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How do I know that a diode drops 0.7 V when it's forward biased?

I've had this problem in this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I know that it depends on the current and it is probably small, but when can I just say (without calculating) that my diode drops close to 0.7 V?

I just don't want to calculate this kind of thing in real life and I don't know when can I ignore calculating something and say "yea, diode has 0.7 V"

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    \$\begingroup\$ Measure it with a volt meter! \$\endgroup\$
    – James
    Aug 27 at 20:11
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    \$\begingroup\$ @James But usually in calculations people says that it is 0,7V or 0,5V and calculates still. But yea here I have 0,29V \$\endgroup\$
    – user331990
    Aug 27 at 20:14
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    \$\begingroup\$ electronics.stackexchange.com/questions/330275/… \$\endgroup\$
    – RemyHx
    Aug 27 at 20:14
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    \$\begingroup\$ See if my non-return valve analogy is any help: lednique.com/what-is-an-led. \$\endgroup\$
    – Transistor
    Aug 27 at 21:24
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    \$\begingroup\$ @user331990 If you want to go crazy with closed solutions, look here. This is just a diode (with an internal bulk resistance) and an external resistor, included. It gets worse with more complex circuits. But have a look. Then settle back and accept approximations, for now. \$\endgroup\$
    – jonk
    Aug 29 at 5:10

4 Answers 4

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"All models are wrong, but some are useful" (George Box)

The idea that a silicon diode drops 0.7V is a model. It's reasonably accurate when the diode is carrying a significant current.

If you want a better model, investigate the diode equation see PVEdulation. That explains why the volt drop is lower at lower currents.

You could read the data sheet for the specific diode you intend to use. Or measure the drop with a voltmeter.

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  • \$\begingroup\$ Hmmm, I thought that it would be always 0,7V or 0,5V. Because I was taught like that in school. And so I thought it would also work here. I was also aware that it is a model. So I was afraid how to use it properly \$\endgroup\$
    – user331990
    Aug 27 at 20:20
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    \$\begingroup\$ You were taught the very simplified model in school -- hopefully it's a technician-level class, or a very low-level engineering one! As stated, diode junctions follow the diode equation -- and real diodes act like a junction in series with some resistance. So if it's really important, you need to calculate, measure, use a data sheet, or use a good simulator with a good diode model. \$\endgroup\$
    – TimWescott
    Aug 27 at 20:23
  • \$\begingroup\$ @TimWescott I was also taught how to connect characteristics using this simple model. Also I was taught in university, about this. \$\endgroup\$
    – user331990
    Aug 27 at 20:24
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    \$\begingroup\$ That's a model I was taught in engineering school -- but I was also taught the diode equation (they left learning about the series resistance to the working world). If you're analyzing circuits "by hand" the constant voltage model is a pretty good starting point -- usually if things are going to break if your diode drop is 0.5V instead of 0.7V, then your circuit was going to fail at some temperature extreme or another anyway. It's good to remember that semiconductor characteristics vary a lot from part to part and temperature to temperature. \$\endgroup\$
    – TimWescott
    Aug 27 at 20:32
  • \$\begingroup\$ @TimWescott, so usually I can use this model 0,7V in calculations but when I see 0,2 V then it must be because of to small current ? Well I thought it would be more simplier. \$\endgroup\$
    – user331990
    Aug 27 at 22:30
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The diode equation is $$ I_{D} = I_{s} \left( e^{\frac {qV}{n\cdot kT}}-1 \right) $$

Plot the I-V graph then pick a range of values of current where the diode voltage value is close enough to 0.7 for you. It depends somewhat on the application as to how important it is. You will find at high current levels that there is an additional voltage drop due to internal diode resistance. 0.7V refers to the diode junction characteristic.

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  • \$\begingroup\$ It might be more accurate to say 0,7V is a common inflection breakpoint on a plot at mid-range of currents relative to max rating than just the "the diode characteristic" as if this was universal, which can vary widely in breakpoints with linear voltage drop from bulk resistance at high currents. e.g. Fig 2 dccomponents.com/upload/product/original/244596315940.pdf (which shows no breakpoints but different voltages at 0.1A @ 25'C) \$\endgroup\$ Aug 28 at 17:06
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All the current answers focus on the diode equation - and that's fine. But if you have a specific diode in mind, you can look in the datasheet - practically all diodes will have the I-V curve in the datasheet. So just look at the graph and check at what current it will drop 0.7 V. Noe that this is still subject to tolerances.

