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In the following question, D1 is operated in forward bias condition. Then the voltage across D1 is 0.7 V.

Then the voltage across D2 also 0.7 V; but no current flows through D2. Then what is the operation condition of D2 – forward or reverse bias?

(If current flows through D2, then the potential drops across 5 ohm resistor and the potential drop across D2 becomes less than 0.7 V.)

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  • \$\begingroup\$ Kinda hard to answer without knowing the diode model you're using. Like you said, if current flows through D2, then the voltage across D2 drops but the predicted behavior really depends on the I/V characteristics you're assuming for D1 and D2. \$\endgroup\$
    – vir
    Commented May 6 at 4:52

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For your first question- you should refer to the definition your instructor has given you for "forward bias".

That's a simple answer, but there is an answer that does not depend on the aforementioned definition but does depend on a more accurate model of a diode. In particular, refer to the Shockley diode equation and look at what happens to the current in a diode for forward voltages (anode > 0V wrt cathode).

Real diodes do not just sharply turn on at 0.7V. Even if the two diodes are quite dissimilar you can predict the sign of current.

If you want to stick with the penultimate simplest model for a diode (0.7V Vf in forward bias) then no current will flow through the 5Ω resistor.

Using a fairly realistic model of real diodes we can simulate the problem set situation and see how close the 0.7V/0.7V 530mA/0mA will be to the more realistic simulation. It's not totally wrong but you can see that it has some problems. In general if the source voltage through a resistor to a diode is much higher than 0.7V so that the exact diode voltage does not affect the current much, the 0.7V estimate can work. If the source voltage is close to 0.7V then it can lead to relatively larger errors. In this case, 0mA rather than ~26mA.

schematic

simulate this circuit – Schematic created using CircuitLab

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