In a full-wave bridge rectifier supplying a DC load, what is the average (DC) current through each individual diode relative to the DC load current?

Introduction / Context:
Bridge rectifiers use four diodes arranged so that two conduct on each half-cycle, delivering full-wave rectification to the load. Understanding diode conduction intervals helps in choosing proper diode current ratings and estimating thermal stress.

Given Data / Assumptions:

• Full-wave bridge rectifier.
• Resistive DC load with average current I_load(dc).
• Ideal diodes for conceptual analysis.

Concept / Approach:
In a bridge, only two diodes conduct at any instant—one from the “top” leg and one from the “bottom” leg—alternating each half-cycle. Each individual diode conducts for half of the total time (i.e., for one polarity of the AC half-cycle). Therefore, the average current per diode over a full cycle is half of the DC load current (ignoring ripple shaping details).

Step-by-Step Solution:
Let I_load(dc) be the average load current.On the positive half-cycle, diodes D1 and D2 conduct; on the negative half-cycle, D3 and D4 conduct.Any single diode (e.g., D1) conducts only during one half-cycle, carrying the instantaneous load current during that interval.Averaged over a full cycle, the DC current through one diode is I_load(dc)/2.

Verification / Alternative check:
Symmetry of the bridge ensures that the conduction duty for each diode is 50%. Therefore, if the load DC current is 1 A, each diode’s average DC current is 0.5 A (though peak and RMS values differ from DC average).

Why Other Options Are Wrong:

• Load current: That would be true only if a single diode conducted continuously, which it does not.
• Twice / one-fourth of load current: No basis in the conduction duty cycle of a symmetric bridge.

Common Pitfalls:

• Confusing average (DC) current with RMS or peak diode current ratings.
• Ignoring that two diodes are always in series with the load at any instant in a bridge.

Final Answer:
half the dc load current