Here is an example of a random diode found online. Looking at the graph, we can see that 1H4-1H5 will drop 0.7V at around 80 mA. (Thanks Tony for providing a better graph). Datasheet link

enter image description here

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  • \$\begingroup\$ Notice how the graph doesn't go down to 50 uA and that an exponential on a log scale shud start as a straight line. Notice how the graticules are not logarithmic. This looks wrong. I would expect better plots from a professional diode company. But I see better results on other parts dccomponents.com/upload/product/original/244596315940.pdf \$\endgroup\$ Aug 28 at 16:58
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    \$\begingroup\$ @TonyStewartEE75 edited to use the datasheet you found. \$\endgroup\$
    – jaskij
    Aug 28 at 17:36
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Well, let's make a mathematical closed solution. I know that this is maybe above the OP's knowledge, but I think it is important to show it in combination with the other answers given.

The Shockley diode equation, gives the relation between the voltage across and the current through a diode:

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\space\Longleftrightarrow\space\text{V}_\text{D}=\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(\frac{\text{I}_\text{D}}{\text{I}_\text{S}}+1\right)\tag1$$

Where \$\text{I}_\text{D}\$ is the diode current, \$\text{I}_\text{S}\$ is the reverse bias saturation current, \$\text{V}_\text{D}\$ is the voltage across the diode, \$\text{q}\$ is the electron charge, \$\text{k}\$ is the Boltzmann constant, \$\text{T}\$ is the temperature and \$\eta\$ is the ideality factor.

So, for your case we can see that:

$$\text{V}_\text{source}=\text{V}_\text{D}+\text{V}_\text{R}\tag2$$

Using Ohm's law, we can see that:

$$\text{V}_\text{R}=\text{I}_\text{R}\cdot\text{R}\tag3$$

Because the diode and resistor are in series we know that:

$$\text{I}:=\text{I}_\text{D}=\text{I}_\text{R}\tag4$$

So, we get:

$$\text{V}_\text{source}=\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(\frac{\text{I}}{\text{I}_\text{S}}+1\right)+\text{I}\cdot\text{R}\tag5$$

Now, we can solve \$(5)\$ for \$\text{I}\$ and plug it into \$(1)\$ in order to find the voltage across the diode.


If you solve my last statement you will find:

$$\text{V}_\text{D}=\text{V}_\text{source}+\text{I}_\text{S}\text{R}-\frac{\eta\text{k}\text{T}}{\text{q}}\cdot\mathcal{W}\left(\frac{\text{q}\text{I}_\text{S}\text{R}}{\eta\text{k}\text{T}}\cdot\exp\left(\frac{\text{q}\left(\text{I}_\text{S}\text{R}+\text{V}_\text{source}\right)}{\eta\text{k}\text{T}}\right)\right)\tag6$$

Where \$\mathcal{W}\left(\cdot\right)\$ is the product log function or Lambert W function.

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    \$\begingroup\$ In industry the forward diode voltage is \$V_F\$ and the very wide tolerance in all datasheets at 25'C is due to Rs which cannot be estimated from the Lambert Function.but can be estimated from the diode power rating.(85'C) Rs=k/Pmax. But if you can derive it mathematically, I may learn something \$\endgroup\$ Aug 27 at 21:35
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    \$\begingroup\$ @TonyStewartEE75 You're right about those but, I think Jan kept it relatively simple for the sake of the OP. After all, delving too deep into non-idealities tends to obscure the underlying principles. \$\endgroup\$ Aug 28 at 6:32
  • \$\begingroup\$ @aconcernedcitizen the underlying question was actually from not understanding the simulator model choice was a poor one. Like using a 25A diode for 50uA \$\endgroup\$ Aug 28 at 12:20
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    \$\begingroup\$ @TonyStewartEE75 If the explanation for such a misunderstanding would encompass all the nitty-gritty details, wouldn't you say it would obscure the understanding, more than not? Yes, Rs is problematic, but until then it's about learning how to use the Shockley equation because, as I understand the OP, it's about mistaking the fact that the diode doesn't have a fixed voltage drop. So, Shockley to the rescue. \$\endgroup\$ Aug 28 at 12:35
  • \$\begingroup\$ It would be better to say it's simply exponential but depends on characteristics of device chosen and reasons for use and expectations. Otherwise this language is foreign. Whereas the problem was simple and the solution is trivial, what does @user think that matters. \$\endgroup\$ Aug 28 at 13:58

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