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 $b_3 b_2 b_1 b_0$ 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 ________.
The circuit is a 4-bit weighted-resistor Digital-to-Analog Converter (DAC) using an inverting Op-Amp. The output voltage ($V_O$) is proportional to the weighted sum of the input bits.
1. Output Voltage Formula
The output voltage $V_O$ of an inverting summing amplifier DAC is given by:
$$V_O = -V_{\text{REF}} \cdot \frac{R_f}{R} \sum_{i=0}^{N-1} b_i 2^{i-N}$$
Alternatively, in terms of the input resistors $R_i$ and feedback resistor $R_f$:
$$V_O = -R_f \cdot V_{\text{REF}} \sum_{i=0}^{N-1} \frac{b_i}{R_i}$$
Here, $N=4$ bits ($b_3$ is MSB, $b_0$ is LSB). The input resistors are $R_3=R, R_2=2R, R_1=4R, R_0=8R$. The feedback resistor is $R_f=R$. $V_{\text{REF}} = 2 \text{ V}$.
$$V_O = -R \cdot V_{\text{REF}} \left( \frac{b_3}{R} + \frac{b_2}{2R} + \frac{b_1}{4R} + \frac{b_0}{8R} \right)$$ $$V_O = -V_{\text{REF}} \left( b_3 + \frac{b_2}{2} + \frac{b_1}{4} + \frac{b_0}{8} \right)$$ $$V_O = -V_{\text{REF}} \left( \frac{8b_3 + 4b_2 + 2b_1 + b_0}{8} \right)$$
2. Calculate Output for Input 1 ($V_{O1}$ for $1110$)
Input $b_3 b_2 b_1 b_0 = 1110$. $V_{\text{REF}} = 2 \text{ V}$.
$$V_{O1} = -2 \text{ V} \left( \frac{8(1) + 4(1) + 2(1) + 1(0)}{8} \right)$$ $$V_{O1} = -2 \cdot \left( \frac{8 + 4 + 2}{8} \right) = -2 \cdot \frac{14}{8}$$ $$V_{O1} = -2 \cdot 1.75 = -3.50 \text{ V}$$
3. Calculate Output for Input 2 ($V_{O2}$ for $1101$)
Input $b_3 b_2 b_1 b_0 = 1101$. $V_{\text{REF}} = 2 \text{ V}$.
$$V_{O2} = -2 \text{ V} \left( \frac{8(1) + 4(1) + 2(0) + 1(1)}{8} \right)$$ $$V_{O2} = -2 \cdot \left( \frac{8 + 4 + 1}{8} \right) = -2 \cdot \frac{13}{8}$$ $$V_{O2} = -2 \cdot 1.625 = -3.25 \text{ V}$$
4. Calculate the Magnitude of the Change in Output Voltage ($\Delta V_O$)
$$\Delta V_O = |V_{O2} - V_{O1}|$$ $$\Delta V_O = |-3.25 \text{ V} - (-3.50 \text{ V})|$$ $$\Delta V_O = |-3.25 + 3.50| \text{ V} = 0.25 \text{ V}$$
5. Convert to $\text{mV}$ and Round
The result is required in $\text{mV}$ ($1 \text{ V} = 1000 \text{ mV}$), rounded off to the nearest integer.
$$\Delta V_O = 0.25 \text{ V} \times 1000 \frac{\text{mV}}{\text{V}} = 250 \text{ mV}$$
Rounding to the nearest integer gives 250.
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