How do I know if the diode has 0.7 V?

Question

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

I've had this problem in this circuit:

^{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"

Answer 1

"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.

Answer 2

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.

Answer 3

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

Answer 4

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.

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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.