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In the I-V characteristics of diodes under forward bias , I have seen that the current starts increasing from zero only after reaching the barrier potential , say 0.7 for silicon. As far I understand , the formation of barrier potential is somewhat like this : when the diode is just formed, due to the carrier concentration differences, diffusion of carries starts to take place. But this creates a depletion region and an opposing electric field. The opposing electric go on increasing, and a potential is generated which increased from 0 to 0.7 for silicon, as time elapsed. At 0.7 V,the diffusion current and drift current due to this 0.7 v gets into equilibrium and net current become zero.

So my question is this : if I apply a 0.1 V as forward bias, won't it reduce the barrier and make it 0.6 V? As the 0.1 V bias continue to exist, I think the voltage barrier should now remain at 0.6 V. At this 0.6 V diffusion component of current should overcome the drift current and a net current flow should be there. This net current should go on increasing continuously on increasing the forward bias, rather than abruptly start increasing at 0.7 V only (no current till 0.7 V). Why isn't it happening?

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    \$\begingroup\$ Have a look at some common diodes datasheets, and you will see they have curves for the forward voltage, as this depends on the flowing current. Or you can measure this yourself, run some small current (like 1µA or less) through a diode and measure its drop. \$\endgroup\$
    – PlasmaHH
    May 17, 2016 at 8:37
  • \$\begingroup\$ The electrons in the semiconductor need a little kick to get moving. There will always be some electrons that have enough energy to jump up over the band gap (thermal energy gives a little push on top of the bias voltage). 0.7V is the point at which the vast majority of electrons have enough energy to jump the gap, some will always make their way over bit most need a little kick to get going. (There's alot more to it than that, but that'd be a much more involved answer) \$\endgroup\$
    – Sam
    May 17, 2016 at 8:46
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    \$\begingroup\$ It all depends on which diode model you use. Many models simplify the diode characteristic so that no current flows until you get to the knee voltage. A model is just that - a simplification - don't confuse it with reality. \$\endgroup\$
    – JIm Dearden
    May 17, 2016 at 8:58
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    \$\begingroup\$ Current THROUGH, voltage ACROSS ... \$\endgroup\$
    – user16222
    May 17, 2016 at 10:52

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Here's what the 1N4148 signal diode characteristic looks like in the foothills: -

enter image description here

As you can see, below ~0.8 volts it's a pretty regular shaped characteristic. The current is approximately falling by a factor of ten for each 100 mV reduction in base-emitter voltage.

On a linear graph it looks like there is a real definable knee voltage: -

enter image description here

But, that linear graph hides the reality/subtlety of things but who cares - most diode circuits are nicely approximated by saying current is zero below 0.6V (or 0.7V) and the diode behaves like a short at this voltage.

There is an equation called the Shockley diode equation that adequately describes current versus voltage based on temperature and reverse leakage - often a sim will use this type of equation but rough and ready calculations on a scrap of paper don't need to.

enter image description here

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