In a digital circuit, a NAND gate has two inputs, A and B . Input A is controlled by a photodiode having a bandgap energy of 2.8 eV . The circuit is designed such that when the photodiode detects light and conducts, Input A becomes logic 1 ; otherwise, it remains at logic 0 . Input B is permanently connected to a 5 V supply (logic 1 ). If light of wavelength 500 nm is incident on the photodiode, what will be the logi

Step-by-step solution

The energy of the incident photons is given by E = hc . Substituting the given values: E = 1240 eV nm 500 nm = 2.48 eV For a photodiode to detect light and conduct, the energy of the incident photons must be greater than its bandgap energy ( E > E_g ). Here, 2.48 eV Input B is connected to a 5 V supply, which corresponds to logic 1 . The final output Y is from a NAND gate with inputs A = 0 and B = 1 . The Boolean expression for a NAND gate is Y = A B . Y = 0 1 = 0 = 1 . Thus, A = 0 and Y = 1 . Answer: A = 0, Y = 1