# Diode Forward Bias behavior?

All the textbooks and articles regarding diodes talk about forward bias or reverse bias and break-down voltage. I don't know why the break-down voltage is never related to the forward bias?

I think the same thing could in reality happen in forward biased mode as it can happen in reverse biased mode. If we apply lot more voltage than 0.7 V in forward direction, this is still bad, and it may lead to a break-down condition.

It's a simple matter of definitions. In either direction, there is a voltage above which the diode begins to conduct a large current for a small increase (or decrease in the reverse case) in voltage. The finer details of the current-voltage function in each direction are somewhat different, but as a first order approximation, above a minimum (reverse breakdown voltage) and below a maximum (forward voltage), a diode does not conduct at all, and at voltages below or above these limits, it conducts a lot. This approximation is sufficient for most engineering purposes.

The reason for the difference in terms is that the underlying physical mechanism is quite different. The forward voltage has to do with the nature of the semiconductor, and for all silicon PN diodes, this will be in the neighborhood of 0.65V. The reverse breakdown voltage additionally depends on the geometry and design of the device, and quite a range of values are attainable, even among silicon PN diodes.

Also, don't let the term "breakdown" suggest that the diode "breaks". What is "breaking down" is the usual state of the diode that prevents reverse current flow. Once the reverse breakdown voltage is exceeded, the diode isn't necessarily damaged. However, a large current will flow, and if this current isn't limited (say, by a series resistor), then the diode will overheat. Then it will be damaged.

Note this isn't really any different from the case when the diode is forward biased. Any attempt to apply significantly more than the forward voltage will result in a very large current which overheats the diode and destroys it. Limiting the current avoids damage.

Ordinary silicon diodes (example, 1n4148) are not often intentionally operated in reverse breakdown. Their behavior in this mode of operation is not usually specified except for some minimum reverse breakdown voltage. There are other diodes, such as Zener diodes, which are usually operated in reverse breakdown (though the physical mechanism is somewhat different). These diodes have more completely specified behavior in this operation, because by virtue of their design, the relevant operational parameters can be more predictable and stable.

It's difficult to apply a forward voltage of much more than 0.7V across a diode, because the diode conducts when forward biased! The graph below shows that the current through a diode increases very rapidly (in fact, exponentially) with increasing forward voltage.

If the forward voltage across the diode were much more than 0.7V, the current flowing through the diode would be far in excess of its rated maximum current, and would cause rapid overheating.

The phrase 'forward breakdown voltage' is sometimes used to refer to the forward voltage of ~0.6V (for a silicon diode) at which, in the simplest model of diode behaviour, the diode 'begins to conduct'.

There is no such phenomena as breakdown in forward bias conditions. Reverse breakdown is a sudden condition on reaching a certain reverse voltage. Forward voltage conduction happens continuously - there isn't a magic 0.7V where diodes start conducting - it's a gradual increase in conduction as soon as any positive voltage is applied - a lot of people say 0.7V because it is a convenient average value to use.

It's a totally different mechanism in reverse bias. In fact there are two mechanisms, one called avalanche breakdown and one called zener breakdown. Again, I reiterate, these are phenomena of a reverse biased diode.

• It's also worth saying that most 1N400x and similar rectifier diodes operate with pretty much a fixed 0.9V-1.1V forward conduction voltage when their load is a smoothing capacitor. The forward diode voltage is well approximated by a rectangular pulse about a volt high :) So, for rectifier calculations, 0.7V is a bit optimistic. A full bridge feeding a capacitor, operating at a "reasonable" RMS current (say 1/4-2/3 of rated current) drops quite close to 2V. If it drops any more than 2.2V, it likely has to be upsized - especially that the forward voltage drops a bit as the diodes get hotter. Commented Mar 13, 2023 at 14:46

Well, yes this could happen. But See, the Breakdown current is due to minority carriers which are very less than majority carriers. Now firstly if you apply a high voltage the current due to majority carriers will be high. Also, at breakdown majority carriers also along with minority carriers will be generated but the current due to majority carriers grew exponentially, whereas in reverse bias the majority carriers were unable to conduct but in forward bias this isn't the case, so, the contribution of minority carriers will still be very less and as a result they won't contribute that much. Also the diode usually burns due to heat generated and the circuit isn't closed anymore.

The key point is current in forward bias grows exponentially and even when no carriers are created their current is negligible due to this exponential growth.

Actually there are some diodes that breakdown in forward bias -- these are called Esaki diodes or tunnel diodes. Basicially, the P-N junction is so heavily doped that the built-in electric field is already close to breakdown without any bias. When a small amount (0.3 V) of forward bias is applied, the device breaks down.

These diodes are not used very much any more, but used to be used in microwave amplifiers and radars before transistor performance was good enough.