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Temperature effects on p-n junction diode

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vsg22

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Hi friends;

A relatively simple question on Barrier Potential of a p-n junction diode:

The equation for the barrier potential is as follows:

Vb = Vt loge(Na*Nd/square(n)) mV
where Vt=KT/e

This equation shows that the barrier potential is directly proportional to the temperature. So as temperature increases the barrier potential should increase.

But in contradiction to this, I have found the following statement:

" With increase in temperature, more minority charge carriers are produced, leading to their increased drift across the junction. As a result, equilibrium occurs at a lower barrier potential. It is found that the Vb decreases by about 2 mV/deegre celcius. "

Please clarify the above, I will be grateful to you.

Thank you
Vaibhav
 
vsg22 said:
Hi friends;

A relatively simple question on Barrier Potential of a p-n junction diode:

The equation for the barrier potential is as follows:

Vb = Vt loge(Na*Nd/square(n)) mV
where Vt=KT/e

This equation shows that the barrier potential is directly proportional to the temperature. So as temperature increases the barrier potential should increase.

But in contradiction to this, I have found the following statement:

" With increase in temperature, more minority charge carriers are produced, leading to their increased drift across the junction. As a result, equilibrium occurs at a lower barrier potential. It is found that the Vb decreases by about 2 mV/deegre celcius. "

Please clarify the above, I will be grateful to you.

Thank you
Vaibhav

I'm not sure and have not looked it up but you might be missing a saturation current term (Is) which, is _usually_ very small to what is being considered and hence ignored. However, it is a much stronger function of temperature that overcomes the directly proportional relationship in the KT/e term.
 
Look at it in this way
Law of mass action states that the product of concentrations of electrons and holes is always constant at a fixed temperature
(ni)^2=n*p where i is subscript
Again this law is applicable to extrinsic semiconductors.
The n in the denominator is actually (ni)intrinsic carrer concentration. Since you are increasing the temperature the value of (ni) will increase more as it is also temperature dependent.
Hence reducing the value of the potential barrier Vb
 
btw if you don't mind me sidetracking a moment. would increasing the temperature of a diode cause it to produce electricity ?
 
Thunderchild said:
btw if you don't mind me sidetracking a moment. would increasing the temperature of a diode cause it to produce electricity ?

If you heat one of the junctions and cool the other then yes
 
then i should connect two diodes together and put one in the cold and one in the hot ? how do i connect them ? pn-pn, pn-np or np-pn ?
 
If you heat one of the junctions and cool the other then yes

No you cannot heat one junction and cool the other. That can be done in metals ( i.e the thermocouple action)

would increasing the temperature of a diode cause it to produce electricity
heating of a semiconductor can cause valence electrons to acquire more energy to break their bonds and enter the conduction band in case of both intrinsic and extrinsic semiconductors. Under open lead condition the direction of flow of electrons or holes is random and hence net current produced is zero. Again no amount of heating can cause a semiconductor to overcome its potential barrier. Too much heat will simply destroy the semiconductor.
But when connected to a proper ckt and proper biasing is done then the current flowing through the diode will depend on the supply and the temperature.
 
electronist said:
If you heat one of the junctions and cool the other then yes

No you cannot heat one junction and cool the other. That can be done in metals ( i.e the thermocouple action)

Except there is devices out there that work on this priciple. BUT the doping area is bigger.

IF you could practically heat one area and cool the other then current would flow. BUT in hte size of a regular PN junction this isn't really possible
 
I just came across the question. (sorry for taking so long to answer)
I have an explanation. Mathematic is not needed here. Just basic semiconductor concept is enough.
Imagine that you have a pn junction and you raise up the temperature so much.
What happen to the p-type and n-type is that they will approach the intrinsic semiconductor.
Imagine that you have intrinsic semiconductor in contact with intrinsic semiconductor....... what do you think the built-in barrier potential?
:)
 
Vsg22,

"Vb = Vt loge(Na*Nd/square(n)) mV
where Vt=KT/e

"This equation shows that the barrier potential is directly proportional to the temperature. So as temperature increases the barrier potential should increase."

Yes, that is true.

"But in contradiction to this, I have found the following statement:

" With increase in temperature, more minority charge carriers are produced, leading to their increased drift across the junction. As a result, equilibrium occurs at a lower barrier potential. It is found that the Vb decreases by about 2 mV/deegre celcius. "

There is no contradiction. You are confusing the barrier potiential or built in voltage across the junction with the Va (applied) voltage. The Vbi is caused by diffusion, and is dependent on the temperature and the relative doping concentrations. It is defined to be valid at thermal equilibrium and zero net current. The Shockley equation defines the relationship between the applied voltage, current, and temperature. That is where the -2mv/°C temperature coefficient somes from. Va and Vbi are not the same.

Varakorn,

"Imagine that you have a pn junction and you raise up the temperature so much.
What happen to the p-type and n-type is that they will approach the intrinsic semiconductor.
Imagine that you have intrinsic semiconductor in contact with intrinsic semiconductor....... what do you think the built-in barrier potential?"

That statement or question is not pertinent to the question the OP asked. You are asking what is the Vbi is when you put two intrinsic semiconductor slabs together. Plug and chug the values into the formula the OP gave and see what the answer is.

Ratch
 
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