Given, A and B are input terminals. Logic 1 is > 5 V, Logic 0 is < 1 V. Which logic gate operation, the following circuit does?

Note: This question was awarded a bonus. C option changed.

1. Both inputs are LOW: $$A = 0\text{ V}$$ (Logic 0), $$B = 0\text{ V}$$ (Logic 0)

Since the inputs are at $$0\text{ V}$$ and the top node is pulled up toward $$V_{cc} = 6\text{ V}$$, both diodes face a positive voltage from anode to cathode. This means both diodes are forward-biased (ON) and act as short circuits to ground. The output node $$V_{\text{out}}$$ is pulled down directly to the input voltage level through the small $$50\text{ }\Omega$$ resistor.

$$V_{\text{out}} \approx 0\text{ V} \implies \text{Logic } 0$$

2. One input is HIGH, one is LOW: $$A = 6\text{ V}$$ (Logic 1), $$B = 0\text{ V}$$ (Logic 0) 

For the terminal at $$6\text{ V}$$ ($$A$$), the cathode is at $$6\text{ V}$$. Since the supply voltage $$V_{cc}$$ is also $$6\text{ V}$$, the anode cannot be higher than the cathode. Diode $$A$$ becomes reverse-biased (OFF). For the terminal at $$0\text{ V}$$ ($$B$$), the diode remains forward-biased (ON). Current flows from $$V_{cc}$$ through diode $$B$$ to ground. The output node $$V_{\text{out}}$$ is still pulled down to the $$0\text{ V}$$ level of input $$B$$.

$$V_{\text{out}} \approx 0\text{ V} \implies \text{Logic } 0$$

3. Both inputs are HIGH: $$A = 6\text{ V}$$ (Logic 1), $$B = 6\text{ V}$$ (Logic 1)

With both cathodes held at $$6\text{ V}$$ (equal to $$V_{cc}$$), neither diode can conduct current forward. Both diodes are reverse-biased (OFF) and act as open circuits. With the path to the input terminals blocked, no current flows through the circuit down to the $$10\text{ k}\Omega$$ pull-down resistor to ground. The output node $$V_{\text{out}}$$ is tied directly up to the supply voltage line.

$$V_{\text{out}} = V_{cc} = 6\text{ V} \implies \text{Logic } 1$$