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When an ideal diode is forward biased, it is replaced by a connecting wire .Does it mean voltage across diode is zero?If yes, then is zero voltage across diode condition for forward bias?

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    \$\begingroup\$ Only for an "ideal" diode (meaning no voltage drop.) Real diodes are different. \$\endgroup\$ Jan 19 '15 at 17:07
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    \$\begingroup\$ In fact, its kind of opposite I believe ; forward bias is a condition for diode to be replaced by short ckt (ideal of course). \$\endgroup\$ Jan 19 '15 at 17:21
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There's not just one "ideal" diode.

The simplest type (what I think you are talking about) has 0V across it if the polarity is such that current would flow in the direction of the arrow, and doesn't conduct at all if the polarity is reversed.

A slightly more complex model has a fixed voltage drop (usually taken to be 0.6V or 0.7V) if it is conducting current, and does not conduct current if the voltage is less than the fixed drop. If you consider the voltage drop to have a temperature coefficient (around -2mV/°C) then you can roughly incorporate temperature effects.

In either of the above two cases, to find out what the diode is doing, remove the diode as a thought experiment and see (it may be obvious or you may need to resort to math) whether the bias would be positive or negative.

An even more complex type of ideal diode obeys the nonlinear and temperature-sensitive Shockley diode equation, with ideality factor 1. Most diodes are not that ideal, and the ideality factor is more like 2. You can also have different Is for positive and negative bias.

Finally, you can include ohmic resistance in the equation (which has a positive temperature coefficient), and you'll be fairly close to modelling diode behavior in most real situations.

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No, a forward biased diode is NOT just like a connecting wire. One specification of a diode is its forward voltage drop. For silicon diodes, that is usually about 500-750 mV across a reasonable current range, but can be as much as 1 to 1.5 V for some diodes at high current. Diodes that can sustain higher reverse voltage tend to have higher forward voltages too. This is due to a tradeoff in how the semiconductor junction has to be made to withstand the high reverse voltage.

Schottky diodes are sortof half of a full semiconductor junction, so have about half the forward voltage, all else being equal. However, all else is almost never equal, so as always, consult the datasheet.

Those are the two most common diode types, but there are several others. Diodes made from germanium instead of silicon have a lower forward voltage drop. These have been largely replaced by Schottky diodes today. Vacuum tube diodes have quite a different voltage to current relationship than semiconductor diodes. It is even possible for such a diode to have effectively 0 forward drop at low currents depending on how hot the cathode is kept and its work function. A vaccuum diode can be tuned on a continuum from having substantial forward voltage drop to actually producing current on its own while being held at 0 V. The latter is essentially a small thermionic generator, with the output power coming from whatever is heating the cathode.

The above was about real diodes that are available today. Theoretical ideal diodes can be found in homework problems when the point of the execise is something else. Such diodes have infinite reverse voltage capability, zero forward drop, and zero reverse recovery time. These can be replaced by a wire when forward biased. Unfortunately, part numbers are hard to find and searches on sites like Mouser or DigiKey don't turn up any.

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    \$\begingroup\$ I think you missed the "ideal" part. \$\endgroup\$ Jan 19 '15 at 17:27
  • \$\begingroup\$ @WhatRoughBeast The O.P. is under some idealization assumptions, and we are trying to figure out what they are. "Forward drop is 0.7V come what may" is an idealization, because it doesn't depend on current, temperature, light. "Forward biased diode is a short circuit" is a different idealization. Which one to use depends on the field of application, or on the assumptions made by the teacher of class. \$\endgroup\$ Jan 20 '15 at 2:23
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It means that there is no 'voltage drop' across diode if it's ideal.The voltage across it will be equal to the source voltage provided that no other impedance is connected to the source.

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No. The model for an ideal diode that is forward biased is a voltage drop of Vt and then a wire. So with a 5Vdc source connected to a diode and then resistor, the diode is forward biased, the resistor sees 4.3V or whatever 5V - Vt is.

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