A contact serves as an interface through which the semiconductor device interacts with its environment. It is needed to apply excitation to the device and to extract signal from it.

A contact should perform this task with the least alteration of the device characteristics.
A Figure-of-Merit that is a measure of this characteristics is the specific contact resistance defined as :
The derivative is defined at 0 = V because no voltage should ideally drop across the junction.
Contacts are formed by depositing metal over the semiconductor. However, a straightforward deposition of a metal such as Aluminum on Silicon will yield only a Schottky diode whose specific contact resistance will be very high.

Example 5.1 : Determine the specific contact resistance of a contact formed by simply depositing Aluminum on N-type Silicon of doping

. Such a contact will be a Schottky contact which will obey the current equation:

Differentiation of the above equation gives:
For a barrier height of 0.7 eV,

Example 5.2 : Such a large contact resistance is completely unacceptable. To understand what would constitute a good value for the contact resistance, let us consider the example of a PN junction diode with a contact area of

. Calculate the contact resistance for contact resistivities of

(b) Calculate also the voltage drop across the contact for a current of 1mA.
(c) Determine the factor by which the diode’s characteristics will get modified as a result of this voltage drop.

Solution :

Because of the exponential nature of diode’s characteristics (~exp(qV/kT)), the modification factor will be

The table below shows the resistance due to contact, voltage drop across it and the modification of diode’s characteristics for various values of specific contact resistances.

This example illustrates that specific contact resistance better than is required.
This implies that through suitable modifications, the specific contact resistance of the metal-semiconductor junction has to be lowered from a value of , a nine orders of magnitude reduction!
This tremendous feat is achieved by increasing the tunneling current in the diode. Normally, this current for doping less than is negligible but as the doping is increased, it becomes increasingly important.
The tunneling probability depends exponentially on the height and the thickness of the barrier:
The transmission probability for the triangular approx. of the barrier is given by:

As doping is increased, the thickness W of the barrier decreases, increasing the transmission coefficient and thereby the current.

Example 5.3 : Determine depletion widths for Schottky barrier diodes made on N-type Silicon with doping values of

Assume that barrier height is 0.7eV .Using the expressions derived earlier, we obtain

The table above shows that for doping

, the barrier width is very narrow making tunneling very significant.

Since the depletion width goes as

the specific contact resistance, being inversely proportional to transmission probability when tunneling becomes the dominant current conduction mechanism can be expressed as:

The expression above shows that a small barrier height and high doping level is required to obtain good contacts.

For

, the specific contact resistance is determined entirely by the tunneling process and decreases rapidly as doping is further increased.

As a result, a contact to an N or P-type semiconductor is made by first heavily doping the semiconductor region under the contact and then depositing a metal such as aluminum over it as illustrated in the Figure.

Although, a contact like this is a good Ohmic contact for electrons, it is however, a poor contact for holes! The reasons for this can be understood by considering a 1-D abstraction of the above schematic as shown below:

The holes travelling from the

see a barrier of height

which obstructs their flow to the contact. As a result, these contacts are called reflecting contacts for the minority carriers.

The current flow under the contacts was assumed to be perpendicular for the cases discussed so far. There are contacts however, where the current flow is lateral as illustrated by the Figure below:

The contact resistance is now determined not only by the specific resistance of the contact but also by the sheet resistance of the semiconductor layer underneath.

These contacts can be modeled using a transmission line model described below:

Definitions:
-

is the sheet resistance of the semiconductor underneath the contact so that

represents the incremental series resistance, where W is the width of the contact perpendicular to the plane of the diagram.

-

where, Rc is the specific contact resistance so that

represents the conductance of the vertical section
Contact resistance:

The following set of equations can be easily written for the transmission line representation

The equations above lead to the following differential equation:

For simplicity we shall take a contact that is long enough in the x-direction and assume that

. If this the case then

Using the other boundary condition that I at x=0 is I(0) , we obtain

The contact resistance in this case scales with only the width of the contact and not its area as for the earlier case. It is therefore specified as

in units of

Example 5.4 : Determine the lateral contact resistance for a contact of area

specific contact resistance of

and a set of doping values given below. The semiconductor under the contact is N-type of thickness 1micro metre.

Solution : The sheet resistance of the semiconductor is given by the expression:

,where t is the thickness of the semiconductor. This equation along with Eq. (69) gives the following values for contact resistance:

The example illustrates that very low values of sheet resistance are required to obtain a small lateral contact resistance and even then it is much higher that the contact resistance

for the above example) when the current flow under the contact is vertical.

Thus a high doping under the semiconductor is needed to lower both the specific contact resistance as well as the sheet resistance.

The importance of Ohmic contact increases as current through the device increases and as the dimensions of the device are scaled to improve its performance.
Since characteristics of most semiconductor devices such as diodes, BJTs, MOSFETs etc are non-linear, while that of a contact linear, the voltage drop across the contact would tend to become comparable and even greater at higher currents as illustrated by the Figure above.
To understand the increased importance of contact resistance as the device is scaled down, let us first consider a BJT and then a MOSFET.

In a BJT let us take the emitter resistance. The current flow through the semiconductor under the contact is perpendicular as illustrated by the Figure below:

Since the I-V characteristics of the emitter-base junction is exponential in nature , the voltage drop should be much smaller than the thermal voltage V

Example 5.5 : For an emitter area of

, specific contact resistivity of

determine the maximum value of emitter current before the contact begins to make appreciable difference to the transistor’s characteristics.

Solution : Taking a maximum contact voltage drop of

so that there is less than 10% change in transistor’s current, we obtain

Suppose now the emitter area is scaled by a factor of 10, while keeping the emitter current constant so as to improve the transistor’s performance (which upto a limit improves as the collector current density increases).

The voltage drop across the contact now becomes 50mV, which is very significant!

If the voltage drop is to be maintained at the earlier 5mV level, the specific contact resistance will have to be lowered to

Consider now a MOSFET and the impact of the source contact resistance:

The current under the source flows laterally so that the expression

is the contact resistance here. The role of the source resistance is that it lowers the transconductance of the device

where

is the transconductance of the transistor in the absence of a source resistor. We would like

to be as small as possible. A simplified expression for the transconductance is

As the channel length and width are scaled by a factor K, the oxide thickness is also reduced so that for a fixed supply voltage, the transconductance increases in magnitude.The resistance due to the contact however, increases by a factor K as the width is scaled.

Therefore, the product

would become larger with scaling making it necessary to reduce the contact resistance with scaling.
For the case, where the supply voltage is scaled also, the transconductance may not increase but due to increase in contact resistance, the problem persists though with lesser impact.

Example 5.6 : Determine the factor by which the transconductance of a MOSFET is reduced as a result of voltage drop across the source resistance . The MOSFET has the following characteristics:

The source contact has the following characteristics:

(b) Redo your calculations for

but W/L = 10 as before. Assume that new ,

For the contact assume that every thing is same except that

.

Solution : Using Eq.(70) and (71), we obtain

= 0.95

(b) As a result of scaling of channel length by a factor of 2, the intrinsic transconductance will increase by a factor of 2 and the source resistance will also increase by the same factor. As a result the

product becomes four times and

= 0.84 now.