Q.Explain with neat circuit diagram working of a binary weighted resistance DAC

A binary weighted resistance DAC feeds each digital bit through a resistor weighted in powers of two into a summing amplifier. …

[!TLDR]

Weighted resistors R,2R,4R,8R feed a summing op-amp; the output is proportional to the binary value.

Circuit: the four digital inputs B3 (MSB), B2, B1, B0 (LSB) are applied through weighted resistors $R, 2R, 4R, 8R$ respectively to the inverting (−) summing node of an op-amp; a feedback resistor $R$ connects output to that node, and the (+) input is grounded. Each bit is at 5 V for logic 1 or 0 V for logic 0.

Working: the inverting node is a virtual ground, so each bit injects a current inversely proportional to its resistor — the MSB (through $R$) injects the largest current and the LSB (through $8R$) the smallest, in the ratio 8:4:2:1. These currents sum and flow through the feedback resistor, giving

$V_o = -R\left(\frac{B_3}{R}+\frac{B_2}{2R}+\frac{B_1}{4R}+\frac{B_0}{8R}\right)$ …