enter image description here

(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)

  • \$\begingroup\$ The diode does not conduct until Vf is 0.7V. So you can ignore the diode. For voltages from negative infinity to +0.7 (using the simplified diode model). +0.5 is in that range. \$\endgroup\$ Sep 3, 2018 at 16:58
  • \$\begingroup\$ I am trying to figure out how simulators do it. So ignoring isnt an option. \$\endgroup\$
    – Etwus
    Sep 3, 2018 at 17:09
  • \$\begingroup\$ What current does your simulator estimate will flow? \$\endgroup\$
    – Transistor
    Sep 3, 2018 at 17:13
  • 2
    \$\begingroup\$ The best book I know of, discussing the details used in simulation, is Laurence W. Nagel's thesis, which later became "Memorandum No. UCB/ERL M520, May 9th, 1975" available from Electronics Research Laboratory, College of Engineering, University of California, Berkeley, CA 94720. You should ask for a copy. Start at page 35 and forward. Then go to page A2.7 and forward. \$\endgroup\$
    – jonk
    Sep 3, 2018 at 17:26
  • 2
    \$\begingroup\$ The D-model includes an Ohmic resistance, \$R_S\$, not shown in your diagram (and which varies with current because the model doesn't normally include high-level injection modeling.) But an approximate method is Current through a resistor with diode and a closed solution is $$I_\text{D}=\frac{1}{R_1}\left[V_\text{T}\cdot\operatorname{LambertW}\left(\frac{R_1\cdot I_\text{SAT}}{V_\text{T}}\cdot e^\frac{V_\text{CC}+R_1\cdot I_\text{SAT}}{V_\text{T}}\right)-R_1\cdot I_\text{SAT}\right]$$ \$\endgroup\$
    – jonk
    Sep 3, 2018 at 17:39

4 Answers 4


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.


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. enter image description here

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.

enter image description here

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

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


You can measure diode current right down to zero volts. However, below the magic figure of 0.7v (for silicon), it tends to be very small.

It depends on the type of diode what the current is. Here is a picture of some measurements I did a while ago to answer a related question. Do click on it if you want to read it. What surprised me is how bad the 3v zener was at stabilising a voltage.

enter image description here

So how do you figure out the diode voltage?

For a first order approximation, we know the diode current is going to be 'very small' (as the voltage is below 0.7v), so we can neglect the resistor voltage drop, and say the diode voltage is about 0.5v.

For a second order approximation, you can see from the figures above the current for my 1N4148 at around 0.5v at whatever ambient temperature it happened to be was about 100uA. That dropped over 100 ohms is 10mV. So with more accuracy we can estimate the diode voltage as 490mV. Interestingly, the figure for a much bigger 1N4006 turns out to be very similar.

There's absolutely no point going to a higher order approximation without specifying the diode, and the temperature.


Diodes have a current-voltage relationship. (see image)

To make a long story short, a diode needs an amount of positive voltage applied to "turn on" the diode and let current flow through it.

In your example the current flow is I=U/R => 0.5/100 = 5mA. The supplied voltage is 0.5V.

There is always some voltage drop across the resistor (but that is negligible) so to answer your question, in this case the voltage drop is 0.5V.

How do I come to this answer? It's by looking at the i-v characteristics:

Diode characteristic (i-v) relationship (Image source: https://learn.sparkfun.com/tutorials/diodes/real-diode-characteristics)

This characteristic is saying the following: In order to let a current flow through a diode, a certain voltage has to be applied to the diode which is in the rule around 0.7V. So you have to look at the voltage and following from that, you can read from the graph the amount of current that can flow through it.

Since in your case only 0.5V is applied, it is all used by the diode and only a small amount of current can flow through it. If the supplied voltage would for example be 5V, then there would be a voltage drop of 0.7V and you would measure 4.3V after the diode. So in summary, the amount of voltage drop that is possible across the diode, determines how much current can flow through it.

There are equations to calculate how much current exactly can flow on certain voltages, but since this is an exponential signal it is hard and not advised to use in real life applications. The reason for this is that if you supply 5V, the voltage is never always 5V; it moves up and down because of noise and of course can be stabilized by capacitors and other circuits, but making it perfect for real life applications is hard and expensive.

I hope I could be of some help for you :)

To help you in the future, LTSpice is a great program for these question to figure out yourself. You should take a look at this program. It is free to download and massively used by engineers


Here is a simulation of diode current from 50mV to 500mV. As you can see, it's a continuous function. I've used a logarithmic Y-axis scale to make it a bit more clear.

enter image description here

Generally the algorithms used in SPICE require the functions to be continuous and differentiable in order to converge. Things like step functions are typically approximated with functions using tanh(), for example.


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