Draw the current-voltage characteristics for the device show in figure between the terminals A and B.
Figure
Answers
In the first figure it is clear that the only one diode is present between A and B. The I-V characteristics will be:
In the second figure, the diode in the upper arm in series with the 10Ω resistor has p-side connected to A and n-side connected to B.
Whereas, the other diode has exactly opposite polarities. When a potential difference is applied across terminals A and B, one of the diodes will become reverse biased and there will be no flow of current through that arm. ∴ I-V characteristics between terminals AB will be same as the previous figure.
Explanation:
(a) Between the terminal A and terminal B, if there is a connection of a battery, with the negative terminal connected to point B and the positive terminal connected to point A, then by the applied voltage, the diode will be forward biased. So, the circuit’s current voltage graph will be the same as that of the forward-biased diode’s characteristic curves.
(b) Between the terminals A and B, if there is a connection of a battery, with negative terminal connected to point B and the positive terminal connected to point A, then there will be a forward bias of the upper diode and there will be reverse bias of the lower diode by the applied voltage. Thus, by an open circuit, there will be a replacement of this lower branch; so, the flow of current via this branch will be zero. Only through the upper diode, the current flows, so on simplification, the circuit will become equal to the part (a) circuit. Hence, for this circuit, the current voltage graph will be the same as that of the forward-biased characteristic curves.
