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The voltage across the diode when forward biased is 0.7 V as written in books, but in datasheets the forward bias voltage is usually rated as 1.1 V for higher currents which is greater than 0.7 V. Why is that?

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  • \$\begingroup\$ In addition, the specific type of diode will have a different forward voltage range. Fast Schottky diodes may only have a drop of 0.3V or even 0.1V, while the forward drop of a Zener may be 0.8V, and a high-voltage diode (kilovolts) may drop 2V or more (all unloaded.) \$\endgroup\$
    – rdtsc
    Feb 22, 2023 at 12:39
  • \$\begingroup\$ ... and LEDs will be all over the map, with huge differences by color. \$\endgroup\$ Feb 22, 2023 at 15:03
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    \$\begingroup\$ 0.7V is a rule of thumb for typical "middle of the road" small signal applications of a silicon diode, with forward currents in the range between 0.1mA and 10mA. It becomes quite inaccurate in pulse current rectification in power supplies, and in low bias small-signal applications - e.g. try pushing 10μA through an 1N4001 and see what forward voltage you get :) The textbooks do everyone a disservice by not being clear in what range of applications this rule of thumb applies. E.g. I do plenty of analog computation circuits where the diode forward voltage of 50mV is perfectly useful. \$\endgroup\$ Feb 23, 2023 at 2:06
  • \$\begingroup\$ You have a number of good answers. Accepting one may be a good idea. \$\endgroup\$
    – Russell McMahon
    Apr 30, 2023 at 12:40

5 Answers 5

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The books just present a simple model.

It really never is 0.7 V and will vary under many conditions, such as current through it and temperature.

Voltage will be higher at higher currents due to resistance.

Voltage will be lower at lower currents due to exponential voltage vs. current characteristic, where the resistance doesn't play a significant role.

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The value 0.7V is usually used as a typical value, for back-of-the-envelope calculations, where the current you expect the diode to be passing is milliamps or tens of milliamps, not amps.

An engineer who expects that a diode will pass amps of current knows it will drop more than that, and may use 1.0V as an estimate, rather than 0.7. Similarly, if the current is in the microamps, 0.6V is a closer estimate.

For approximations there's no point using more than a single digit of precision, and it's very often sufficient to get a really good idea of the behaviour of the circuit. On top of that, doing calculations in your head using values with only 1 significant figure is easy.

In the final analysis, though, it may be necessary to get precise, in which case you may need to use the good old diode equation, and/or refer to the device's datasheet.

The diode equation is some variation on:

$$ I = I_S\left(e^{\frac{qV}{kT}}-1 \right) $$

This is a continuous relationship between voltage \$V\$ across the diode, and current \$I\$ through it. If you plot that relationship, there won't be a sudden corner, flat-lining the voltage at 0.7V when current is positive:

enter image description here

As you can see, it's flat-ish for significant positive currents, somewhere around 0.7V, but that value is just a ballpark figure, for typical signal currents, and for convenience.

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Other answers have already stressed the fact that voltage in a diode depends on the current and so the customary 0.7 volts is not the universal answer for all diodes in all conditions. But you might have glanced at the characteristic of a diode on a curve tracer, or on the screen of an oscilloscope attached to an octopus. And the shape of the characteristic, in this case, seems to leave no doubt about where the knee is located.

W2AEW video on diodes
(Source: W2AEW video 308 Back to basics: diodes)

Moreover, considering how steep the curve appears to be, any value above the knee (the bias voltage you refer to in your question) seems bounded to be approximately equal to the knee voltage.

I want to show you that, even for the same diode in the same environmental conditions, there is no knee. It's just an illusion, whose position is based on the vertical scale on which the figure is shown. This is a consequence of the self-similarity property of the exponential function.

The following plots show very different threshold locations and can lead one to think these are different functions:

simple exponential

The first seems to have a knee at about \$x=90\$; the second at \$x=45\$; the third at about \$x=22\$ and the fourth at about \$x=6\$. But in reality all four plots are plots of the very same function:

$$y = e^x.$$

It is just plotted on four different vertical scales.

