# PN Junction Current: Understanding Dependence on Parameters

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• daudaudaudau
In summary, the current increases when you decrease ##N_a## because this decreases the built-in potential.
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?

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There is still ##n_i^2##.

mfb said:
There is still ##n_i^2##.
##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?

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

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.

phyzguy said:
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.

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...

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## 1. What is a PN junction current?

A PN junction current refers to the flow of electric current in a semiconductor device created by the junction of two differently doped regions. It is the result of the movement of majority charge carriers (electrons or holes) across the PN junction due to the presence of an electric field.

## 2. How does the PN junction current depend on the parameters?

The PN junction current is affected by several parameters such as the doping concentration, the temperature, and the applied voltage. The current increases with increasing doping concentration and temperature, and also with increasing forward bias voltage. On the other hand, the current decreases with increasing reverse bias voltage.

## 3. What is the relationship between PN junction current and doping concentration?

The PN junction current is directly proportional to the doping concentration of the semiconductor material. This is because a higher doping concentration results in a higher number of majority charge carriers, making it easier for current to flow through the junction.

## 4. How does temperature affect PN junction current?

Temperature affects the PN junction current in two ways. Firstly, it affects the number of majority charge carriers in the semiconductor material, with higher temperatures resulting in more charge carriers and therefore a higher current. Secondly, temperature affects the mobility of charge carriers, with higher temperatures resulting in increased mobility and therefore a higher current.

## 5. What is the difference between forward and reverse bias in a PN junction?

In a forward bias, the P-side of the junction is connected to the positive terminal of a voltage source, while the N-side is connected to the negative terminal. This results in a current flow across the junction. In a reverse bias, the P-side is connected to the negative terminal and the N-side to the positive terminal, creating a barrier to current flow. However, a small reverse current can still flow in the opposite direction due to minority charge carriers.

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