# Why is diffusion current due to p side in a diode inversely proportional to doping concentration of p side in the case of forward bias?

In a diode at equilibrium, there are Pn holes on the n side of the junction. These are maintained by the dynamic equilibrium of diffusion due to holes = drift due to holes. When a forward bias is applied, the amount of holes diffusing increases. That is an extra amount of holes $$\ P_n . e^{V_d/V_t}\$$ found on the n side of the junction. Here $$\ P_n = n_i^2 / N_d \$$ and depends on N side doping.

Why is it that the extra holes that diffuse from P-N is independent of the doping of on the P side(majority hole concentration on p side). If P side hole concentration/cm3 is higher, shouldn't the concentration gradient be higher and hence diffusion current increase. However it is dependent on the N side hole concentration (minority concentration at n side) instead. Similarly the number of electrons diffusing from N-P region depends on the concentration of electrons on P side rather than N side.

In a nutshell, I'm trying to understand the physical intuition of why the equation $$\ P_e = P_n . (e^{V_d/V_t} - 1) \$$ depends on Pn and not hole concentration on P side of junction.
Here,
$$\ P_e\$$ = excess holes flowing due to forward bias on n region
$$\P_n\$$= holes on n side at equilibrium (no voltage applied)
$$\N_d\$$ = N side doping concentration
$$\n_i\$$ = intrinsic concentration
$$\V_d\$$ = Forward Bias Voltage
$$\V_t\$$ = Thermal Voltage

• It is a common assumption that both layers are uniformly doped so Nd would function for the doping concentration of both regions(N and P), can you confirm if the book you got the formula is making this assumption or not?
– Juan
Commented Apr 13, 2023 at 7:20
• @Juan Are you asking if Nd=Na. This assumption is not made. However it can be assumed that Nd != Na but both are uniformly doped (concentration of majority carrier is same throughout the n,p region) Commented Apr 13, 2023 at 14:09