The same is true with diode characteristics. Assuming the well known Shockley equation is a faithful model for the real device (there might be secondary effects at very low and very high currents), the diode characteristic is just a translated exponential function:

$$I_d = I_S(e^{\frac{V_d}{\eta \, V_{Th}}} - 1).$$

Let's put some random values to get pretty pictures for values of voltage and current in the ballpark of typical electronics applications. Using \$I_S = 15 \times 10^{-12} A\$, \$\eta = 1.4\$, \$V_{Th} = 26 mV\$ and we get the following plots at different vertical ranges

Shockley scaled

So, where is the voltage threshold of this diode?
Is it \$1.1V\$, \$0.9V\$, \$0.7V\$ or \$0.6V\$?
The answer is that the threshold depends on what current you consider the diode conducting, while the bias voltage across the diode depends on what your range of admissible currents happens to be.

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  • \$\begingroup\$ IMHO this is the most important answer, as it points out the fact that scales on an exponential function graph define the look of the function, due to the self-similarity property. So, the threshold voltage of the diode depends only on the forward current range of interest. Diodes biased with small currents will have small forward voltages - way below 0.7V, as you point out! \$\endgroup\$ Feb 23, 2023 at 2:01
  • \$\begingroup\$ @Kubahasn'tforgottenMonica "There is no spoon." \$\endgroup\$ Feb 23, 2023 at 2:13
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The voltage across a diode does vary with load. But because the relationship between voltage and current is exponential, the changes are often small enough that its adequate to model the diode as having a constant voltage. 0.7V is often used as a convenient approximation for the voltage of a typical rectifier diode. 0.7V is not meant to accurately reflect the voltage of any real diode, it's just sort of an average for a silicon diode.

For a more precise model we can use the Shockley diode equation.

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

  • Id is the current through the diode.
  • Vd is the voltage across the diode.
  • Vt is the "thermal voltage", approximately 26mV at room temperature.
  • n is an "ideality factor" that is unique for each diode.
  • Is is the reverse saturation current, which is unique for each diode.

"n" and "Is" are often not given in diode datasheets, but they can be figured out using a little algebra if you know two points on the voltage/current curve.

Note that even the Shockley equation has limitations. For example, Vt varies with temperature. So when you begin using the diode, Vd is going to slowly change as the diode heats up.

The Shockley also ignores any internal resistance in the diode in the leads and other connection points. This resistance, while often small, becomes important at higher currents.

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In the greater scheme of things 0.7V and 1.1V (or 0.5V) are often about the same. In cases where that is not true (the difference is material to the purpose of the calculation) you need a more accurate model of the diode. One that includes current, temperature, ideality factor (emission coefficient), resistance etc., not to mention AC characteristics.

For a rectifier diode such as a 1N4004 in a capacitive filter supply (no inductance other than transformer leakage inductance), 1V is probably a better approximation to use, so 2V drop for a bridge rectifier.

If it's a 1N4148 used in a small-signal circuit at 100uA, it's probably more like 0.5V at room temperature.

Both are silicon PN junction diodes (albeit with quite different characteristics- trr is about 3 orders of magnitude higher for the 1N4004). If it's some other kind of diode then it might be very different (or only a little bit different).

Both those are just examples of "rule of thumb" numbers that I have found useful in certain classes of situations. You would also generally want to err on the conservative side. Even with DC numbers there are subtleties such as the opposing temperature coefficients of resistance and diode junction components of the drop. The -2mV/°C tempco is another "rule of thumb" and you need to know when to question its use. It's still very useful, and simple models often aid in understanding even if they are not as accurate as more complex nonlinear ones used in simulations. Understanding is crucial to coming up with "out of the box thinking". As British statistician George Box wrote:

Since all models are wrong the scientist cannot obtain a "correct" one by excessive elaboration. On the contrary following William of Occam he should seek an economical description of natural phenomena. Just as the ability to devise simple but evocative models is the signature of the great scientist so overelaboration and overparameterization is often the mark of mediocrity.

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