Basic electronics 2a 2006

8/9/2019 Basic electronics 2a 2006

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Chapter 2. Diode Circuits

Now we form our first useful semiconductor device, a pn junction in a bar of Si. At first we assume that we

have a step junction i.e. the concentration changes abruptly from p to n at the pn junction interface as shown

below

In real devices we never have a perfectly abrupt junction, but this complicates our analysis of the pn junction

needlessly, since we only really need to know what is happening right around the metallurgical junction

(where the concentrations are equal).

We start our analysis of the pn junction using the Poisson equation from physics

Eq. 5.10

sK

E= which simplifies in 1-D to

Eq. 5.20

sKdx

Ed=

EE 329 Introduction to Electronics 38


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where Ks is the semiconductor dielectric constant and e0 is the permittivity of free space, here is NOT

the resistivity but the charge density (charge / cm3). Assuming total ionization we get

Eq. 5.3 )( AD NNnpq +=

Remember that charge density = 0 in a uniformly doped semiconductor at equilibrium. Also note that is

proportional to dE / dx in 1-D.

Qualitatively we will look at the pn junction problem and find the appropriate band diagram to represent the

situation. Begin with the dopants

Now we draw a band diagram for each side assuming that they are not in atomic contact.



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As we move the p and n sides closer together what will happen? First the EF lines up and causes all the other

bands to shift up or down in energy accordingly. We then need to connect up Ec, Ev and Ei and we also

make the assumption that the variation is monotonic in nature with zero slope at both ends. The exact nature

of the bands is not known but the figures below appear to be good representations.



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This then is the equilibrium band diagram of a pn junction diode. Now look at what is happening inside the

semiconductor. Remember from Ch. 3 that we can sketch the voltage potential in the semiconductor simply

by inverting one of the bands. This leads to

We arbitrarily set V = 0 on the p side of the junction, and we note that there is some type of potential across

the pn junction region.

Now we want to sketch the electric field in the device. Remember that the electric field is proportional to the

derivative of the voltage with respect to x in 1-D



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Eq. 5.4dx

dVE =

and it is proportional to the integral of the charge density as we saw before since electric fields are caused by

separation of charges and we note that the derivative of the electric field (with respect to x 1-D) is

proportional to the charge density

Eq. 5.20

sKdx

Ed=

We can see these relationships when we apply Gausss law to our situation



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The electric field begins where the bands are bent (i.e. where we have non-zero charge density), with its

MAGNITUDE steadily increasing to a maximum value at the metallurgical junction then decreasing back

down to zero at the end of the band bending on the other side of the pn junction.

Since we are at equilibrium, we know that the net charge distribution = 0, so the area under each region in

the above figure is the same. Not only are the areas the same, note that they are of opposite sign so that the

net positive charge = net negative charge in the device, as we would expect at equilibrium.



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How does this charge distribution result? We start by bringing the p and n regions into atomic contact as

shown below

Here we have drawn both the e- and their positively charged donor atoms on the n side and the h+ and their

negatively charged acceptor atoms on the p side. As we would expect we have charge neutrality holding on

both isolated sides. As soon as we make atomic contact between the p and n sides suddenly there is a large

concentration gradient of h+ and e-. Initially, diffusion drives the h+ into the n region and simultaneously e-into the p region.



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Note that the charged donor and acceptor atoms are NOT mobile as they are bonded into the Si lattice. The

majority mobile carriers on each side of the junction diffuse into each others region and we see that the arearight around the metallurgical junction is swept free of MOBILE carriers due to this diffusion. But

remember that the charged donor and acceptor ions in that region are still there, so we end up with charge

separation and therefore an electric field centered around the metallurgical junction. Since this region has so

few MOBILE carriers it is called the depletion region or more accurately the space charge region (scr - since

the area is only depleted of MOBILE carriers). What we end up are p and n regions as shown above and

which leads to our charge density distribution shown below



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Another way to look at the pn junction formation, is that as soon as the diffusion of MOBILE carriers begins,

we get charge separation which leads to an electric field.

Now we know that the electric field is going to try and restore the equilibrium situation where we had more

e- in the n region and more h+ in the p region. This is in direct opposition to what diffusion is trying to



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accomplish. So after a short amount of time, we reach a new equilibrium where net diffusion is balanced by

net drift (due to the electric field). See figure below.

If the e- and h+ were uncharged particles then diffusion would drive the new equilibrium to the point where

the concentration of e- and h+ would be equal across the entire pn junction. Instead we get diffusion out of a

small region around the metallurgical junction while a drift current works in opposition. Even though there



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is still a concentration gradient, no additional MOBILE carriers can diffuse across the junction due to the

electric field present. We are left in a dynamic equilibrium drift and diffusion are occurring on both sides

of the junction but all are balanced with each other so we have no NET charge flow, therefore no NET

current, ie

Jn|drift = Jp|drift AND Jn|diffusion = Jp|diffusion AND Jn = -Jp therefore net J = 0

We now look at Vbi in more detail.

Lets see how big Vbi is for a typical case; NA = ND = 1e15 at 300K -> kT / q = 0.0259 and Vbi 0.6V. Innon-degenerately doped diodes the Vbi < EG / q which is simply the bandgap converted to volts. Therefore

Vbi < 1.12V.

The next step is to determine quantitatively how many carriers we have. But in order to do this we have to

solve Poissons equation which in 1-D is



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Eq. 5.14 ( )ADosos

NNnpK

q

Kdx

Ed+==

When we try to solve this equation for(x) (or n(x) and p(x) ) we assume we know the distribution of

NA(x) and ND(x) but we find we also need to know n(x) and p(x) in the scr, but they are functions of theelectric field that we are trying to solve. The electric field depends on the distribution of charge carriers, but

the charge carrier distribution depends on the electric field. Therefore solving becomes a complex task of

iterations.

