In a p-n junction diode that is forward biased, the direction of the applied voltage allows current to flow through the diode. Let's discuss the movement of the charge carriers (electrons and holes) that contribute to the current when the p-n junction is forward biased.
When the p-n junction is forward biased, the p-type semiconductor is connected to the positive terminal of the battery, and the n-type semiconductor is connected to the negative terminal of the battery. This results in the following effects:
The forward bias causes the following movements of charge carriers:
As electrons flow from the n-type material into the p-type material, they move through the external circuit towards the positive terminal of the battery, creating a current in the external circuit. Similarly, holes move from the p-type material to the n-type material and also contribute to the current. Thus, the movement of both electrons and holes results in a continuous current flow in the circuit.
In a forward-biased p-n junction, the current is produced by the movement of electrons from the n-type material to the p-type material, and the movement of holes from the p-type material to the n-type material. The flow of these charge carriers (electrons and holes) across the junction leads to a continuous current in the external circuit.
Assuming in forward bias condition there is a voltage drop of \(0.7\) V across a silicon diode, the current through diode \(D_1\) in the circuit shown is ________ mA. (Assume all diodes in the given circuit are identical) 


For the given logic gate circuit, which of the following is the correct truth table ? 