Question

A 4-bit weighted-resistor DAC with inputs \( b_3, b_2, b_1, \) and \( b_0 \) (MSB to LSB) is designed using an ideal opamp, as shown below. The switches are closed when the corresponding input bits are logic ‘1’ and open otherwise. When the input \( b_3b_2b_1b_0 \) changes from 1110 to 1101, the magnitude of the change in the output voltage \( V_o \) (in mV, rounded off to the nearest integer) is _________. 

 

Hint:

In a weighted-resistor DAC, each input bit is associated with a weighted resistor, and the output voltage is determined by the sum of the voltages across these resistors. The change in output voltage can be found by calculating the output voltages for the two given input states and subtracting them.

Answer 1

Correct Answer: 250

Step 1: DAC Output Formula
The output voltage for a weighted-resistor DAC is: \[ V_O = -V_{{REF}} \left( \frac{b_3}{1} + \frac{b_2}{2} + \frac{b_1}{4} + \frac{b_0}{8} \right) \] {Step 2: Calculate \( V_O \) for tt{1110}}
For input tt{1110} (\( b_3b_2b_1b_0 \)): \[ V_O = -2 \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{0}{8} \right) = -2 \times 1.75 = -3.5 \, {V} \] Step 3: Calculate \( V_O \) for tt{1101}}
For input tt{1101}: \[ V_O = -2 \left( 1 + \frac{1}{2} + \frac{0}{4} + \frac{1}{8} \right) = -2 \times 1.625 = -3.25 \, {V} \] Step 4: Compute Change in \( V_O \)} \[ \Delta V_O = | -3.25 - (-3.5) | = 0.25 \, {V} = \boxed{250} \, {mV} \]