Topic: Digital Circuits
An 8-bit Digital-to-Analog converter (DAC) using two identical 4-bit DACs with equal reference voltage is shown in Figure. If $b_0$ represents LSB, $b_7$ MSB and the op-amp is ideal, to obtain correct analog values corresponding to an 8-bit DAC at the output $V_o$, what should be the value of resistor R?
(Circuit: $V_{o1}$ from LSB 4-bit DAC feeds into a summing op-amp via a 16 k$\Omega$ resistor. $V_{o2}$ from MSB 4-bit DAC feeds in via resistor R. Feedback resistor is 16 k$\Omega$.)
- A. 8 k$\Omega$
- B. 0.25 k$\Omega$
- C. 1 k$\Omega$
- D. 0.5 k$\Omega$
Answer: C. 1 k$\Omega$
Explanation: The output of the summing amplifier is given by $V_o = -\left(V_{o1} \frac{16k\Omega}{16k\Omega} + V_{o2} \frac{16k\Omega}{R}\right)$.
For an 8-bit DAC, the weight of the least significant bit of the MSB nibble ($b_4$) must be $2^4 = 16$ times the weight of the LSB of the LSB nibble ($b_0$).
This means the gain for $V_{o2}$ (the MSB DAC) must be 16 times the gain for $V_{o1}$ (the LSB DAC).
Gain for $V_{o1}$ is 1. Thus, gain for $V_{o2}$ must be 16.
$\frac{16k\Omega}{R} = 16 \Rightarrow R = 1k\Omega$.