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Consider the circuit below:

enter image description here

I'm assuming ideal diode model(\$V_{ON} = 0V\$). If \$V\$, \$A\$ and \$B\$ are all 5V, the diodes will be reversed biased and no current will flow through the circuit. Therefore, \$V_{OUT}\$ = 5V.

However, for either of the diodes, the voltages across them will be

\$V_{A} + V_{D} = V_{OUT}\$.

Since \$V_{A} = V_{OUT}\$, \$V_{D}\$ will 0. Since we're considering an ideal diode, wouldn't that mean that the diode is actually forward biased? What am I missing here?

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    \$\begingroup\$ You've said it yourself: the voltage across the (ideal) diode is 0. Therefore the diode is neither forward or reverse biased, but has zero bias. \$\endgroup\$ – henros Sep 17 '18 at 9:05
  • \$\begingroup\$ Diode can have (positive voltage or high impedance) or GND. If both are positive or HiZ then the output is true. If any of the diode is tied to the GND, then output is false. AND logic isn't it? \$\endgroup\$ – Marko Buršič Sep 17 '18 at 9:06
  • \$\begingroup\$ @henros I was under the impression that current starts flowing the diode at 0V i.e. diode is forward biased at V = 0V. \$\endgroup\$ – Rish Sep 17 '18 at 9:35
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    \$\begingroup\$ How is current going to flow if there is no voltage across the diode? Sure, for very small forward voltages, you will get a correspondingly small forward current, until you hit the cut-in voltage. But in your scenario there is exactly 0V across the diodes, so no current flows through the diode. \$\endgroup\$ – Andrew Guy Sep 17 '18 at 10:35
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    \$\begingroup\$ You've already stipulated that the voltage across the diode is zero. What possible difference could it make whether the diode is in a conducting state or a non-conducting state? \$\endgroup\$ – Dave Tweed Sep 17 '18 at 11:09
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If the voltage across the diode is equal to its \$V_f\$ (what you call \$V_{ON}\$) and no current is flowing through it then the diode is on the knife edge between forward and reverse bias. For forward bias we usually assume a positive current flowing through the diode and for reverse bias we assume that \$V_D < V_f\$.

Since the initial conditions (\$V_A = V_{SOURCE}\$ and \$V_{OUT} = 0V\$) have the diode reverse biased I would probably call it reverse biased even when the output voltage rises to the source voltage.

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