0
\$\begingroup\$

If the diode's dynamic resistance was zero , will not we be able to increase the voltage across the diode above 0.7 v ? (I am assuming forward bias )

I am asking this question because my book ( electronic devices by floyd ) says that the increase in forward voltage across the diode above the barrier potential is due to the voltage drop across the internal dynamic resistance of the semiconductive material .

\$\endgroup\$
  • \$\begingroup\$ 1) Real-world diodes never have a dynamic resistance of zero. 2) if the voltage increase above 0.7 V was purely resistive based, what would the I-V curve of a diode look like, especially around Vd = 0.7 V? And what does an actual diode curve look like? \$\endgroup\$ – Bimpelrekkie Feb 28 '19 at 9:18
2
\$\begingroup\$

Your question can be paraphrased as ...

If we take a model of a diode that contains a certain parameter, and set that parameter to zero, then please confirm whether or not this consequence will occur.

It's sort of implied in your question that you find this predicted behaviour incompatible with what you'd expect from a real diode. It is.

If you set up a model, which you might expect to be valid (ie match reality within a certain error) for a certain range of parameters, and then set one parameter outside that range, then you can reasonably expect to get incorrect predictions from your model.

The 'barrier potential' + 'dynamic resistance' model of a diode is very simple, though better than the 'voltage drop = 0.7v' model which works just fine for 95% of applications. There are better models, if you're having trouble with the accuracy of these two.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I am just asking , will the voltage increase above 0.7v \$\endgroup\$ – user207332 Feb 28 '19 at 10:13
  • 2
    \$\begingroup\$ @user207332 will the voltage increase above 0.7v Yes/No answers are "too easy" and serve no purpose. What do you think is the answer and why? I gave some hints in my comment above. I know I'm nasty and trying to get you to think about it but come on, it's not that hard. \$\endgroup\$ – Bimpelrekkie Feb 28 '19 at 10:18
0
\$\begingroup\$

the increase in forward voltage across the diode above the barrier potential is due to the voltage drop across the internal dynamic resistance.

There are 2 components to VI curve:

a) The intrinsic exponential current of If vs Vf
b) a bulk series resistance, Rs where the curve turns almost linear from ~ 1/3 to 100% rated current @ 25'C.

While the If increases exponentially with Vf, is evident on all diode VI curves in datasheets This incremental linear slope of the curve is due to the fixed interface resistance of electrodes and the crystal interface. This is inversely related to size and thus power rating of a PN junction or LED.

I have found by comparing hundreds of datasheets that diodes fall into groups of constant k= Rs*Pd(max rated at some temp) = depending on the design and chemistry.

This means for part of the rated current you can see an exponential rise to 0.7V for Silicon diodes then after this the slope follow a more linear resistor Rs=k/Pmax

For example in epoxy encapsulated LEDs k = 1 +50% /-25% ( e.g. 65mW 5mm LEDs have incremental Rs near 15 Ohms, while SMD power LEDs are closer to k=0.5 and in high power diodes k = 0.25 nom. New LED's tend to have tighter tolerances such as +25%/-13% on Vf typ. due to design/process improvements.

A 100W silicon diode will be < 10 mOhm but > 2mOhm for the Rs value in the upper current range.

So a small signal diode rated for 100mW might have an Rs of 2~10 Ohms so after 0.7V including this resistance so you can extrapolate the pulsed voltage knowing this.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ @ Sunnyskygyt Might you mean "exponential region" instead of quadratic or logarithmic? \$\endgroup\$ – analogsystemsrf Feb 28 '19 at 13:47
  • \$\begingroup\$ I certainly did, TY , must have too late or senior's moment. @analogsystemsrf \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 28 '19 at 16:11
  • \$\begingroup\$ but I could not remember $$V_d\approx n\frac{k\:T}{q}\cdot \operatorname{ln}\left(\frac{V_{f}-V_d}{Rs\cdot I_f}\right)$$ was correct \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 28 '19 at 16:19
  • 1
    \$\begingroup\$ You say "If increases quadratically with Vf"; I think you probably meant exponentially there too. \$\endgroup\$ – Hearth Feb 28 '19 at 16:22
  • \$\begingroup\$ haven't had my 3rd cup of cafe yet \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 28 '19 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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