Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. Built in potential in each diode is 0.7 V. For the input voltages shown in the figure, the voltage (in Volts) at point A is __________.

Solution

Physics sir · 3 steps

Find which diode actually conducts; then A sits exactly 0.7 V below that diode's input.

1. Left diode. Its anode is at the 12.7 V input and its cathode is at A, which connects to ground through the resistor. So it is forward biased and conducts. A forward-biased diode drops its built-in 0.7 V, and its resistance is negligible.

   V_A = 12.7 - 0.7 = 12 V

2. Right diode. Its anode is at 4 V and its cathode is at A, which is now at 12 V. The cathode is higher than the anode, so this diode is reverse biased and carries no current. It does not change V_A.
3. So. Only the left diode conducts, and the voltage at A is 12 V.

The common slip: 12.7. Forgetting the 0.7 V drop across the conducting diode gives 12.7, not 12.

The idea to keep. With two diodes feeding one node, guess which one conducts, find the node voltage, then check the other diode is reverse biased. Always subtract the 0.7 V drop across a conducting silicon diode.

Answer from the NTA final answer key. Solution by PrepWiser.