I was wondering about the voltage drop in real diodes (0.7 V in Si diodes, 0.3 in Ge diodes etc.). According to my experience this drop is linked to the used material, but the shockley equation which should describe a diode never uses any material constant. So, how is the material constant introduced in a diode equation for modelling the additional voltage drop compared to an ideal diode?


2 Answers 2


Shockly diode equation is given by: $$\style{}{I=I_S(e^{V_D/nV_T}-1)}\tag1$$


  • \$I_S\$ is the reverse bias saturation current.
  • \$V_D\$ is the voltage across the diode
  • \$V_T\$ is the thermal voltage
  • \$n\$ is the ideality factor, also known as the quality factor or emission coefficient.

I see two material dependent parameters here:

  1. The reverse saturation current, \$I_S\$ depends on the material.
  2. The parameter \$n\$ depends on fabrication process and semiconductor material.

From equation (1), $$V_D = n \cdot V_T \ln\left(\frac{I}{I_S}+1\right) \approx n \cdot V_T\ln\left(\frac{I}{I_S}\right)$$ $$V_D \approx n \cdot V_T \cdot \ln10 \cdot \log_{10}\left(\frac{I}{I_S}\right)$$

Assuming room temperature and \$n=1\$, $$V_D \approx 0.05916 \cdot \log_{10}\left(\frac{I}{I_S}\right)\tag2$$

Typical values of the saturation current at room temperature are:

  • \$I_S = 10^{-12}\$ for silicon diodes;
  • \$I_S = 10^{-6}\$ for germanium diodes.

For a current of 1.0 mA:

  • \$V_D \approx 0.53 V\$ for silicon diodes (9 orders of magnitude)
  • \$V_D \approx 0.18 V\$ for germanium diodes (3 orders of magnitude)

For a current of 100 mA:

  • \$V_D \approx 0.65 V\$ for silicon diodes (11 orders of magnitude)
  • \$V_D \approx 0.30 V\$ for germanium diodes (5 orders of magnitude)

Further increase in current won't cause much increase in \$V_D\$ (0.05916V per decade is the rate of change of \$V_D\$ with respect to \$I\$). Hence in the common cases (current in the range of mA), the voltage drop remains constant around 0.6V for silicon diodes.

And hence values of 0.6 or 0.7 Volts are commonly used as voltage drop for silicon diodes and 0.3 for germanium diodes.

source: wikipedia

  • 2
    \$\begingroup\$ The questioner did ask for the "additonal voltage drop compared to an ideal diode". The above equation (Shockley) applies to an ideal voltage only and does not take into consideration the voltage drop across the path resistance. \$\endgroup\$
    – LvW
    Jul 10, 2014 at 15:15
  • \$\begingroup\$ I already know the Shockley equation, but if I want to calculate a real diode with it, I get into trouble... \$\endgroup\$
    – arc_lupus
    Jul 10, 2014 at 20:05
  • \$\begingroup\$ @arc_lupus please see the edited answer. \$\endgroup\$
    – nidhin
    Jul 11, 2014 at 19:20
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    \$\begingroup\$ @arc_lupus, what do you mean with "get into trouble"? As I have explained in my answer (below) there is a linearizing effect caused by the diode´s ohmic path resistance. Thus, the voltage drop does NOT remain constant around 0.6V (as mentioned before). Rather, a tangent drawn at the linear part of the I-V characteristic crosses the voltage axis at a voltage of app. 0.65 volts. A corresponding verification can be found in many textbooks. \$\endgroup\$
    – LvW
    Jul 12, 2014 at 8:22
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    \$\begingroup\$ @arc_lupus, one more information (perhaps it helps): It is easy to construct the I-V characteristics of a real diode. At first, draw the ideal exponential curve for the pn junction. At second, draw a straight line with the slope 1/Rpath (ohmic law for the path resistance). At third, add both curves horizontically (because both voltage drops within the diode are added). As a result, you get the linearized diode characteristics. \$\endgroup\$
    – LvW
    Jul 12, 2014 at 10:54

The Shockley equation does NOT describe the equation of the part called "diode". It establishes the voltage-current relation of the pn junction only. This exponential function is linearized (for increasing currents) caused by the ohmic path resistance.

EDIT: The known exponential function has a factor Io (reverse saturation current) which is different for Si and Ge. This factor contains the carrier concentration factor which is proportional to the material-dependent potential gap (1.1 for Si and app. 0.6 for Ge).

  • \$\begingroup\$ Thus I have to add a constant value (how do I get this?) to the result of my shockley equation? (And why is the shockley equation called "diode equation"?) \$\endgroup\$
    – arc_lupus
    Jul 10, 2014 at 12:06
  • \$\begingroup\$ But a schottky diode is made of metal-semiconductor rather than semiconductor-semiconductor (P-N) doped regions. Does it still comform to this equation? \$\endgroup\$
    – sherrellbc
    Jul 10, 2014 at 12:31
  • \$\begingroup\$ @arc_lupus: For the pn junction the exponential function is exp(Vj/Vt) with Vj: Voltage across the junction and Vt: thermal voltage. For the whole diode you have to add the ohmic voltage drop across the path resistance Rp. This can be expressed as exp(Vd-I*Rp) with Vd: external diode voltage. Note that I=f(Vd). This causes the effect of linearizing the exponential function. \$\endgroup\$
    – LvW
    Jul 10, 2014 at 13:02
  • \$\begingroup\$ Anything wrong with my answer (seems to be "not useful")? \$\endgroup\$
    – LvW
    Jul 10, 2014 at 15:26
  • \$\begingroup\$ No, thus I gave you an upvote. And thanks for your help, too! \$\endgroup\$
    – arc_lupus
    Jul 12, 2014 at 13:22

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