So once again we simplify things by make what is known as the DEPLETION APPROXIMATION. This

allows us to make good approximations without having prior knowledge of the carrier concentrations.

What the depletion approximation does is to make a simplification to the actual charge distribution in the scr

and two things are assumed;

1. the carrier concentrations in the scr are assumed to be negligible with respect to the net doping

concentrations outside the scr

2. the charge density outside the scr is assumed to be 0

These are summarized in the figure below



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Graphically this means that in a step junction as shown below

What we now have is a simple rectangular region of charge on either side of the metallurgical junction. Our

1-D Poisson equation simplifies to



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Eqs. 5.15a,b

( )

np

npAD

os

xxA Nxx

xxxNN

K

q

d x

Ed

0

We see that except for the values of xn and xp, the charge density is totally specified by invoking the

depletion approximation, and the charge density will have exactly the same functional form as ND-NA in the

scr.



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5.0

2

+

=+ DADAbios

pn NN

NN

q

VK

xxW

Now that we have solved for all the key parameters in a pn junction diode at equilibrium, lets add an

external (applied) voltage. We know from Kirchoffs voltage law that when there is a closed loop electrical

system, the voltage drops around the entire circuit must sum to zero. So if we apply an external voltage of

VA, this voltage must be dropped somewhere within our circuit.

First we assume negligible voltage drops in the external wires and in the diode contacts to the outside world.

Therefore all of the drop must occur somewhere within our pn junction diode. When we add the constraint

of low-level injection (remember this means that we are not producing new charge carriers in excess of the

majority carrier concentration), then we find that the no voltage drop occurs in the quasi-neutral n and p

regions away from the scr. This leads us to the conclusion that all of the voltage drop (VA ) must occur

across the scr.

When VA > 0, the voltage on the n side of the junction is lowered relative to the p side. And the opposite

occurs when we make VA


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Now let us look at what is happening with respect to the band diagram. We show the biasing cases in the

figure below. Begin with the bands at equilibrium as shown in the left column



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Lets consider what is going on inside the diode. Initially there are an excess number of h+ on the left side

and e- on the right. So an e- wants to diffuse to the p side but it sees an energy barrier at the scr due to the

electric field previously created due to charge separation. Also note that any e- that reaches the scr on the p

side will be quickly swept down the energy ramp to the n side (drift). So with VA = 0 we have reached an

equilibrium. Note that everything for the h+ is the same just different polarity and directions of movement.

Now when VA>0, we are making the end of the p region more positive and this potential attracts e-. This

lowers the potential on the n side and on our band diagram the energy barrier from the p to the n side for both

e and h is reduced as shown in the center column of the previous figure. Another way of looking at forward

bias is to think of holding both ends of the band diagram in your two hands and under forward bias, while

holding the p side steady (i.e. your reference voltage), raise the n side up with respect to e- energy.

Now what happens to our drift and diffusion components? The reduction in the energy barrier primarily

affects diffusion and since the barrier for diffusion is reduced we get increased diffusion. The drift

component does not change much since it depends on the electric field due to charge separation. The

drift component does not change much since an e- on the edge of the p side is still swept into the n side, but

we know that there are not many e- there in the first place because they are minority carriers there, therefore

the magnitude of the diffusion current changes dramatically under applied bias but the drift component

changes very little, and this is called forward biasing of the diode. We will see how this affects current

flow in the device in the next chapter, but we can guess that forward biasing will dramatically increase the

current flow through the device (and it does exponentially, see figure below).



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So now what happens when VA < 0? Now we are putting a negative potential on the end of the p side so we

are less likely to attract e-. This increases the potential on the n side and on our band diagram the energy

barrier from the p to the n side for both e- and h+ is increased as shown on the right column in the previous

columns.

Now what happens to our drift and diffusion components? The increase in the energy barrier again

primarily affects diffusion and since the barrier is increased we get less diffusion. The drift

component also does not change for the same reasons mentioned above, therefore the magnitude of thediffusion current changes dramatically but little change in the drift component and this is called

reverse biasing of the diode. We can guess that reverse biasing will dramatically decrease the current flow

through the device.



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Now lets look at what is happening quantitatively. The only change we have to make from our previous

derivation is that we now have VA, and since we assume that all of VA is dropped across the scr, all we have

to change in our previous equations is to substitute Vbi- VA everywhere Vbi occurred.

What we end up with are the equations on p. 216 leading up to

Eq. 5.38{ }

5.0

2

+=

DA

DAAbios

NN

NN

q

VVKW

We see that the amount and polarity of the applied bias will affect the size of the scr width.

Now look at how , the electric field and voltage have changed for the various bias conditions. First look

at Eq. 5.38 for W. We note that W (xn and xp) increases with reverse bias and decreases with forward bias.

Now for the case where we have a heavily doped p side and lightly doped n side, see below.



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Another thing to note in this figure is that the scr width is predominantly on the lighter doped side, i.e. to

insure charge neutrality we have to deplete farther in the lightly doped n side than on the more heavily doped

p side (to get equal number of charges).

Now what consequence does this have on the charge density ( )? The area of each of the charge regions

decreases with forward bias because we have uncovered fewer dopant atoms. Therefore we get Figure 5.11b

above.

Now what does this mean for the electric field? In forward bias we have less charge on each side, therefore

there is less total charge separation so we expect a reduced electric field as shown in Figure 5.11c above.

Finally what about the voltage? Reducing the electric field, reduces the voltage drop so we expect a reduced

voltage across the scr in forward bias as in Figure 5.11d above.

Also remember that this voltage plot is simply an inverted form of the energy band diagram and under

forward bias we saw that there was LESS band bending, therefore less voltage.

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Basic electronics 2a 2006

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