Question
A silicon diode has a forward voltage drop of at the operating current. It is connected in series with a resistor and a battery of . (a) Find the current in the circuit. (b) If the battery polarity is reversed, what is the current (assume reverse leakage of )? (c) What is the power dissipated in the diode in forward bias?
Solution — Step by Step
Apply Kirchhoff’s voltage law: battery voltage = drop across diode + drop across resistor.
In reverse bias, the diode blocks almost all current. Only the tiny reverse saturation current flows. Given as .
So .
Power dissipated = voltage drop current.
Forward current , reverse current , .
Why This Works
In forward bias, the diode behaves like a small constant-voltage source (0.7 V for Si, 0.3 V for Ge). The series resistor sets the current. The diode does not “know” the source voltage — it just maintains its drop.
In reverse bias, the depletion region widens and only thermal-generation carriers cross. This leakage current is roughly temperature-dependent and very small.
Alternative Method
Load-line approach: plot the diode’s - characteristic and the line . Their intersection gives the operating point. For Si at room temperature, this lands close to — same answer, useful when the diode is non-ideal.
Common Mistake
Treating the diode as an ideal short circuit in forward bias (). Students then compute , which overshoots by 16%. The 0.7 V drop is the whole point of using Si over Ge — it provides better noise immunity and temperature stability.