Voltage drop on diode

(0.5V ---- 100R resistor --- diode --- gnd)

How do you figure out voltage drop on this diode? (Lets assume forward voltage of this diode is 0.7V)

Answer

To answer your question, what is the diode voltage drop with Vin=0.5V with 100Ω in series?

My answer

• 430mV with 70mV drop on the resistor from 700uA for an ideality factor of 1.

You cannot figure it out without some assumptions on the Ideality Factor or knowing the current in the diode.

more

In addition to other fine answers, looking at common Silicon diodes with Ideality Factors other than 1. (@analogsystemsrf previously illustrated if Ideality Factor=1, then the diode voltage rises 58mV/decade in current @ 25'C.

The important thing to remember is that it is continuously logarithmic over at about 4+ decades of current. For simple applications, this log. characteristic may be limited by noise at the lower limit and linear behavior from internal resistance, Ri at the rated current, although not often stated in datasheets.

From the datasheet curves, the log Vf/If slope and Ri may be computed near rated DC current.
After repetition, you remember for future use he values for 2 common diodes;

1N4448 rated for 0.1 Adc @ 1.0V ... Vf/If= 113 mV/ decade (If), Ri= 600 mΩ
1N4005 rated for 1.0 Adc @ 1.0V ... Vf/If= 140 mV/ decade (If), Ri = 66 mΩ

Notice that the current rating and Ri product are similar. This is not coincidence. (meaning high the current rating , lower the Ri internal resistance.

The higher current diode here also has a higher Vf/If slope.

Now if you put a sawtooth current source into a diode to plot voltage by using a large series resistor and voltage, you get this.

So in short, we use 0.7V for diode drop for convenience.

But it is hardly a hard switch, but still very effective.