PN junction diode

1. Aug 2, 2006

jinyong

For a PN junction diode, the total potential across the semiconductor equals the built-in potential minus the applied voltage.

Phi = Phi(built-in)-Va(voltage applied)

My question is what happens when the Va exceeds the Phi(built-in)? Will the Phi = negative? But how come they always assume drop of 0.7V for diodes.

2. Aug 2, 2006

chroot

Staff Emeritus
When the applied voltage exceeds the junction's built-in potential, the diode begins conducting. It is said to be "forward-biased."

0.7V is a rough approximation for the built-in potential of normally-doped silicon diodes. Other semiconductors, or unusual doping, will result in a different built-in potential.

- Warren

3. Aug 2, 2006

jinyong

Hmm my semiconductor book defines as
Phi = Phi(built-in)-Va(voltage applied)
And Va as the voltage applied. If it is the way u described shouldn't it be Phi=Va-Phi(built-in) and if Phi > 0 then it's forward biased?

The book also defines when Va is about Phi(built-in) then it's considered high level injection. But what happens when Va exceeds Phi(built-in)?

4. Aug 3, 2006

Staff: Mentor

$$I = I_s (e^{\frac{qV}{kT}} -1)$$

5. Aug 3, 2006

Corneo

I though the term forward biased refered to the configuration when the positive connection is applied to the p side of the diode and the negative on the n side.

6. Aug 3, 2006

7. Aug 7, 2006

D_Dean

**Disclaimer: There is a ton of extra information here, mostly on the functionality of the diode and if you know most of it you can skip to the end. In all honesty I needed to go over most of it to remember how diodes worked and so I thought I'd type it out**

It has been a little while but it seems like what you are asking is what happens when you increase the forward bias or make it more forward bias. When the device has no bias there is a Vbi barrier that limits the P-side hole and N-Side electron majority carriers this is seen in the website you gave us in figure 4.2.3. Thus the main form of current is due to recombination and generation (which gives the slight negative current). As you increase the voltage, the barrier for those majority carriers decreases from Vbi to Vbi - Va, this gives rise to the diffusion current (which is the current that is due to the excessive holes on the P-side and electrons on the N-side flowing across the junction). A linear change in the Phi yields an exponential change in the diffusion current (as can be seen in berkeman's equation above). When Vbi = Va there is no electrostatic barrier and the P-side holes and N-side electrons can move freely across the junction and will move due to the diffusion principle.

This then brings us values of Va which are greater than Vbi. Ideally (according to simplifications made to form most of the equations) the current increases exponentially to infinity as the voltage increases linearly, this is because as you increase the voltage the fermi levels increase and cause there to be more and more majority carriers and therefore more and more current. Picture it like this, in figure 4.2.3 of the site you gave us the device is in equalibrium. if you fix the p-side and move the n-side up, you will eventually get two parallel bands (valence and conductance) with two different fermi levels. As you move the n-side up the majority carriers can still freely cross the junction (actually more easily) because it will "want" to move to a lower potential due to electrostatic principles. The p-side holes will "want" to go to the n-side because the n-side has many more lower energy positions. Same with the electrons vice-versa.

** Start here if you don't need/want to know about the diode functionality**
In the real world problems arise in three ways.
1. At lower forward biases (kT/q < Va < .4v) the R-G current (recombination and generation) still plays a role causing a different equation than the ideal diode equation. Specifically I = Is(exp(qVa/2kT)-1).

2. As you increase the voltage across the junction passed the ideal region (Va > 1v, IIRC) a phenomenon called high-level injuection takes place. Basically, like 1 above some of the assumptions and simplifications taken earlier cause the ideal diode equation to be invalid. And coincidentally the relationship between current and voltage is roughly the same as in 1.

3. As you increase the current running through the semiconductors, the series resistance of the metals begins to play a larger role. I think something like Rs = 1 ohm. So the actual voltage across the junction is Vj = Va - IRs (this is true always but only really matters when the current gets high enough to effect Va). This causes the current to start to plateau as you increase the voltage.

In conclusion.
Yes, Phi does become negative.

The assumed drop of .7 is a rough estimate of the diode in its working condition in the forward bias. A slightly lower voltage will yeild a significantly low current and will therefore not be on, and a slightly higher voltage will yeild (or need to draw) a significantly higher current. Also if you have the current capacity to run at these higher voltages it will run into the problems above.

Again, It's been a little while so I welcome any questions, comments, or criticisms.

D Dean

8. Aug 8, 2006

jinyong

Thank you for replying and writing such a detailed explanation. I just have a couple of questions that I still don't get.

I'm not sure what this means. The diagram indicates for the electrons to wander to the p side it's a higher potential. And for the holes to wander to the n side it's a lower potential(figure 4.2.3). I thought the reason w/ forward bias the electrons and holes can diffuse easier is b/c you have lowered the energy barrier Phi = Phi(built-in)-Va so if Va is positive then Phi will be lower which allows carriers to cross easier.

So if Phi becomes negative(i.e. let's assume Phi(built-in)=0.7V and Va=0.8V, then Phi=0.7-0.8=0.1V right?) why it is that we assume about a 0.7V voltage drop across the diode. If Phi is defined as the total potential across the semiconductor when why we assume 0.7V and not -0.1V or any other negative number? And people have said When the applied voltage exceeds the junction's built-in potential, the diode begins conducting. It is said to be "forward-biased."

