# Homework Help: Breakdown voltage

1. Apr 12, 2008

### Fermi_98

Hi,
Being a computer science and eng first year student, just trying to understand one artical regarding Breakdown voltage of the pn junction.Could anybody explain me, which diode out of the following have higher breakdown voltage if they have same design parameters such as doping and thickness and why ?
p+-n diode
n+-p diode

Last edited: Apr 12, 2008
2. Apr 13, 2008

### maverick280857

I am not sure I understand the symbols in your question, but I am assuming you are referring to forward and reverse biased situations. Is this correct?

The breakdown voltage is indeed a function of the doping. But, I am not sure what you're referring to as the breakdown voltage in your question. Breakdown voltage is typically a reverse biased voltage.

The following link might be of interest to you:

http://people.seas.harvard.edu/~jon...de_characteristics/diode_characteristics.html

3. Apr 13, 2008

### Fermi_98

1. Diodes are fabricated by diffusing either n type dopant into p-type silicon or p-type dopant into n-type silicon. So, in the former case, it is called as a n-p diode and p-n diode in the latter case.

2. Breakdown voltage of a diode is a case when it is reverse biased. which means when the n- side of the junction is connected to the positive terminal of the battery grounding p-side, in the case of a p-n diode. When p side is connected to the negative terminal of the battery and n-side to ground, it is called as an n-p diode.

3. The sign following the character shows that that type of the dopant is in abundant in the respective type of the silicon. e.g. p+ n means that p type of the dopant are in abundant in the n-type silicon and vice versa.

thanks

4. Apr 13, 2008

### maverick280857

Interesting, despite being an EE student, I have never encountered that convention :tongue:

Which college are you from? And could you figure out which of the two diodes--the one being operated normally in FW bias condition and the one being operated normally in RW bias condition, has the higher breakdown voltage?

5. Apr 15, 2008

### Fermi_98

Hi,
I suppose, you have not understood the question properly.I know that once the diode is on, it supports only a voltage equal to built-in-potential (0.7V) as large current flows in this situation.On the other hand, it is only in the reverse biased condition, the diode allows very little current to flow (leakage current) due to the large opposing electric field at the junction.

2. As far as your familiarity with the symbol is concerned, i suppose, definitely, you are unaware of.

3.Thirdly,i am talking about the reverse biased condition, no forward bias case at all in both kind of diodes. p+n and n+p.

4. For your help, there is a term called multiplication coefficient, you might want to have a look, as it might surprise you that multiplication coefficient is different for holes and electrons.So, this might create a difference in the breakdown voltage of the two diodes (p+n or n+p)

Last edited: Apr 15, 2008
6. Apr 20, 2008

### maverick280857

For your information, holes and electrons have different mobilities. I suppose you're talking about the avalanche multiplication factor. I don't know if there's something else by that name. And judging from your statements, there seems to be some confusion about multiplication, breakdown and normal reverse biased operation. I suggest you have a look at the web site I quoted in an earlier post. These are different phenomena...I hope you're clear about that.

To clear things up a bit, you should read about reverse biased operation in more detail without first worrying about all these factors. A qualitative understanding of the effect on the pn junction as the depletion width increases will help you understand how different KINDS of diodes will respond to reverse bias. Then, you can read about breakdown, and avalanche breakdown, which is a multiplicative effect.

If you're into the physics aspects, then you can refer to "Physics of Semiconductor Devices" by Sze. There are of course other books too...I'm not sure what your level is. Are you a freshman?

EDIT: Just noticed your first post. Well, since you're a freshman, I would suggest looking at a book by Linda Edwards Shea. It will help you transition to a more advanced text like Sze.

Last edited: Apr 20, 2008
7. Apr 20, 2008

### maverick280857

8. Apr 21, 2008

### dashkin111

In general, two processes (avalanching and the Zener Process) can cause the breakdown current to occur. The breakdown voltage is also expected to progressively increase changing materials from Ge, to Si, to GaAs.

The doping dependence doped below 10^17/cm3 is roughly:

$$V_{BR}=\frac{1}{N_{B}^{0.75}}$$

Where Nb is the doping on the lightly doped side of the junction.

Another dependence is also worth mentioning. The Vbr due to avalanching is found to increase as the temperature increases. If you think hard, you'd also notice lattice scattering increases as the temperature increases. Increasing lattice scattering means a smaller mean free path, a larger critical electric field for avalanching, and hence a higher breakdown voltage.

Carrier mobility, in semiconductor work, is a measure of the ease of carrier motion in a crystal. A lot of lattice scattering would make a low carrier mobility, huh? So what types of carriers would have higher mobilities?

Hopefully this might get you thinking in the right area.

9. Apr 22, 2008

### Fermi_98

Firstly thanks.
Secondly, i suppose, whatever you have said, is right. However, this is not the answer to my question. I do understand that the electron mobility is approximately three times as that of hole.
Reiterating the question once again. With regards to my diode/rectifier case, i have been talking about the avalanching case in both type of the diode/rectifier (p+nn+ and n+pp+).
These sorts of the design of the diode is usually meant for blocking high voltage . The high voltage could be as high as 10000V and as low as 10V. So, usually middle region (n in the case of p+nn+ and p in the case of n+pp+) supports the voltage in the reverse direction. I am not talking about the zener or any other breakdown mechanism. I am talking particularly about avalanching case, if that helps you.

10. May 1, 2008

### maverick280857

You know if you can point us to a textbook or paper or source where you got this problem from, we might be able to look in the right direction to answer your question.

In your first post you said you were reading about this from an article which led you to this question. Perhaps you can point us to that article..

11. Jul 10, 2008

### backtothefutu

Do we already know the answer to Fermi's question , and trying to reason it out or are we in the process of finding the answer ,

What is the role of n+ or p+ region in pnn+ diode or npp+ diode afterall and what significance will it have on the breakdown voltage?

These type of diodes are generally used in power electronics , where recovery time is not the major issue like high frequency diodes , the important thing is here is the breakdown voltage ,and since high doping leads to high electric feilds in depletion region and hence lover avalanche voltage , we use n or n- in between, while the n+ or p+ region comes into picture only when width of depletion region in lightly doped n or p region reaches the n+ or p+ boundary , after which it doesn't increase by much in either of the cases , so any more voltage applied across the diode will just increase the electric feild across the depletion region which is mainly in the lightly doped middle region , so in my opinion we can neglect the n+ or p+ region from the analysis and what remains is a pn and np diode , which if simmilarily doped will have same avlanche breakkdown voltage.

thanks

12. Nov 22, 2008

### StgIIIOvrDvn

Breakdown voltage is not persay a basis of doping flavor but of doping strengths and the effects on junction depletion region thickness. A reverse bias field causes an inflation of the depletion region until a point is reached where the depletion region encompasses the entire depth of one or both elements in the pn junction. Lower dopings in both elements shift toward the higher breakdown voltages and higher intrinsic resistances in the elements of the junction.
As for avelanche and zenering, avelanche occours when zener currents are sufficiently high as to escelate generation of secondary charge carriers to a point where currents from the secondary carriers sustain the creaton and generation of additional carriers. A junction in avelanche will exhibit a terminal voltage signifigantly below its initial zener voltage.