# Intrinsic current of the diode, what is its current?

I'm having some difficulty finding the diode's intrinsic current, I applied some formulas, but I don't know if it's correct. Could anyone tell me if this is correct or why it is wrong?

If this is right, I would like to understand why in the electric field we have to measure it in meters, but when I applied it in centimeters it went wrong, since the structure is in centimeters. Do you have to apply in meters anyway? E = v/d

answer below:

• Are you sure it's a diode? Commented Mar 25 at 22:58
• @JohnD yes, What there was? Commented Mar 25 at 23:10
• It seems to be doped with only boron, and no other dopant. So, ti doesn't seem to be a PN diode. Commented Mar 26 at 1:40
• @MathKeepsMeBusy I agree with you, so much so that I insert the calculations according to the data I have but also insert the material type n with the same value as type P. I still don't know if the calculations are still correct, I put what I had in the statement Commented Mar 26 at 12:19

## 2 Answers

It's not a diode. Just doped semiconductor (silicon.)

I'll be using the problem value for $$\q\$$. Yours is more precise. But I want to keep it simpler, for now.

### getting an answer

The value of $$\n_i=1\times 10^{10}\:\frac{\text{e}^-}{\text{cc}}\$$ is typical for pure intrinsic silicon at $$\T=300\:\text{K}\$$. (GaAs has $$\E_g\approx 1.4\:\text{eV}\$$ and so would be much, much fewer conduction band electrons, for example.) So I conclude this is a silicon cube.

Without getting into the whys and wherefores about the mass-action law, you know that $$\n\:p=n_i^2\$$, given that pretty much all donors/acceptors are all ionized at $$\T=300\:\text{K}\$$, given that boron is an acceptor, and given that $$\\left(p=1\times 10^{15}\right)\gg \left(n_i=1\times 10^{10}\right)\$$, you can find that $$\n=\frac{n_i^2}{p}=1\times 10^5\$$.

Since $$\\left(p=1\times 10^{15}\right)\gg \left(n=1\times 10^{5}\right)\$$, ignore $$\n\$$ and just focus on $$\p\$$. So just use $$\\mu_p\$$ and ignore $$\\mu_n\$$: $$\\sigma=q\,p\,\mu_p= 0.08 \:\frac{\mho}{\text{cm}}\$$. That, times the applied volts, times the cross-section area, divided by the length gives you the current: $$\800\:\mu\text{A}\$$.

(The part associated with $$\q\,n\,\mu_n\$$ is so small as to be ignorable -- about $$\160\:\text{fA}\$$.)

### looking over your work

When you tried to compute $$\\sigma\$$ you were, instead, computing the current density, $$\J=\sigma\,\mathscr{E}\$$.

You are supposed to use $$\p\$$ and not $$\n_i\$$. Another mistake.

You also screwed up on the units. It is true that $$\\mu_p=500\:\frac{\text{cm}^2}{\text{V}\,\cdot\,\text{s}}\$$. But note the use of $$\\text{cm}\$$. Meanwhile, you applied $$\\mathscr{E}=1\times 10^6\:\frac{\text{V}}{\text{m}}\$$. But that's using $$\\text{m}\$$ and not $$\\text{cm}\$$. So you mixed up your units, even for computing $$\J\$$.

So even if you knew you were computing $$\J\$$ (which you don't appear to have known), you still got it wrong.

$$\J=800\:\frac{\text{A}}{\text{cm}^2}\$$.

Using $$\J\$$ instead, just to take a different tack, then $$\I=A\,J=\left(10\:\mu\text{m}\right)^2\cdot 800\:\frac{\text{A}}{\text{cm}^2}=800\:\mu\text{A}\$$.

### summary

I think the answer should be $$\I=800\:\mu\text{A}\$$. (Give or take a little.)

• Thanks for your help, but I still have some questions. There are still some points that left me with doubts. First regarding the calculation, I converted micrometers to centimeters because of the formula σ= 1/p = ((IL)/(VA)), to find it I needed to convert. Before the formula σ=q*(pμp+µnn)*E. I ignored n, because I didn't have the information in the statement, but you demonstrated to me, I should have calculated and discovered the value of n=1*10^(5). Commented Mar 27 at 0:14
• With the data in hand, I recalculated and the result matched, but I did not use the value of the electric field (E) in the calculation. σ=1.602*10^(−19)*(1*10^(5)*1000+1*10^(15)*500) ▸ 80.1E−3 80.1E−3= ((I*1.E−3)/(10*1.E−6)) i=((0.0801*10*1.E−6)/(0.001)) ▸ i=801.E−6 A I know that I must calculate the electric field, ([σ=q*(pμp+µnn)*E ]), but if I insert the electric field data it won't match the value 800 μA, I don't know what is happening, I see that it was just a coincidence calculation. Still, I need to know why. Commented Mar 27 at 0:14
• Second: I hadn't realized that the electric field density could be used in the formula. I=J*A But I realized that you calculated using the micrometer multiplied by centimeter, to work correctly I must convert the micrometer value to centimeters, thus achieving 800 microamperes, which is why I previously converted everything into centimeters. Commented Mar 27 at 0:14
• @LUFER I just tossed $n$ away. It's way too small to care about. You can include it if you want. But I've nothing more to add about that bit. $\mathscr{E}$ is a million volts per meter, just as you computed. So I don't disagree with that. (It's also 10,000 volts per cm.) And it computes correctly when doing $I=A\cdot\sigma\cdot\mathscr{E}$. I've no problem reaching the right result so I'm not sure what the problem is. 0.08 siemens/cm * 1e6 volt/m * (10 micron)^2 = 800 uA. It just works. Commented Mar 27 at 0:25
• Sorry, when you demonstrated the calculation you simplified it, I like to make it clear where you use the values for J (sorry). Regarding the calculation using the electric field according to the formula σ=q*(µpp+nμn)*E. σ= 1/p = ((IL)/(VA)) See the 3 resolutions: 1° Not complicating the electric field (I know that n is negligible, but I still want to use it). σ=1.602*10^(−19)*(1*10^(5)*1000+1*10^(15)*500) ▸ 80.1E−3 (without electric field) i=((0.0801*10*1.E−6)/(0.001)) ▸ i=801.E−6 A (Here you get your answer) Commented Mar 27 at 12:36

I think OP is after the current that flows through an intrinsic semiconductor with an applied electric field.

We measure electric field EITHER in V/m or V/cm. (Or V/nm if you are perverse). If you start with a true equation and carry the units along with the numbers you must get the correct result WITH the correct units. This will be easiest if you use consistent units (all m for distance, etc.).