In a p-n junction diode under forward bias, the charge carriers responsible for current flow in the external circuit are free electrons. These electrons are present in the n-type material, and they move towards the p-type material. The free electrons from the n-side travel through the external circuit to the p-side, where they recombine with holes. This flow of free electrons constitutes the current in the connecting wire.
In a forward-biased p-n junction diode, the electrons from the n-side and the holes from the p-side move toward the junction. However, in the external connecting wire, the current is due to the movement of \(\textbf{free electrons}\), as electrons are the actual charge carriers in metal conductors (wires). Holes contribute to current only within the semiconductor, not 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 ? 
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2