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I am reading in a book about the forward bias of a diode. It mentions that in order for a diode to be in the forward mode and its current to rise exponentially:

$$V_{D}\geq + 4V_{T}=0.1V$$

where \$V_{D}\$ is the diode's voltage and \$V_{T}\$ the thermal voltage (approximately 0.025V as a typical value).

My problem is that from what I know and from what I read in another section of the same book and also in other books, then for a diode to be in forward bias:

$$0.5V\leq V_{D}\leq 0.8V$$

If we use the equation of a diode with \$V_{D}=0.1V\$ then:

$$I=e^{0.1/0.025}\approx 53\cdot I_{S}$$

where \$I_{S}\$ is the reverse saturation current.

But with \$V_{D}=0.5V\$ then:

$$I=e^{0.5/0.025}\approx 485165194\cdot I_{S}$$

Clearly with \$V_{D}=0.1V\$ the current through the diode is very small compared to the current that will pass if the voltage is \$V_{D}\geq 0.5\$, as most sources mention as the necessary threshold for a diode to be in forward bias.

Am I missing something?

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  • \$\begingroup\$ I didn't carefully follow to see what you are trying to say with "485165194", but whatever it is you can't possibly know the value to 9 digits! -1 for the absurd precision. \$\endgroup\$
    – Olin Lathrop
    Feb 29, 2016 at 21:34
  • \$\begingroup\$ @OlinLathrop It is not precision it is just $e^{V_{D}/V_{T}}=e^{0.5/0.025}=e^{20}=485165194$. You could ask before downvoting... \$\endgroup\$
    – Adam
    Feb 29, 2016 at 21:37
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    \$\begingroup\$ No need to ask since it's clearly silly. Even if it is a purely mathematical value, it still makes no sense to express it to 9 digits. This is just plain bad engineering. \$\endgroup\$
    – Olin Lathrop
    Feb 29, 2016 at 21:49
  • \$\begingroup\$ @OlinLathrop All of these come from a book that just tries to explain the phenomenon theoretically through equations. It is not supposed to be about a real diode. \$\endgroup\$
    – Adam
    Feb 29, 2016 at 21:53
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    \$\begingroup\$ @OlinLathrop as a physicist (and I am sure the same applies to engineers), I strongly disagree that this is bad practice. Yes, it looks unusual, and many people may complain about it. But there is really nothing logically or mathematically wrong with it. On the contrary, what is really bad practice is to try to infer something about the precision of a value based on the way it is written. "Significant figures" are simply not an accurate or useful way to express the concept of variance or error. A rather long but convincing diatribe here. \$\endgroup\$
    – Oleksandr R.
    Mar 1, 2016 at 2:36

2 Answers 2

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Ideality factor is more like \$2\$ on a real diode so your 2nd number is high by \$4\$ or \$5\$ orders of magnitude.

With \$V_{T} = 0.025\$ and \$V_{F}= 0.1\$ or \$0.5V\$ the current will generally be pretty low (nA or tens of uA respectively).

The diode is forward biased when \$V_{F} > 0\$. It starts to conduct substantially (for many purposes anyway) when the current exceeds roughly \$0.5V\$.

They are saying that the diode equation works better when \$V_{F} > 0.1V\$, or at least you can start to ignore that pesky \$-1\$ in the equation.

Edit: Shockley diode Equation: \$I = I_\text{S}( e^\frac{V_\text{D}}{n V_\text{T}}- 1)\$

"\$n\$ is the ideality factor, also known as the quality factor or sometimes emission coefficient" to quote Wikipedia.

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  • \$\begingroup\$ Thank you for your answer. One question though. What factor do you mean in the beginning of your answer? I mean when you say "factor is more like 2 on a real diode". \$\endgroup\$
    – Adam
    Mar 1, 2016 at 19:32
  • \$\begingroup\$ The "ideality factor". It's 1.00 for ideal diodes, but for real diodes such as 1N4148 or 1N914 it is more like 2. See above edit. \$\endgroup\$
    – Spehro Pefhany
    Mar 1, 2016 at 19:37
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Forget turning this into a Master's thesis in electrical engineering. Answer is 0.7V for a silicon diode and 0.3V for a Germanium diode.

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