Characteristics of P-N Junction Diodes

Hall Effect

If a metal or a semiconductor carrying a current I is placed in a transverse magnetic field of a flux density B, an electric field is developed along a direction perpendicular to both B and I. This phenomenon is known as Hall Effect. This is used for:-

Determining the type of semiconductor p or n-type

Finding out the carrier concentration

Measuring the conductivity of the material

Determining the carrier mobility

Detecting and measuring magnetic field 1000000 times smaller than that of the earth(with Hall-effect magnetic meters)

Under the influence of an applied magnetic field of a certain force electrons end to crowd towards one side of the semiconductor and consequently gives rise to a potential gradient. This prevents additional electrons from accumulating there. And it reaches a steady state when the two oppositely acting magnetic and electric forces become equal.

Volt - Ampere characteristics

For a p-n junction, the current I is related to the voltage V by equation

I = Io [e ( v / ξVT) -1](3.3)

Where,

I = diode current

Io = diode reverse saturation current at room temperature

V = External voltage applied to the diode

ξ = a constant, 1 for germanium and 2 for silicon

VT= KT/q = T/11600, volt equivalent of temperature i.e. thermal voltage;

where K is the Boltzmans constant (1.38066-23 J/K)

q is the charge of electron (1.6021910-19C)

T is the temperature of diode junction (oK)

At from temperature (T=300oK), VT = 26mv. Substituting this value the equation we get

I = Io [e ( 40 v / ξ )-1]

Therefore for germanium diode

I = Io [e ( 40 v )-1] (3.4)

and for silicon diode it is

I = Io [e ( 20 v )-1] (3.5)

If the value of applied voltage is greater than unity, then the equation of diode current for germanium becomes

I=Ioe40v and

for silicon it is

I = Ioe20v.

When the diode is reverse biased, and ê V ê is several times VT, I ~ - Io. The reverse current is therefore constant, independent of the applied reverse bias. Consequently, Io is referred to as the reverse saturation current.

EXAMPLE 7.0

When a reverse bias is applied to a germanium PN junction diode, the reverse saturation current at room temperature is 0.3m A. Determine the current flowing in the diode when 0.15v forward bias is applied at room temperature.

Given Io = 0.3 * 10-6A and VF = 0.15v

The current flowing through the PN diode under forward bias is

I = Io (e 40VF -1) for germanium diode

= 0.3 * 10-6 (e 40*0.15 -1)

= 120.73m A.

 

Diode Resistance

An ideal diode should offer zero resistance in forward bias and infinite resistance in reverse bias. But in practice, no diode can act as an ideal diode.

The static resistance R of a diode is defined as the ratio V/I of the voltage to the current. The static resistance varies widely with V and I is not a useful parameter.

For small signal the dynamic or incremental resistance r is an important parameter, and is defined as the reciprocal of the slope of volt ampere characteristics, r =dv / dI. The dynamic resistance is not a constant, but depends upon the operating voltage. The dynamic resistance varies inversely with current i.e., rf = ξ VT / I, where VT = T/11600, the volt equivalent of temperature (T) of the diode junction (0K) and ξ is a constant where value is I for Ge and 2 for Si diodes. The dynamic resistance varies inversely with current; at room temperature and for ξ =1, r=26/I where I is in milliamperes and r in ohms. For a forward current 26mA, the dynamic resistance is 1W . Although rf varies with current, in a small signal model, it is reasonable to use the parameters rf as a constant.

The resistance offered by the p-n junction diode under reverse bias condition is very large compared to forward resistance, which is in the range of several mega ohms.

Transition or Depletion region Capacitance (CT)

The parallel layers of oppositely charged immobile ions on the two sides of the junction from the transition capacitance given by

CT = e A / W (3.6)

Where e is the permittivity of the material.

A is the cross sectional Area of the junction

W is the width of the depletion layer over which the ions are uncovered.

The depletion width is given by

W = ( ((2e oe r (Vo-V) ) / q )(NA + ND / NA ND ))1/2 (3.7)

where the forward bias +V is applied, the effective barrier potential VB = [Vo-(+V)], is lowered, and hence the width of the depletion region W decreases and CT increases. Under reverse bias condition, the majority carriers move away from the junction, thereby uncovering more immobile charges. Now the effective barrier potential VB = [Vo - (-V)] is increased and hence, W increases and CT decreases correspondingly. The value of CT ranges from 5 to 200pF, the larger values being for high power diodes. This property of voltage variable capacitance with reverse bias appears in varactors, vari-caps or volta-caps.

Diffusion Capacitance (CD)

The capacitance that exists in a forward bias junctions is called a diffusion or storage capacitance CD, whole value is usually much larger than CT which exists in reverse bias junction. This is also defined as the rate of injected charge with applied voltage i.e.,

CD = dQ / dv (3.8)

Where dQ represents the change in the number of minority carriers stored outside the depletion region, when a change in voltage dv is applied across the diode. Diffusion capacitance CD is proportional to forward current I. There fore

CD = y I / x VT(3.9)

where y is the mean life time of holes and electrons.

The value of CD ranges from 10 to 1000pF.

Effect of Temperature on PN junction Diodes

The rise in temperature increases the generation of electron hole pairs in semiconductor and increases their conductivity. As a result the current through PN junction diode increases with temperature as given by the diode current equation.

I = Io [e(V/x VT) -1] (3.10)

The reverse saturation current Io of the diode increases ap[proximately 7 percent / 0c for both germanium and silicon. Since (1.07)10 ~ 2, reverse saturation current approximately doubles for every 10oc rise in temperature. Hence if the temperature is increased at fixed voltage, the current I increases. To bring the current I to the original value, the voltage V has to be reduced. It is found that at room temperature for either germanium or silicon, dv/dt = -2.5 mv/oc.

In order to maintain the current I to a constant value. The current may be expressed as

Io2 = Io1 *2 (T2-T1) / 10 (3.11)

where Io1 is the saturation current at at T1

and Io2 is the saturation current at T2.

EXAMPLE 8.0

A silicon diode has a saturation current of 7.5 m A at room temperature 300oK. Calculate the saturation current at 400oK.

Io1 = 7.5 * 10-6A at T1 = 300K = 27oc and T2 = 400oK. = 127oC

\ The saturation current at 400oK is

Io2 = Io1 * 2(T2-T1) / 10 = 7.5 * 10-6 * 2 (127-27) /10

= 7.5 * 10-6 *210

= 7.68m A