this method, you can calculate all the combination. Yes you can add more bits, say you put one more 20K resistor and use it as LSB instead, you get a 4 bit DAC.
The limitation of this method is you ability to get precision resistors. Normal resistor is like 1%, so you really cannot get anything better than 6 bit. For 10 bits, you need something like 0.025% or better. Remember you need to be accurate to +- half LSB, so even the resistor need to be 0.5% accurate in order to get to that as the input. The worst part is the Feedback resistor(2K in your circuit), that is another source of error. So the total error is twice the accuracy of one resistor. That is the reason I spec 0.25%. But the feedback resistor contribute to FULL SCALE ERROR, not the bit error. That is the reason if you read data sheet, they always specified separate full scale error and bit error.
Hope this help.
The second amp is nothing more than to invert to 0 to +7V output.
h
