# I have fundamental confusions of PN junction diode.

Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Jan 5, 2016.

1. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.

For silicon PN junction, built-in voltage is ~ 0.7 V. I believe it means voltage measured on the silicon diode without biasing should give us 0.7 V in a way that n is higher than p in voltage.

However, we normally say that forward-bias voltage of silicon diode for good conduction is also ~ 0.7 V. Previously, I simply thought 0.7 V forward-biasing (p is higher than n in voltage) is necessary to suppress original built-in voltage but...if this is right, we have to say that silicon diode starts to conduct when 0 V is across the diode!

There must be something I'm confusing right now.

Another confusion is Shockley diode equation. It looks this equation is widely used however, it doesn't include any built-in potentials for different material. It means, this is only accurate for high voltage across the diode, high enough that built-in voltage is ignorable. Is my conclusion right?

2. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
89
I am not getting you.

3. ### Dodgydave AAC Fanatic!

Jun 22, 2012
5,159
772
Yes your totally Barking mad, silicon diodes have a forward voltage drop of approx 0.7V, whereas a Shockley, or ultrafast diode has 0.2V.

4. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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5. ### kubeek AAC Fanatic!

Sep 20, 2005
4,691
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Diodes don´t produce any voltage on their own. The foward drop acts like a barrier that you have to overcome.

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6. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Am...I'm sorry but I need to clarify more.

You mentioned the diode doesn't produce any voltage on its own. But I know the presence of built-in voltage in zero bias. The zero bias means we didn't do anything on the diode. So built-in voltage is the voltage we must be able to measure without any electrical play on it. Which step was I wrong?

7. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
89
what you want to do?
don't make your own theory it is not easy.

8. ### RRITESH KAKKAR Senior Member

Jun 29, 2010
2,831
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So built-in voltage is the voltage we must be able to measure without any electrical play on it.

It mean where the +ve and -ve ion meet it make barrier voltage like in resistance.

9. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Thanks to comment me, however, I still don't get your point. I'm sorry.

When you look at 4th image of figure B in https://en.wikipedia.org/wiki/P–n_junction
you can easily see what I mean, clearly N side voltage is higher than P side without biasing due to built-in voltage. That figure rose this question to me.

10. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
I'm sorry people. I found what I was stuck in.

As applied voltage increases, depletion region gets narrower and the voltage across the region also becomes weaker. When the voltage is close to the original built-in voltage of the junction, the voltage over the region is so small so we can ignore it.

1st question is solved. But there is still 2nd question; Shockley diode equation (also called ideal diode equation) doesn't include this feature so it only roughly describe exponential increase of the current so it is no more than educational usefulness right?

11. ### anhnha Active Member

Apr 19, 2012
776
48
Actually when diode is forward biased, the voltage across diode will be about 0.7V. The reason is that measuring built-in voltage requires that metal wires be attached from the device to a meter. It can be shown that the sum of the contact potentials at the interfaces from metal to p and at n to metal exactly cancel built-in voltage.
http://electronics.stackexchange.co...ential-difference-across-a-disconnected-diode

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12. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Thanks to give me detailed comments.

The contact potential across the metal to semiconductor...It seems little more complicated than I expected.

I've read articles you cited but still no reveals clear explanation of what this is.

Could you give me how this potential arises or good article to explain this?

Anyway, I found very interesting picture in https://learn.sparkfun.com/tutorials/diodes/real-diode-characteristics

The multimeter has special function of measuring diode voltage so clear built-in potential is revealed! I don't know how it is possible.

13. ### kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
Shockley equation https://en.wikipedia.org/wiki/Diode#Shockley_diode_equation describes the relarion between applied external voltage and flowing current.
The parameter that sets the forward voltage is n - the ideality factor, also known as the quality factor or sometimes emission coefficient along with Is, which set where the knee of the I-V curve happens.
See http://www.wolframalpha.com/input/?i=plot+y=4.37e-9*(e^(x/(1.9*0.025))-1)+0+to+1 for a model of 1N4148 with the parameters filled in, X axis is voltage in volts, Y is current in amps.

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14. ### anhnha Active Member

Apr 19, 2012
776
48
I am not sure exactly what is your confusion now.
From the link above, here is what I summarize:
Let's call the contact voltages as follows:
metal-p semiconductor junction: V1
p-n junction: Vbi (built in voltage)
n semiconductor -metal junction: V2
We have: V1 + Vbi + V2 =0. Therefore, we can't measure Vbi by using a normal voltmeter.
So if you forward bias the pn junction the forward voltage Vf:
The voltage across pn juction is -(V1 + Vbi + V2) + Vf = Vf that is about 0.7V.

You can google more about contact potential and built-in voltage in PN junction.
Here is one:
web.mit.edu/6.012/www/SP07-L5.pdf

Last edited: Jan 5, 2016
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15. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.

I'm actually wondering what is the mechaism to make contact potential (How junction between metal and semiconductor makes the potential across its interface). Is it also due to diffusion process as PN junction?

16. ### anhnha Active Member

Apr 19, 2012
776
48
I don't think that I can explain it better than this.
https://en.wikipedia.org/wiki/Metal–semiconductor_junction

17. ### crutschow Expert

Mar 14, 2008
13,509
3,385
The diode potential is a virtual voltage that cannot be measured without applying a current through the junction.
That's what the ohmmeter does in the article you referenced in your post #12.
It applies a current to the device and measures the voltage generated from that current.

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18. ### hp1729 Well-Known Member

Nov 23, 2015
2,105
235
You are not wrong, just incomplete. A diode starts conducting long before you say. Remember diodes are used to detect a low power RF signal in a crystal radio. May I suggest a diode exercise. See attachments comparing a silicon diode (1N4148) with a germanium diode (1N270) and a Schottky (1N5819). The exercises are a simple circuit of a diode through a resistor to +V. Using various resistors measure diode voltage and calculate currents and effective resistance of the diode. Graph the voltage.

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• ###### 1N5819.pdf
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Last edited: Jan 7, 2016