With the help of a neat diagram, explain the working of a weighted-resistor D/A converter.
Ans
Weighted Register D/A Converter:
Digital input that has 4 bits is applied to a register network throughout electronic switch. Such electronic switch generates current I at MSB (consequent to Logic 1), I/2 at the subsequent lower significant position. The whole current produced will be proportional to digital input. Such current can be converted to consequent voltage by using an op-ampere. Circuit of this convertor is referred to as weighted register converter because the resistance values are weighted in accordance along with the binary weights.
The current Ii is specified by
Ii = I_{N-1} + I _{N-2 }+ ........+ I_{0} where I_{N-1} = V_{N-1}/ R, I_{N-2} = V_{N-2}/ 2R, I_{N-3} = V_{N-3}/4 R and also V(O) if bn = 0 and V_{N} = V(1) if bn = 1
For straight binary inputs V(0) = 0 and V(1) = - VR and the output voltage is specified by:
V_{0} = -[-V_{R}] [(R_{F}/R) b_{n-1} +(R_{F}/2^{n} ) b_{n-1} +(R_{F}/2^{2}R) b_{n-1} +_______________(R_{F}/2^{n-1}R) b_{0}]