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What is the "cut-in" voltage or knee voltage of a diode or transistor?

In a lot of books I read that the cut-in or knee voltage of a diode (or transistor) is when the diode starts conducting or when the diode starts allowing significant of current or the voltage at which current in diode starts increasing rapidly with in small increase in voltage accross it.

What is the that significant current? Is there any value for that significant current and a precise defintion for "cut-in" voltage or is it just a vague definition?

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    \$\begingroup\$ Exponential curves look linear when examined close enough. And everywhere. Meaning that if you look closely enough at 10 mA it looks linear and if you look closely enough at 1 nA it looks just as linear (but different slope.) The issue is that we generally don't care about 1 nA. But do care about 10 mA. So if you take a step back and look at many orders of magnitude, then there is a place where the change in current for changes in voltage we care about yield almost crazy changes in current. And before that point, almost negligible changes. That's the knee. \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 14:02
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    \$\begingroup\$ Another way to look at this is that for every \$\Delta V\$ change in voltage there is a \$10\times\$ change in current. Step by \$\Delta V\$ and you go from 1 nA to 10 nA. And you don't care. Step again by \$\Delta V\$ and you go from 10 nA to 100 nA. And you don't care. Step again by \$\Delta V\$ and you go from 100 nA to 1 uA. And you don't care. Etc. But step by \$\Delta V\$ and you go from 10 uA to 100 uA. And now you start to care. Another \$\Delta V\$ and you are into milliamps. Etc. It doesn't take long before it really matters a lot. So going from 100 mA to 1 A really matters!! \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 14:07
  • \$\begingroup\$ The knee occurs where you decide things matter and where they don't matter. (With multiples of 10 piling up fast, there is most definitely a knee somewhere.) Of course, it's really still everywhere just the same self-similar exponential equation. \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 14:09
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    \$\begingroup\$ Regardless, the knee moves only tiny distances in terms of the voltage -- stepping only a small bit by \$\Delta V\$. Suppose \$\Delta V=100\:\text{mV}\$ (as in the 1N4148 diode case.) So if 10 mA matters to you but 1 mA matters less then perhaps the knee occurs at 630 mV. If 10 uA matters to you but 1 uA matters less, then the knee occurs at 330 mV. It works like that. But since the voltage only changes a little, it's common to see 330 mV as not so different from 630 mV. And to just say that the knee occurs somewhere around that region. \$\endgroup\$
    – periblepsis
    Commented Apr 25 at 14:32
  • \$\begingroup\$ The "knee" on all exponential curves is an illusion - If you label your axis using a different order of magnitude, the knee shifts accordingly. The knee's location has no mathematical significance. But for practical purposes, the X axis is always in milliamps or amps, not in gigaamps, so the "knee" of a diode curve is around 0.7 V consistently. \$\endgroup\$
    – 比尔盖子
    Commented Apr 25 at 17:26

1 Answer 1

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'Cut in' voltage and 'knee' voltage for diodes are Lies to children.

A rather good model for diode conduction is the Shockley Equation, which as you can see allows you to estimate the current for any given voltage for a diode.

$$ I_D = I_S (\frac{V_D}{e^{nV_T}}-1)$$

For the range of 'useful' currents that students and hobbyists are likely to encounter, say the 10 uA to 100 mA range, the voltage across a room temperature silicon diode will be in the 400 to 700 mV range. The voltage is 'almost constant', less than a factor of 2 range, for 4 orders of magnitude range in current. That's where we put the 'knee' for students and hobbyists.

Given the exponential nature of the VI relationship, depending on how you scale the graph axes, you can move the 'knee' to just about anywhere.

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  • \$\begingroup\$ Sometimes, the so called "knee" voltage is defined as the voltage that can be found at that point where a tangent to the "linear" part of Shockleys equation crosses the horizontal axis. This is because, in reality, there will be small ohmic path resistance which linearizes the e-function for larger currents. However, there is nothing like a "precise" definition for this "knee voltage". \$\endgroup\$
    – LvW
    Commented Apr 25 at 16:41
  • \$\begingroup\$ @LvW can you explain your definition of clearly and provide me the source of ur definition \$\endgroup\$
    – Qwe Boss
    Commented Apr 30 at 18:03
  • \$\begingroup\$ There is no "clear" or "precise" definition of this prameter. This is because the corresponding function i=f(U) has no typical or characteristic point which could be the basis for such a definition. \$\endgroup\$
    – LvW
    Commented Apr 30 at 18:33

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