That's what confuses me the most.

Thank you for your help again.

9. Aug 8, 2006

D_Dean

It was probably wrong of me to say potential, I didn't mean it as voltage, I'll explain here. The top band (Valance band) represents the electron potentials. As you move further and further up away from the valance band, the charge from the nuclei of the silicon atoms (or other semiconductor and dopants) is weaker. Therefore the electrons "want" to be as close as possible to the valance band. The bottom band (conductance band) works as an x axis flip of the valance band. The holes "want" to be as close as possible to the conductance band. So as the barrier decreases the buildup of majority carriers will cross the junction. That said, your last statement above is correct. The reason they do wander is because of the extra electrons on the N-side and Holes on the P-side due to the power source feeding the extra carriers to the respective sides and not vice versa (the Va is not feeding holes to the N-side and electrons to the P-side). Thus the imbalance and diffusion across the junction.

First of all Va is the voltage drop across the diode, not Phi. The voltage drop is defined as the difference in the two fermi energies (N-side minus P-side) times q [I may have gotten that backwards (P-side minus N-Side) I always forget]. Phi is the difference between the two bands; at equilibrium (Va = 0) Phi = Phi(built in) which is an intrinsic property of the doping levels, while at Va = V(builtin), Phi = 0 and you have a flatband situation but the fermi levels are different.

Again, you can force Va to greater values than .7 but it requires exponentially more current to reach those higher levels, and the diode characteristic equation is no longer valid in those higher voltage cases due to the reasons I stated in my last post.

All that said, for most practical cases Phi is roughly zero plus or minus a few tenths of a volt in forward bias. And yes, when the voltage is roughly .7 the diode is conducting (allowing electrons to go from N-side to P-side and holes from P-side to N-side) because of the lack of a barrier across the junction.

If you look at a graph of the ideal diode characteristic (V-I) plot you can really see how once you get a positive voltage of around .7v the current sharply increases. Of course V(bi) is different for different semiconductors and Si happens to be .7v whereas Vbi of Ge is around .2v. In this article from wiki you can see a a V-I plot of an ideal diode. Notice that the current shoots up quickly as you reach Vbi. Also, at extreme negatives the current shoots down quickly due to some other phenomenon (mainly avalanching).

http://en.wikipedia.org/wiki/Diode

If you have any more questions feel free to ask, I am enjoying actually remembering all of this stuff.

D Dean

10. Aug 9, 2006

jinyong

I believe it's the top band that's the conduction band and bottom band is valance band.

I see. If Va is the voltage drop across the diode(like when you solve a circuit problem you assume 0.7V drop) then why do they define the Phi variable as the drop across the semiconductor? I think that's what confuses me the most.

Right, different materials have different turn-on voltages. So these so called estimated turn-on voltages are actually ~V(bi) and not the value of Phi=Phi(built-in)-Va. Is this Phi just some variable to describe the energy band diagrams??

Thanks again.

11. Aug 9, 2006

D_Dean

You are correct, I hope that I haven't made too many errors involving the semantics. I appologize, though the main points are still valid.

Now I see the issue. Phi is indeed the drop across the semiconductor junction, but since Vbi is going to be present regardless of the Va, Vbi doesn't play a role in the potential seen in other devices. To put it in another way, the inherent potential drop Vbi does not act like a battery when you hook it up to resistors. It is an intrinsic property of the device and therefore is not included in the drop seen by other devices. Therefore the only drop seen is the Va (or -Va). This explains why when the device is on we see .7 volts rather than 0 volts. It is Va not Phi which is seen by the other devices and which effecting its surroundings. I'll think about this some more and try to get a better explanation later.

Yes, the device turns on when the applied voltage is roughly equal to the built in voltage. This causes a flat band and carriers can run across the junction easily. Since Phi has the built in drop in it, it is basically a variable to describe the bands. Vbi is a barrier and doesn't actually measure a voltage across it and therefore Phi is the difference between this this barrier (which is intrinsic to the semiconductor) and the applied voltage (which is seen from the outside).

I feel like I am a little incomplete in my description and I would like to do some research on this in the not too distant future. Unfortunately, I am rather busy tonight. I will get back to you eventually, in the meantime if anyone has anything to add I would appreciate it. Also, I think I was a little sloppy and incomplete with my wording in the previous posts, I appoligize again for that.

D Dean

12. Aug 11, 2006

jinyong

What you have said makes a lot of sense. However do you know why Va is the only parameter that appears to the "outside" world? Why doesn't the Phi(built-in) kind of cancel the Va since Phi is the drop across the deletion region under low injection conditions.

Let me know what you think,

Thanks.

13. Aug 12, 2006

D_Dean

Actually, I'm having a little problem conceptualizing it right now to be quite honest. I'll have to think back. Basically, it has to do with the fact that at steady state (when you have no Va and the diode is just sitting there with no external components connect) the voltage measured across the diode is zero (rather than Vbi or -Vbi) even though there is a potential of Vbi across it. I think maybe some of the problem comes with definitions. The potential used for Vbi is not the same as the potential used for Va (if that makes any sense). I'll get back to you on that in the not to distant future.

D Dean