The p and ntype silicon or germanium can be obtained by adding appropriate acceptor or donor impurity into Si or Gemell while growing a crystal.
These crystals arc cut into thin slices called ‘Wafers’. Semiconductor devices are usually made on these wafers.
These aremany methods of making a pn junction as given below.
(a) ‘Alloying’ technique
(b) Diffusion technique
(c) Vapour deposition (epitaxial growth) and
(d) Rate growth junction.
The donor ion is indicated schematically by a plus sign ( + ) because, after this impurity atom “donates” ah electron it becomes a positive ion. The acceptor ion is indicated by a minus sign
() because, after this atom “accepts” an electron, it becomes a negative ion.
Initially, there are nominally only ptype carriers to the left of the junction and only nlypc carriers to the right.
Because, there is a density gradient across the junction holes will diffuse to the right across the junction and electrons to the left.
Asa result of the displacement of these carriers, an electric field will appear across the junction. Equilibrium will be established when the field becomes large enough to restrain the process of diffusion.
The un neutralized ions in the neighborhood of the junction are referred to as “uncovered charges”. Since the region of the junction is depleted of mobile charges, it is called the “depletion region, the spacecharge region, or the transition region”.
The thickness of this region is of the order of 104 cm = 106 m = 1 micron
The general shape of the charge distribution may be as shown in figure 1.1b.
The electric field intensity in the neighborhood of the junction is indicated in fig 1.1c.
The electrostaticpotential variation in the depletion region is shown in fig l.ld.
This variation constitutes a potential energy barrier against the further diffusion of holes across the barrier.
The form of the potential energy barrier against the flow of electrons from the nside across the junction as shown in fig l.le.
The potential bamer (or field across the junction) and the depletion layer width (or junction width) depends upon the doping concentration on the two sides.
At equilibrium, there is no net current flowing across the pn junction.
The current due to the drift of carriers in the electric field must exactly cancel the diffusion current. J_{p}(drift) + J_{p}(diffusion)  0
J_{n}(drift) + J_{n}(diffusion)  0
Since the net hole current density is zero, the negative of the hole diffusion current must equal the hole drift current.
The potential barrier V0 called the contact, or diffusion, potential is of the order of magnitude of a few tenths of a volt.
V^{2}V = d^{2}V/dx^{2} =  ρ/ε → Poisson equation is satisfied inside the junction.
V^{2}V = d^{2}V/dx^{2} = 0 → Laplace equation is satisfied outside the junction.
A schematic diagram of a p  n junction, including the charge density, electric field intensity, and potential  energy barriers at the junctions, (not drawn to scale.)
Diode permits the easy flow of current in one direction but restrains the flow in die opposite direction.
Forward Bias:
Suppose we apply a voltage V such that nside is negative and pside is positive.
The applied voltage V (or bias V) is opposite to the junction barrier potential V
The consequences of this are:
(i) The effective barrier potential becomes (V_{o} — V) and hence the energy barrier across the junction decreases.
(ii) more majority carriers will be allowed to flow across the junction.
(iii) the junction width decreases.
The current flow is principally due to majority charge carriers and is large (mA).
Reverse Bias:
The applied voltage V on the nside is positive and is negative on the pside.
The applied bias V and the barrier potential V# are in the same direction making the effective junction potential as V + V_{o}. As a result, the junction width will increase.
The higher junction potential would restrict the flow of majority carriers to a much greater extent.
However, such a field will favour the flow of minority carriers (as they have opposite charges).
So, the reverse bias current will be due to the minority carriers only.
Since, the number of minority carriers is very small as compared to the majority carriers, the reverse bias current is small (≈ μA).
This energy E_{0} represents the potential energy of the electrons at junction.
Band diagram for a pn junction under opencircuit conditions. This sketch corresponds to fig (le) and represents potential energy for electrons. The width of the forbidden gap is E_{g} in electron volts.From figure,
E_{F } E_{vp}  1/2 E_{G}  E_{1} and E_{Cn}  E_{F}  1/2 E_{G}  E_{2} E_{G} = forbidden energy gap
By adding the above equations
E_{0} = E_{1} + E_{2} = E_{G}  (E_{Cn}  E_{F})  (E_{F}  E_{Vp})
E_{G} = kT In (N_{C}N_{v}) / ni^{2}] ....(1)
E_{cn } E_{F} = kT ln (N_{C}/N_{D}) ....(2) ∴ KT = 26 meV
E_{F } E_{v}_{p} = kT ln (N_{v}/N_{A}) ....(3)
The E’s are expressed in electron volts andk has the dimensions of electron volts per degree Kelvin. The Contact difference in potential V_{0} is expressed in volts and is numerically equal to E_{0} and it is depends only upon the equilibrium concentrations, and not at all upon the charge density in the transition region(space charge region).
