A 3-bit DAC is implemented using ideal opamp and switches as shown in the figure. Each of the switches gets closed when its corresponding digital input is at logic 1. For a digital input of 110, the resistance \( R_{in} \) seen from the reference source and the current \( I \), are
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In an inverting opamp-based DAC, the inverting terminal is a virtual ground. The current from the reference voltage through the selected resistors sums at this virtual ground and flows through the feedback resistor.
Step 1: Identify closed switches.
For the digital input 110 (\(D_2=1, D_1=1, D_0=0\)), the switches connected to the \(3 k\Omega\) and \(6 k\Omega\) resistors are closed. The switch connected to the \(12 k\Omega\) resistor is open. Step 2: Calculate the equivalent input resistance (\(R_{eq}\)).
The closed resistors (\(3 k\Omega\) and \(6 k\Omega\)) are in parallel as seen from the virtual ground at the opamp's inverting input.
$$\frac{1}{R_{eq}} = \frac{1}{3 k\Omega} + \frac{1}{6 k\Omega} = \frac{2 + 1}{6 k\Omega} = \frac{3}{6 k\Omega} = \frac{1}{2 k\Omega}$$
$$R_{eq} = 2 k\Omega$$ Step 3: Determine the input resistance seen by the reference source (\(R_{in}\)).
The reference voltage source is connected to this equivalent resistance \(R_{eq}\). Therefore, \(R_{in} = R_{eq} = 2 k\Omega\). Step 4: Calculate the current flowing into the virtual ground.
Using Ohm's law, the current from the reference source into the inverting input is:
$$I_{in} = \frac{V_{ref}}{R_{eq}} = \frac{6 V}{2 k\Omega} = 3 mA$$ Step 5: Determine the current through the feedback resistor (\(I\)).
For an ideal opamp with infinite input impedance, all the current entering the inverting input must flow through the feedback resistor \(R_f\). Thus, \(I = I_{in} = 3 mA\). Step 6: Match with the options.
The calculated values \(R_{in} = 2 k\Omega\) and \(I = 3 mA\) match option (A).
