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Resistance of a semiconductor diode

I was reading this question about the resistance in a diode. I have a question regarding the resistance of the depletion region. I know that the conductivity of depletion region is infinite but due to presence of electric field there's a dynamic resistivity thus a diode is represented by an EMF with a equivalent resistor. Does this resistor dissipate heat?

My second question, the recombination of electrons and holes does dissipate heat?

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Tonylb1
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  • I really suggest you [split your question](http://meta.stackexchange.com/questions/246328/dealing-with-bundle-omnibus-list-of-questions-question-that-consists-of-rather-d). How are you going to decide the best answer? Stephen answered the first question while Olin answered the 2nd. And the 2nd question is easily answered (correctly) on https://en.wikipedia.org/wiki/Carrier_generation_and_recombination – Fizz Nov 15 '15 at 17:48

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As I said in my answer to that question, a forward-biased diode (or an IGBT or BJT for that matter) can be approximated as a fixed voltage drop in series with a resistor. This "resistor" does dissipate heat; a current flowing through a voltage drop always means energy is either being stored or dissipated, and diodes don't store substantial energies.

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Stephen Collings
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  • yes but here this resistor does not mean the conductivity of the depletion region. Thus there's an electric field due to separation of charge in the depletion region. And the depletion region has a very high conductivity, thus this resistance is equal to V/I in the depletion region. That's why I am asking if it does dissipate heat – Tonylb1 Nov 15 '15 at 14:46
  • http://www2.pv.unsw.edu.au/nsite-files/pdfs/UNSW_Understanding_the_p-n_Junction.pdf here's an example. In page 21, It is said for a forward bias diode, the electric field of the source opposes the electric field inside the depletion region thus lowering its resistance. Here than the resistance is due to electric field and not the conductivity like in a normal resistor. – Tonylb1 Nov 15 '15 at 16:48
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Recombination of electrons and holes doesn't really dissipate heat. Keep in mind that "holes" are really a handy abstraction for thinking about semiconductors. They don't really exist as physical entities other than a place where a electron could be but isn't.

Electrons aren't completely free to move around. Their movement requires some power to sustain inside a conductor. A inaccurate but still useful mental picture is that there is "friction" between the electrons and the molecules they move between.

Olin Lathrop
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  • I am talking about the characteristic voltage-current of the depletion region and its equivalent resistance. – Tonylb1 Nov 15 '15 at 14:32
  • Tony, the diode and its pn junction have a nonlinear V-I characteristic. For this reason, you must clarify what you are interested in: DC resistance R=V/I or dynamic resistance r=dV/di. Both are different and depending on the selected operating point. – LvW Nov 15 '15 at 14:46
  • Yes I know this. But both, and for any selected operating point, the resistance here is different than a conductor. My point is, does this resistance dissipate heat or just there's an electric field thus a potential drop? ( I know also that in a resistor there's an electric field but the difference here that R does not depend on conductivity of depletion region) – Tonylb1 Nov 15 '15 at 14:52
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    "Recombination of electrons and holes doesn't really dissipate heat." is wrong. Unless you're talking about an LED, the energy released as the electron drops from conduction band to valence band will end as heat. – The Photon Nov 15 '15 at 18:18
  • I was thinking the same. The magic words being "phonon" and [SRH](https://en.wikipedia.org/wiki/Carrier_generation_and_recombination#Shockley.E2.80.93Read.E2.80.93Hall_.28SRH.29_process). I think the reason a junction without an externally applied E field doesn't warm up is [instead] that there's equilibrium of generation and recombination, so the thermal effects of these cancel out in this case. – Fizz Nov 15 '15 at 18:31
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Like all semiconductors, they have a dynamic conductance which is dependent on the applied voltage (electric field) and load applied. All diode datasheets have V-I curve which is basically giving you its "resistance". Since R = V/I. You can draw a load line on this curve. The greater the current the larger the voltage that develops across it.

Thus P = VI, and P = I^2 R or V^2/R so yea your dam right it dissipates heat.

Also its dangerous to think of quantities as infinite. Nothing is infinite.

crowie
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  • Nothing is infinite ... except the universe and man's capacity for stupidity, except I'm not sure about the universe (Albert Einstein). – Neil_UK Nov 15 '15 at 15:11
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The dynamic resistance does dissipate heat, as does the current flowing through the 'fixed' V in series with it.

Imagine a diode with 1 mA, and 0.726 V. this is 726 uW total power. The small signal dynamic equivalent model is 700 mV in series with a 26 ohm resistor.

The power is 1 mA * 700 mV + 1mA^2 * 26 ohm = 700uW + 26 uW = 726 uW -- same as before.

This model remains valid for small excursions around 1 mA (e.g. 0.1 mA). Changing the current to 1.1 mA increases power to 1.1 mA * 700 mV + 1.1mA^2*26 = 770u+31.5u = 801.5 uW.

Now, I know that the voltage on a diode increases by about 2.6 mV when the current increases by 10 % (from the exponential behavior). Therefore, the 726 mV increases to 728.6 mV. 1.1 mA in 728.6 mV is also 801.5 uW (basically because of the same reason that 10 % gives 2.6 mV).

Recombination of electrons and holes does dissipate heat, just as separating them across a potential barrier does. While separated, the energy stored in them is similar in nature to the energy stored in a capacitor (although there are complications analogous to the small-signal resistance with the capacitor also; not least of which is that the capacitance is not constant with voltage).

jp314
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  • The dynamic resistance is like a normal resistor (depending of conductivity of the material etc..)? – Tonylb1 Nov 15 '15 at 21:01