To quantitatively describe the situation,

the major change of

the energy band structure under reverse bias or

forward bias is the change in the band bending.

So, at equilibrium, your band bending is characterized by the built-in potential V sub i,

on the reverse or forward Bias,

your band bending is reduced or increased by the amount of your applied bias.

So, you simply replace V sub i with V sub i minus V_a,

then use the same equation that we derived for the equilibrium situation.

So, for example, if you want to calculate the depletion region width,

all you have to do is change V sub i with V sub i minus V_a.

Forward bias means V sub a is positive,

reverse bias means V sub a is negative.

Likewise, inlet maximum electric field which is the electric field at

the junction is given in terms of the depletion region width,

and so you just again substitute V sub i minus V_a for V sub i,

you get this equation,

and this is the maximum electric field on the bias.

Now, under reverse bias as I mentioned a few minutes ago,

you don't expect much current,

you don't get a lot of current.

So, the response of the diode pn junction is primarily capacitive.

It increases and decreases the stored charge across the junction,

this is the main response of the diode under reverse bias.

By definition, your capacitance is the change in charge,

dQ, in response to the change in applied voltage.

Now, the charge stored at the junction is the space charge within the depletion region,

the ionized donor and ionized acceptors within the depletion region.

So, the change in storage charge is due to

the increase and decrease of

the depletion region in response to your bias voltage change.

So, say that this was the depletion region width here,

X sub n at certain voltage V sub R, reverse bias voltage,

if you increase it by dV_R,

then the depletion region width on the n side will increase by this much,

dx sub n. Likewise,

the depletion region width on the p side will slightly

increase as well and we call that dx_p.

So, if you know how much change in depletion region width

on either side in response to the change in the reverse bias voltage,

then you can calculate your capacitance.

So, we already have an expression for

the depletion region width x sub n and x sub p. And once again,

you substitute V sub i minus V_a for V sub i in the equilibrium equation,

then you get the equations for x sub n and

x sub on the bias and you take the derivative with respect to the applied voltage.

Then, from the definition of your C,

capacitance in the previous slide,

you derive this equation here and it

simplifies into this very simple expression of epsilon s ,

the permitivity of the semiconductor divided by x sub d,

the total depletion region width.

So, this capacitance looks very similar to

the parallel plate capacitor separated by distance x sub d and

filled with a dielectric material with a dielectric constant epsilon sub s. Now,

the capacitance, C, is proportional to the square root of one over the applied voltage.

So, if you plot the applied voltage with respect to one over C squared,

then you get a straight line curve as shown here in the red curve,

and the slope of this curve is related to the doping density,

and the x intercept is the built-in potential.

So, this is a very standard technique

that are used the capacitance voltage characteristics,

is a standard technique used to characterize the semiconductor material in diode,

and is commonly used to determine the built-in voltage and the doping density.

Now, in this case,

the doping density is kind of convoluted,

you get one over N sub A plus one over N sub D. So,

you don't get the individual doping densities on the n-side and p-side,

but you get something that contains both.

If you make a one sided junction, that is,

if the doping density on one side is much greater than the other side, for example,

if your acceptor density is much greater than the donor density,

then the depletion region width on the p-side,

the heavily doped side is negligible,

the entire depletion builds on the lightly doped side, n-side,

and your capacitance depends solely on the doping density of the lightly doped side.

So, in this case, the slope,

if you plot one over C squared versus the applied voltage,

then the slope is inversely proportional to the doping density of the lightly doped side,

and once again, the x-intercept is the built-in potential.