Other expressions for E_{0} are
....(4)
Rewriting the above equation (4) in another form
p_{p}0 = p_{n}0 e^{v}0/^{v}_{T}
n_{n}0 = n_{p0 e}^{v}_{0/}^{v}_{T ....(5)}
_{}
_{}
The hole and electroncurrent components vs, distance in a pn junction diode. The spacecharge region at the junction is assumed to be negligibly email
I = I_{pn}(0) + I_{np}(0)
I_{pp}(x) = I  I_{np}(x)
P_{n}(x) = P_{no} + P_{n}(0)e^{x/L}p ....(6)
Defining the several components of hole concentration in the n side of a forward  biased diode.
Where the parameter L_{p} is called the diffusion length for holes in the n material.
And the injected or excess concentration at x = 0 is
P_{n}(0) = P_{n}(0)  P_{no}
The diffusion hole current in the nside is given by
junction
....(7)
Taking the derivative of equation (6) and substituting in equation (7) we obtain
.... (8)
This equation verifies that the hole current decreases exponentially with distance.
Where q = charge of an electron
For a pn junction, the current I is related to the voltage V by the equation
I = I_{0}(e^{v/ηv}T 1) ....(13)
η ≅ 1 for Ge
η ≅ 2 for Si
V_{T} = T/11,600 = 26 m V at T = 300°K.
At room temperature,
(a) When the voltage V is positive and several times V_{T} , the unity in the parenthesis of equation (13) may be neglected, i.e., e^{v/ηv}T >>1,
∴ l = l_{0} e^{v/ηv} T
Accordingly except for a small range in the neighborhood of the origin, the current increases exponentially with voltage.
(b) When the diode is reversebiased and V is several times VT, I ≈  I_{0}. The “reverse
current” is therefore constant, independent of the applied reverse bias. Consequently, I_{0} is referred to as the “reverse saturation current”
At a reverse biasing voltage V_{z}, the diode characteristic exhibits an abrupt and marked departure from (13). At this critical voltage a large reverse current flows, and the diode is said to be in the “break down region”.
The voltampere characteristic for a germanium diode
The Cutin Voltage, Offset, breakpoint, or threshold voltage (V_{y}):
Below V_{y} the current is very small (say, less than 1 per cent of max. rated voltage).
Beyond V_{y} the current rises very rapidly. V_{y} = 0.2 V for Ge, 0.6V for Si.
The reverse saturation current 10 in a Ge is normally larger by a factor of about 1000 than the 10 in a silicon diode of comparable ratings. Since n = 2 for small currents in Si, the 1 increases as e^{V}^{/2V}T only at higher voltages.
The resistance R, cutin voltage Vy, power dissipation and noise margin of Germanium is less— than the silicon. But Germanium is not used in switching characteristics.
The temperature dependence of pn diode characteristics:
The dependence of 10 on temperature T is given approximately by
I_{0} = kT^{m} e^{v} Go^{/ηvT}
Where k is constant and qV_{GO} (q is the magnitude of the electron charge) is the forbidden gap energy in Joules:
For Ge : n = 1 m = 2 V_{GO} = 0.785V
For Si: n = 2 m = 1.5 V_{GO = }1.21V
Taking the derivative of the logarithm of above equation, we find
At room temperature, d(l_{n}l_{0})/dT = 0.08 /°C for Silicon and 0.11 / °C for Germanium.
The 10 increases approximately 7 percent / °C for both Si and Ge. Since (1.07)10 ≈ 2.0, we conclude that the “reverse saturation current(Io)” approximately doubles for every 10°C rise in temperature.
Consider a diode operating at room temperature (300°K) and just beyond the V_{y}.
Then we find,
Since these data are based on “average characteristics”, it might be well for conservative design to assume a value of
dV / dT = 2.5 mV/°C ....(14)
for either Ge or Si at room temperature.
For Germanium, an increase in temperature from room temperature (25°C) to 90°C increases the 10 to hundreds of microamperes, although in Silicon at 100°C the 10 has increased only to some tenths of a micro ampere.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
10 docs5 tests
