# I PN junction current

1. Jun 12, 2016

### daudaudaudau

Hi. I'm trying to understand how the current in a PN junction depends on various parameters. I have found this formula in Ashcroft and Mermin. It looks like the current will increase if we decrease the density of donors and/or acceptors, Na and Nd. Is this correct?

Last edited: Jun 12, 2016
2. Jun 12, 2016

### Staff: Mentor

There is still $n_i^2$.

3. Jun 13, 2016

### daudaudaudau

$n_i^2$ is the intrinsic carrier concentration, which doesn't depend on $N_a$ or $N_d$, so how could that affect the result?

4. Jun 13, 2016

### Staff: Mentor

Ah wait, you have the doping in the denominators, not the actual number density of electrons/holes. Hmm, then it looks strange.

5. Jun 13, 2016

### phyzguy

The saturation current is caused by the thermal generation of carriers in the depletion region, which are then swept out of the depletion region by the built-in field. As the doping concentration(Na and Nd) goes down, the depletion region gets larger, so there is a larger volume of depleted semiconductor where generation takes place. I think the equation is correct.

6. Jun 13, 2016

### daudaudaudau

Actually this particular derivation in Ashcroft and Mermin doesn't include any generation in the depletion region, but anyway you're right that the formula does make sense when thinking about a reverse biased diode. If we let $p_{no}$ denote the equilibrium density of holes on the N side, and since $n_i^2/N_d=p_{no}$, we can write for the hole current under reverse bias

$$J_p=en_i^2\frac{D_p}{L_pN_d}=e\frac{D_p}{L_d}p_{no}$$

So the reverse hole current is proportional to the density of holes on the N side, which makes sense.

When the diode is forward biased I would like the hole current to depend on the density of holes on the P side. This is of course possible. If $V_{bi}$ is the built-in potential, $V_T$ the thermal voltage, and $p_{po}$ the density of holes on the P side, we have

$$p_{no}=p_{po}e^{-V_{bi}/V_T}$$

Which leads to a hole current that is proportional to $p_{po}=N_a$ and decreases as the built-in potential increases.

$$J_p=e\frac{D_p}{L_d}N_ae^{-V_{bi}/V_T}(e^{V/V_T}-1)$$

So the answer to my initial question must be that the current increases when you decrease $N_a$ because this decreases the built-in potential

$$V_{bi}=V_T\log{\frac{N_aN_d}{n_i^2}}$$

WELL, still puzzling that a low carrier density gives the highest current....

Last edited: Jun 13, 2016