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Calculating the built in potential

To calculate the built in potential of device we must know the following things:

Then infinite recombination velocity on the contacts is assumed.

Figure 4: Band structure of device in equilibrium.
Image bands



The left hand side of the device is given a reference potential of 0 V. See figure 4. We can then write the energy of the LUMO and HOMO on the left hand side of the device as:

$\displaystyle E_{LUMO}=-\chi$ (1)

$\displaystyle E_{HOMO}=-\chi-E_{g}$ (2)

For the left hand side of the device, we can use Maxwell-Boltzmann statistics to calculate the equilibrium Fermi-level ($ F_i$ ).

$\displaystyle p_{l}=N_v exp \left(\frac{E_{HOMO}-F_p}{kT} \right)$ (3)

We can then calculate the minority carrier concentration on the left hand side using $ F_i$

$\displaystyle n_{l}=N_c exp \left (\frac{F_n-E_{LUMO}}{kT} \right)$ (4)

The Fermi-level must be flat across the entire device because it is in equilibrium. However we know there is a built in potential, we can therefore write the potential of the conduction and valance band on the right hand side of the device in terms of $ phi$ to take account of the built in potential.

$\displaystyle E_v=-\chi-q\phi$ (5)

$\displaystyle E_v=-\chi-E_g-q\phi$ (6)

we can now calculate the potential using

$\displaystyle n_{r}=N_c exp \left (\frac{F_n-E_{LUMO}}{kT} \right)$ (7)

equation 5.

The minority concentration on the right hand side can now also be calculated using.

$\displaystyle p_{r}=N_v exp \left (\frac{E_v-F_{HOMO}}{kT} \right)$ (8)

The result of this calculation is that we now know the built in potential and minority carrier concentrations on both sides of the device.


next up previous
Next: Transport Up: Electrical model Previous: Electrical model
rod 2015-01-07