1
\$\begingroup\$

In datasheet I observed that forward voltage of diode given as between two values (min and max). In my application the current though the diode is not constant it switches every 15 min(may be more or less).

Is it that the forward voltage remains constant even if the forward current changes ? Does current has any affect on forward voltage?

\$\endgroup\$
  • 2
    \$\begingroup\$ Have you ever had a look at a diodes datasheet? \$\endgroup\$ – PlasmaHH Sep 20 '17 at 12:52
  • 1
    \$\begingroup\$ It could be the temperature. If you are passing huge current without much heat sinking, the diode will get hot and might lead to a decrease in current. \$\endgroup\$ – Whiskeyjack Sep 20 '17 at 13:00
  • \$\begingroup\$ Out of interest, what are the currents? \$\endgroup\$ – Trevor_G Sep 20 '17 at 13:30
6
\$\begingroup\$

Let's have a look to the current versus voltage of a real diode (1N4148):

enter image description here

The voltage varies almost linearly with the LOG of ID for ID < 10-30 mA (then series resistance dominates, and there is an almost linear relation). If you have 1mA there will be a drop of about 0.6V. This rises to about 0.8V@10mA. At 100mA, it's about 0.9V.

So a quick answer: no the forward voltage is not constant, but for small variations of the current, it can be considered constant.

In fact, in (manual) circuit analysis it's usual to replace, as a first-order approximation, a ON-state diode with a constant voltage generator, that has a 0.7-V value (or 0.35 for Schottky diodes, or 0.2 V for germanium diodes).

schematic

simulate this circuit – Schematic created using CircuitLab

But this is a really crude approximation. A much better approximation is to replace the ON-state diode, with the series of a voltage generator and a resistor. In some cases the source voltage is decreased (because there is already the resistor).

schematic

simulate this circuit

A much better approximation is given by the Shockley equation:

$$I=I_S(e^{\frac{V_D}{nV_T}}−1)$$

Still, this is an approximation as it does not take into account the Shockley-Hall-Read generation-recombination at low forward levels in the space charge regions, nor it takes into account the breakdown at high negative bias.

In any case, since \$V_T=\frac {KT}{q}\$, where T is the absolute temperature in Kelvin, the current also depends strongly on the temperature.

This is in fact confirmed on real diodes:

enter image description here

Finally, device-parameter dispersion will make the forward voltage of each diode slightly different from the other.

| improve this answer | |
\$\endgroup\$
  • 3
    \$\begingroup\$ To anyone who thinks "0.7V" is Bob's own truth, try measuring across a big power diode while heavy currents are running through it. I've seen 1.1V before. (Bob Pease, of course.) \$\endgroup\$ – Warren Young Sep 20 '17 at 13:36
  • 1
    \$\begingroup\$ I totally agree. When doing ball-park calculations I always use 1V or larger (or what's written on the datasheet) for power diodes @ high current. Also look at the 1n4148 @ 100mA: it already goes up to almost 1V... \$\endgroup\$ – next-hack Sep 20 '17 at 13:41
  • 1
    \$\begingroup\$ Shockley; not Schockley (but Schottky; not Shottky) \$\endgroup\$ – Curd Sep 20 '17 at 13:57
5
\$\begingroup\$

Diode current has a small effect on the forward voltage, as given by the Shockley diode equation

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

For most forward voltages, and currents where the residual resistance of the diode is not important, \$V_D\$ is proportional to the natural log of the current flowing. n is normally in the range 1 to 2, and \$V_T\$ is around 25mV at room temperature, so the forward voltage increases by a few tens of mV when the current increases by a factor of 3.

\$I_S\$ is a function of the diode junction area and doping. It's in the nA range for diodes like 1N4148, smaller for low leakage diodes, and larger for power diodes and schottkies.

As \$V_T\$ is a function of temperature, if the change in current changes the temperature, that will cause a a further change in diode forward voltage.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Shockley; not Schockley (but Schottky; not Shottky) \$\endgroup\$ – Curd Sep 20 '17 at 13:57
  • \$\begingroup\$ @Curd whatever, I'm glad somebody's awake! Fixing. It's very inkonsideratte of them to have such schimilar names. \$\endgroup\$ – Neil_UK Sep 20 '17 at 15:07
  • 1
    \$\begingroup\$ at least they could have choosen not to work both in semiconductor physics \$\endgroup\$ – Curd Sep 20 '17 at 15:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.