This makes that transistor T C does not receive enough potential and it will be in OFF condition. Then, current starts to flow to the ground via this path and the +5V will be dropped down at resistor R. Here, the supply voltage will find a path to the ground terminal through both the transistors T A and T B or either of them. This makes the transistors T A and T B or either of the transistor to be in an ON state. When both the input pins X and Y or either of the inputs is applied with high logic (+5V) which means the base terminals of both the transistors have +5V voltage. Ideally, the transistor T C has some voltage drop (0.6V – 0.7V) across it and this voltage appears at the output which is considered as 0. As the third transistor T C is in ON condition, it performs as short-circuited device and the entire +5V gets dropped down to the ground and output appears to be logic LOW (0V). Here, +5V voltage moves to the ground through R1 and transistor T C. This makes the transistor T C get the required amount of potential and it will be in an ON state. At this condition, the supply voltage of +5V will not have a path to pass to the ground via any one of the transistors T A and T B. When the input pins X and Y are provided with ‘0’ Volts or when both the pins are grounded, then the transistors T A and T B move into OFF condition correspondingly. Below is the circuit diagram of a two-input OR logic gate using a transistor. The operation and construction of an OR logic gate can also be derived using a transistor. When both the inputs are grounded or provided with 0V, then there will be no voltage at the output and considered as logic LOW. Then the 4.4V voltage appears at the output. When both the inputs are provided with a supply voltage of +5V, both the diodes will move into forward-biased state. This 4.4V or 4.3V is ideally taken into consideration as logic HIGH. In practical scenarios, the entire supply voltage +5V will not be present at the output, some level of voltage drop (0.6V – 0.7V) happens at the diode and the rest of the voltage level (4.3V or 4.4V) passes to an output pin. So, the +5V voltage appears at output Z which means the output is logic HIGH. Whereas when either of the inputs is provided with a supply voltage of +5V, then the corresponding diode moves into forward biased condition and operates as a short, circuited device. As per the above circuit diagram, when the inputs X and Y are provided with 0V, then the output X shows no voltage.
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